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40d09c4aa432a4aace30c0b830ecc9fa | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_5 | How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$ | Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.
There are $7!$ ways to order $7$ objects. However, since there's $3!=6$ ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and $2!=2$ ways to order the green tiles, we have to divide out these possibilities: $\frac{7!}{3! \cdot2}=\boxed{420}$ | B | 420 |
40d09c4aa432a4aace30c0b830ecc9fa | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_5 | How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$ | We can choose a different frame to solve this problem. Our tile combination can be written as $Y, Y, Y, G, G, B, P.$ We can focus on $Y, Y, Y, G, G$ first, which gives us $\binom{5}{3}=10.$ Now we can insert our brown tile into this which only has $6$ choices(like $Y,B, Y, Y, G, G$ and $Y, Y, Y, G, G, B$ etc.), then insert purple tile which only has $7$ choices(like $B,Y, P, Y, Y, G, G$ and $B, Y, Y, Y, G, G, P$ etc.). Multiply them together we get $10\cdot 6\cdot 7=\boxed{420}$ | B | 420 |
40d09c4aa432a4aace30c0b830ecc9fa | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_5 | How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$ | Let $B$ be brown, $P$ be purple, $G$ be green, and $Y$ be yellow. Then, we are just ordering $Y$ $Y$ $Y$ $G$ $G$ $B$ , and $P$ . Hence, $\frac{7!}{3! \cdot 2!} = \boxed{420}$ | B | 420 |
39a604ab82f19405c40da69abfbddc54 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_6 | Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$ -hour period?
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70$ | In order to get the smallest palindrome greater than $15951$ , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.
So we raise $9$ to the next largest value, $10$ , but obviously, that's not how place value works, so we're in the $16000$ s now. To keep this a palindrome, our number is now $16061$
So Megan drove $16061-15951=110$ miles. Since this happened over $2$ hours, she drove at $\frac{110}{2}=\boxed{55}$ mph. ~quacker88 | B | 55 |
d544c65134a6fcd33af4cb8c1bb7eaf4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_7 | How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$ | Any even multiple of $3$ is a multiple of $6$ , so we need to find multiples of $6$ that are perfect squares and less than $2020$ . Any solution that we want will be in the form $(6n)^2$ , where $n$ is a positive integer. The smallest possible value is at $n=1$ , and the largest is at $n=7$ (where the expression equals $1764$ ). Therefore, there are a total of $\boxed{7}$ possible numbers.-PCChess | A | 7 |
d544c65134a6fcd33af4cb8c1bb7eaf4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_7 | How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$ | A even multiple square of $3$ can be represented by $3^2 \cdot 2^2 \cdot x^2$ , where $3^2$ is the multiple or $3$ and $2^2$ makes it even. Simplifying we have $36^2 \cdot x^2$ . We can divide $2020$ by $36$ (floor) and get $56$ see the result. We can then see that there are $7$ different values for $x$ . It can't be larger or else $x^2 > 56$ . And thus $\boxed{7}$ | A | 7 |
d544c65134a6fcd33af4cb8c1bb7eaf4 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_7 | How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$ | It can be seen that the problem is just asking for squares that are multiples of six. Thus, all squares of multiples of six can be listed out: $6^2$ $12^2$ $18^2$ $24^2$ $30^2$ $36^2$ , and $42^2$ . Since $48^2=2196 > 2020$ , there are $\boxed{7}$ valid answers.
~airbus-a321, November 2023 | A | 7 |
e0fb72b6e1b8e3785100a2de5fa7715f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_8 | Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:
[asy] /* Made by MRENTHUSIASM */ size(250); pair O, P, Q, R1, R2, R3, R4, R5, R6, R7, R8, I1, I2; O = (0,0); P = (-4,0); Q = (4,0); R1 = (-4,3); R4 = (4,3); R5 = (-4,-3); R8 = (4,-3); path C; C = Circle(O,4); R3 = intersectionpoints(C,R1--R4)[0]; R2 = intersectionpoints(C,R1--R4)[1]; R6 = intersectionpoints(C,R5--R8)[0]; R7 = intersectionpoints(C,R5--R8)[1]; I1 = intersectionpoint(R2--R6,P--Q); I2 = intersectionpoint(R3--R7,P--Q); markscalefactor=0.0375; draw(rightanglemark(R1,P,Q)^^rightanglemark(R2,I1,Q)^^rightanglemark(R3,I2,P)^^rightanglemark(R4,Q,P),red); draw(Circle(O,4),dashed); draw(R1--R5^^R4--R8^^R2--R6^^R3--R7^^P--Q); dot(O,linewidth(4)); dot("$P$",P,1.5W,linewidth(4)); dot("$Q$",Q,1.5E,linewidth(4)); dot("$R_1$",R1,1.5NW,blue+linewidth(4)); dot("$R_4$",R4,1.5NE,blue+linewidth(4)); dot("$R_5$",R5,1.5SW,blue+linewidth(4)); dot("$R_8$",R8,1.5SE,blue+linewidth(4)); dot("$R_2$",R2,1.5NW,blue+linewidth(4)); dot("$R_3$",R3,1.5NE,blue+linewidth(4)); dot("$R_6$",R6,1.5SW,blue+linewidth(4)); dot("$R_7$",R7,1.5SE,blue+linewidth(4)); dot(I1,linewidth(4)); dot(I2,linewidth(4)); Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15)); draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy]
We apply casework to the right angle of $\triangle PQR:$
Together, there are $\boxed{8}$ such locations for $R.$ | D | 8 |
e0fb72b6e1b8e3785100a2de5fa7715f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_8 | Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ $Q$ , and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
Without the loss of generality, let $P=(-4,0)$ and $Q=(4,0).$ We conclude that the $y$ -coordinate of $R$ must be $\pm3.$
We apply casework to the right angle of $\triangle PQR:$
Together, there are $2+2+4=\boxed{8}$ such locations for $R.$ | D | 8 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$ . Then, notice that $x$ can only be $0$ $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$ , which is impossible as $y$ must be real. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$ $(1,1)$ $(-1,1)$ , and $(0,2)$ gives a total of $\boxed{4}$ ordered pairs. | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Bringing all of the terms to the LHS, we see a quadratic equation \[y^2 - 2y + x^{2020} = 0\] in terms of $y$ . Applying the quadratic formula, we get \[y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.\] In order for $y$ to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$ must be nonnegative. Therefore, \[4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.\] Here, we see that we must split the inequality into a compound, resulting in $-1 \leq x \leq 1$
The only integers that satisfy this are $x \in \{-1,0,1\}$ . Plugging these values back into the quadratic equation, we see that $x = \{-1,1\}$ both produce a discriminant of $0$ , meaning that there is only 1 solution for $y$ . If $x = \{0\}$ , then the discriminant is nonzero, therefore resulting in two solutions for $y$
Thus, the answer is $2 \cdot 1 + 1 \cdot 2 = \boxed{4}$ | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $|x^{2020}| \leq 1$ , because the discriminant must be positive.
Then $x=-1,0,1$ . Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: \[y^2-2y+1=(y-1)^2=0\] This has only has solutions $1$ , so $(\pm 1,1)$ are solutions.
Next, if $x=0$ \[y^2-2y=0 \Rightarrow y(y-2)=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$
These are the only 4 solutions, so our answer is $\boxed{4}$ | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$
Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$
If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$
When $y=1$ , the right side become $1$ , therefore $x=1,-1$
Our solutions are $(0,2)$ $(0,0)$ $(1,1)$ $(-1,1)$ . These are the only $4$ solutions, so the answer is $\boxed{4}$ | D | 4 |
9bbdcbd5c80022c82c6dc3437a5bf9b3 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_9 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$
For $y=0$ $x$ can only be $0$
For $y=1$ $x^2=1$ so $x=1, -1$
For $y=2$ $x$ can only be $0$ as well.
This gives us the solutions $(0, 0)$ $(1, 1)$ $(-1, 1)$ , and $(0, 2)$ . These are the only solutions, so there is a total of $\boxed{4}$ ordered pairs. | D | 4 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$ | We have
\[\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}\]
Now we do some estimation. Notice that $2^{20} = 1024^2$ , which means that $2^{20}$ is a little more than $1000^2=1,000,000$ . Multiplying it with $10^{20}$ , we get that the denominator is about $1\underbrace{00\dots0}_{26 \text{ zeros}}$ . Notice that when we divide $1$ by an $n$ digit number as long as n is not a power of 10, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\underbrace{00\dots0}_{26 \text{ zeros}}$ , there are $\boxed{26}$ zeros in the initial string after the decimal point. -PCChess | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$ | First rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$ . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$
$\log{5^{20}} = 20\log{5}$ and memming $\log{5}\approx0.69$ (alternatively use the fact that $\log{5} = 1 - \log{2}$ ), $\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14$ digits.
Our answer is $\boxed{26}$ | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$ | Just as in Solution $2,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the decimal point is before the $9.$ Dividing the number again by $10$ twenty-six more times allows a string of $\boxed{26}$ zeroes to be formed. -OreoChocolate | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$ | We see that $\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}$ . We see that this has $27-1=26$ zeros after the decimal point before coming to $9$
Therefore, the answer is $\boxed{26}$ | D | 26 |
302eaa9d47b6a11581c03e0b6113982e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_12 | The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$ | \begin{align*}|\lceil \log \dfrac{1}{20^{20}} \rceil| &= |\lceil \log 20^{-20} \rceil| \\ &= |\lceil -20 \log(20) \rceil| \\ &= |\lceil -20(\log 10 + \log 2) \rceil| \\ &= |\lceil -20(1 + 0.301) \rceil| \\ &= |\lceil -26.02 \rceil| \\ &= |-26| \\ &= \boxed{26} | D | 26 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$ | Note that cycles exist initially and after each round of erasing.
Let the parentheses denote cycles. It follows that:
Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: \[(12\underline{425}3415251).\] Therefore, the answer is $4+2+5=\boxed{11}.$ | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$ | After erasing every third digit, the list becomes $1245235134\ldots$ repeated. After erasing every fourth digit from this list, the list becomes $124235341452513\ldots$ repeated. Finally, after erasing every fifth digit from this list, the list becomes $124253415251\ldots$ repeated. Since this list repeats every $12$ digits and $2019,2020,2021$ are $3,4,5$ respectively
in $\pmod{12},$ we have that the $2019$ th, $2020$ th, and $2021$ st digits are the $3$ rd, $4$ th, and $5$ th digits respectively. It follows that the answer is $4+2+5= \boxed{11}.$ | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$ | Note that cycles exist initially and after each round of erasing.
We will consider one cycle after all three rounds of erasing. Suppose this cycle has length $L$ before any round of erasing. It follows that:
The least such positive integer $L$ is $\operatorname{lcm}(5,3,2,5)=30.$ So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: [asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,4)--(i+0.5,4)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,4)--(i+0.5,4)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,4)--(8,4)--(8,0)--cycle,green); fill((18,0)--(18,4)--(19,4)--(19,0)--cycle,green); fill((27,0)--(27,4)--(28,4)--(28,0)--cycle,green); for (real i=1; i<5; ++i) { for (real j=0; j<30; ++j) { label("$"+string(1+j%5)+"$",(j+0.5,i-0.5)); } } for (real i=0; i<30; ++i) { label("$\vdots$",(i+0.5,-1/3)); } add(grid(30,4,linewidth(1.25))); [/asy] As indicated by the white squares, each group of $30$ digits on the original list has $\frac25\cdot30=12$ digits remaining on the final list.
Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: [asy] /* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green); fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green); fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green); for (real j=0; j<30; ++j) { label("$"+string(1+j%5)+"$",(j+0.5,0.5)); } draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow); draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow); draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow); add(grid(30,1,linewidth(1.25))); [/asy] Therefore, the answer is $4+2+5=\boxed{11}.$ | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$ | As the LCM of $3$ $4$ , and $5$ is $60$ , let us look at a $60$ -digit block of original numbers (many will be erased by Steve). After he erases every third number $\left(\dfrac{1}{3}\right)$ , then every fourth number of what remains $\left(\dfrac{1}{4}\right)$ , then every fifth number of what remains $\left(\dfrac{1}{5}\right)$ , we are left with $\dfrac{2}{3} \cdot \dfrac{3}{4} \cdot \dfrac{4}{5} \cdot 60=24$ digits from this $60$ -digit block. $2019 \equiv 3 \pmod {24}, 2020 \equiv 4 \pmod {24}, 2021 \equiv 5 \pmod {24}$ . Writing out the first few digits of this sequence, we have $\underbrace{1}_{\#1}, \underbrace{2}_{\#2}, \cancel{3}, \underbrace{4}_{\#3}, \cancel{5}, \cancel{1}, \underbrace{2}_{\#4}, \cancel{3}, \cancel{4}, \underbrace{5}_{\#5}, \dots$ . Therefore, our answer is $4+2+5=\boxed{11}$ | D | 11 |
053dff916f2f7418ffea03505877273e | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 | Steve wrote the digits $1$ $2$ $3$ $4$ , and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$ rd, $6$ th, $9$ th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$ th, $8$ th, $12$ th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$ | Lemma: Given a sequence $a_1, a_2, a_3, \cdots$ , and an positive integer $k>2$ . If we erase every $k$ th item in this sequence, and we name $b_1, b_2, b_3, \cdots$ as the remaining sequence.
Then we have \[b_{(k-1)m+1}=a_{km+1}, b_{(k-1)m+2}=a_{km+2}, \cdots, b_{(k-1)m+k-1}=a_{km+k-1}.\]
Proof: For $a_{km+j}$ with some $j, 1\le j\le k-1$ , we will have $m$ items removed before this item, so it becomes $b_{(k-1)m+j}$ in the new sequence. Hence, we have $b_{(k-1)m+j}=a_{km+j}$
If we start with $a_1, a_2, a_3, \cdots,$ and let $b_1, b_2, \cdots$ be the sequence after removing all the indexes that are multiples of $3$ . Then, we have $b_{2n+1}=a_{3n+1},b_{2n+2}=a_{3n+2}$
Similarly, if we start with $b_1, b_2, b_3, \cdots,$ and remove all multiples of $4$ , and get $c_1, c_2, \cdots$ We have $c_{3n+1}=b_{4n+1},c_{3n+2}=b_{4n+2}, c_{3n+3}=b_{4n+3}$
If we start with $c_1, c_2, \cdots$ and $d_1, d_2, \cdots$ are remove all multiples of $5$ , we get \[d_{4n+1}=c_{5n+1}, d_{4n+2}=c_{5n+2}, d_{4n+3}=c_{5n+3}, d_{4n+4}=c_{5n+4}.\] Therefore, \begin{align*} d_{2019}&=d_{4\cdot504+3}=c_{5\cdot 504+3}=c_{2523}=c_{3\cdot 840+3}=b_{4\cdot 840+3}=b_{3363}=b_{2\cdot 1681+1}=a_{3\cdot 1681+1}=a_{5044}=a_4=4, \\ d_{2020}&=d_{4\cdot504+4}=c_{5\cdot 504+4}=c_{2524}=c_{3\cdot841+1}=b_{4\cdot841+1}=b_{3365}=b_{2\cdot1682+1}=a_{3\cdot 1682+1}=a_{5047}=a_2=2, \\ d_{2021}&=d_{4\cdot505+1}=c_{5\cdot505+1}=c_{2526}=c_{3\cdot841+3}=b_{4\cdot841+3}=b_{3367}=b_{2\cdot1683+1}=a_{3\cdot1683+1}=a_{5050}=a_5=5, \end{align*} and the answer is $4+2+5=\boxed{11}$ | D | 11 |
667c9a200bc443791b4e290668337093 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_17 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$ | Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.
Let us use casework on the number of pairs that form a diameter of the circle.
Case 1: $0$ diameters
There are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs with $1$ $2$ pairs with $3$ , etc.
Case 2: $1$ diameter
There are $5$ possible diameters to draw (everyone else pairs with the person next to them).
Note that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.
Case 3: $3$ diameters
There are $5$ possible sets of $3$ diameters to draw.
Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have $5*2=10$ possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $5$
Note that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.
Case 4: $5$ diameters
There is only $1$ way to do this.
Thus, in total there are $2+5+5+1=\boxed{13}$ possible ways.
- Minor edits by Pearl2008 | C | 13 |
667c9a200bc443791b4e290668337093 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_17 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$ | If each person knows exactly $3$ people, that means we form " $4$ -person groups". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\dbinom{4}{2}=6$ . The $2$ nd pair is just $\dbinom{2}{2} =1$ . We need to multiply these together since these are $1$ group. The $3$ rd pair would be $\dbinom{4}{2}=6$ . The $4$ th pair is $\dbinom{2}{2}=1$ . We multiply these $2$ together and get $6$ . The final group would be $\dbinom{2}{2}=1$ . So we add these up and we have $6 + 6 + 1 = \boxed{13}$ possible ways. | C | 13 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.
Now, suppose that we have this deck again, with only one black card. Each time we pick a red ball, we place a card above the black card, and each time we pick a blue ball, we place a card below the black card. It is easy to see that the probability that the card is inserted into the top part of the deck is simply equal to the number of red balls divided by the total number of balls, and the probability that the card is inserted into the bottom part of the deck is equal to the number of blue balls divided by the total number of balls. Therefore, this is equivalent to inserting the card randomly into the deck.
Finally, four more red cards will be inserted into the deck, and so the black card can be in five possible positions. Only one corresponds to having three balls of each type. Our probability is thus $\frac{1}{5}$ , and so the answer is $\boxed{15}$ | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Let $R$ denote the action where George selects a red ball and $B$ denote the action where he selects a blue one. Now, in order to get $3$ balls of each color, he needs $2$ more of both $R$ and $B$
There are 6 cases: $RRBB, RBRB, RBBR, BBRR, BRBR, BRRB$ (we can confirm that there are only $6$ since $\binom{4}{2}=6$ ). However we can clump $RRBB + BBRR$ $RBRB + BRBR$ , and $RBBR + BRRB$ together since they are equivalent by symmetry.
$\textbf{CASE 1: }$ $RRBB$ and $BBRR$
Let's find the probability that he picks the balls in the order of $RRBB$
The probability that the first ball he picks is red is $\frac{1}{2}$
Now there are $2$ reds and $1$ blue in the urn. The probability that he picks red again is now $\frac{2}{3}$
There are $3$ reds and $1$ blue now. The probability that he picks a blue is $\frac{1}{4}$
Finally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\frac{2}{5}$
So the probability that the $RRBB$ case happens is $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{2}{5} = \frac{1}{30}$ . However, since the $BBRR$ case is the exact same by symmetry, case 1 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.
$\textbf{CASE 2: }$ $RBRB$ and $BRBR$
Let's find the probability that he picks the balls in the order of $RBRB$
The probability that the first ball he picks is red is $\frac{1}{2}$
Now there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\frac{1}{3}$
There are $2$ reds and $2$ blues now. The probability that he picks a red is $\frac{1}{2}$
Finally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\frac{2}{5}$
So the probability that the $RBRB$ case happens is $\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{30}$ . However, since the $BRBR$ case is the exact same by symmetry, case 2 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.
$\textbf{CASE 3: }$ $RBBR$ and $BRRB$
Let's find the probability that he picks the balls in the order of $RBBR$
The probability that the first ball he picks is red is $\frac{1}{2}$
Now there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\frac{1}{3}$
There are $2$ reds and $2$ blues now. The probability that he picks a blue is $\frac{1}{2}$
Finally, there are $2$ reds and $3$ blues. The probability that he picks a red is $\frac{2}{5}$
So the probability that the $RBBR$ case happens is $\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{30}$ . However, since the $BRRB$ case is the exact same by symmetry, case 3 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.
Adding up the cases, we have $\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\boxed{15}$ ~quacker88 | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | We know that we need to find the probability of adding 2 red and 2 blue balls in some order.
There are 6 ways to do this, since there are $\binom{4}{2}=6$ ways to arrange $RRBB$ in some order.
We will show that the probability for each of these 6 ways is the same.
We first note that the denominators should be counted by the same number. This number is $2 \cdot 3 \cdot 4 \cdot 5=120$ . This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from.
The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally.
The same goes for the blue ones. The numerator must equal $(1 \cdot 2)^2$
Therefore, the probability for each of the orderings of $RRBB$ is $\frac{4}{120}=\frac{1}{30}$ . There are 6 of these, so the total probability is $\boxed{15}$ | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\frac{1}{2}$ ), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance $\frac{2}{3}$ ), in which case he must choose two blues to get three of each, with probability $\frac{1}{4}\cdot\frac{2}{5}=\frac{1}{10}$ or a blue for two blue and two red in the urn, with chance $\frac{1}{3}$ . If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a $\frac{1}{2}\cdot\frac{2}{5}$ for total of $2\cdot\frac{1}{5}=\frac{2}{5}$ . The total probability that he ends up with three red and three blue is $2\cdot\frac{1}{2}(\frac{2}{3}\cdot\frac{1}{10}+\frac{1}{3}\cdot\frac{2}{5})=\frac{1}{15}+\frac{2}{15}=\boxed{15}$ | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Let the probability that the urn ends up with more red balls be denoted $P(R)$ . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is $1-2P(R)$ $P(R) =$ the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, $P(\text{no more blues}) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1}{5}$
The second case, $P(\text{1 more blue}) = 4\cdot\frac{1\cdot1\cdot2\cdot3}{2\cdot3\cdot4\cdot5} = \frac{1}{5}$ . Thus, the answer is $1-2\left(\frac{1}{5}+\frac{1}{5}\right)=1-\frac{4}{5}=\boxed{15}$ | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Here $X$ stands for R or B, and $Y$ for the remaining color.
After 3 rounds one can either have a $4+1$ configuration ( $XXXXY$ ), or $3+2$ configuration ( $XXXYY$ ). The probability of getting to $XXXYYY$ from $XXXYY$ is $\frac{2}{5}$ . Observe that the probability of arriving to $4+1$ configuration is \[\frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}\] $\frac{2}{3}$ to get from $XXY$ to $XXXY$ $\frac{3}{4}$ to get from $XXXY$ to $XXXXY$ ). Thus the probability of arriving to $3+2$ configuration is also $\frac{1}{2}$ , and the answer is \[\frac{1}{2} \cdot \frac{2}{5} = \boxed{15}.\] | B | 15 |
bb239d071470a1086a438c49de275807 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_18 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | We can use dynamic programming to solve this problem.
We let $dp[i][j]$ be the probability that we end up with $i$ red balls and $j$ blue balls.
Notice that there are only two ways that we can end up with $i$ red balls and $j$ blue balls: one is by fetching a red ball from the urn when we have $i - 1$ red balls and $j$ blue balls and the other is by fetching a blue ball from the urn when we have $i$ red balls and $j - 1$ blue balls.
Then we have $dp[i][j] = \frac{i - 1}{i - 1 + j} dp[i - 1][j] + \frac{j - 1}{i - 1 + j} dp[i][j - 1]$
Then we start can with $dp[1][1] = 1$ and try to compute $dp[3][3]$
\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 1 & 2 & 3\\ \hline\hline 1 & 1 & 1/2 & 1/3\\ \hline 2 & 1/2 & 1/3 & 1/4\\ \hline 3 & 1/3 & 1/4 & 1/5\\ \hline \end{array}\] The answer is $\boxed{15}$ | B | 15 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | $158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}$
We're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\binom{52}{10}$
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
We need to get rid of the multiples of $3$ , which will subsequently get rid of the multiples of $9$ (if we didn't, the zeroes would mess with the equation since you can't divide by 0)
$9\cdot5=45$ $8\cdot6=48$ $\frac{51}{3}$ leaves us with 17.
$\frac{52\cdot\cancel{51}^{17}\cdot50\cdot49\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}\cdot44\cdot43}{10\cdot\cancel{9}\cdot\cancel{8}\cdot7\cdot\cancel{6}\cdot\cancel{5}\cdot4\cdot\cancel{3}\cdot2\cdot1}$
Converting these into $\pmod{9}$ , we have
$\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9}$
$4A\equiv8\pmod{9} \implies A=\boxed{2}$ ~quacker88 | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | We're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\binom{52}{10}$
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
And after simplifying, we get $26\cdot17\cdot7\cdot47\cdot46\cdot5\cdot11\cdot43$ .
Now, if we examine the number $158A00A4AA0$ , we can notice that it is equal to some number $n$ times 10.
Therefore, we can divide 10 from the aforementioned expression and find the units digit, which will be $A$
Now, after dividing ten, we will have $26\cdot17\cdot7\cdot47\cdot23\cdot11\cdot43$ .
We can then use modulo 10 and find that the unit digit of the expression is $\boxed{2}$ ~lucaswujc | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43$
Since this number is divisible by $4$ but not $8$ , the last $2$ digits must be divisible by $4$ but the last $3$ digits cannot be divisible by $8$ . This narrows the options down to $2$ and $6$
Also, the number cannot be divisible by $3$ . Adding up the digits, we get $18+4A$ . If $A=6$ , then the expression equals $42$ , a multiple of $3$ . This would mean that the entire number would be divisible by $3$ , which is not what we want. Therefore, the only option is $\boxed{2}$ -PCChess | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | It is not hard to check that $13$ divides the number, \[\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43.\] As $10^3\equiv-1\pmod{13}$ , using $\pmod{13}$ we have $13|\overline{AA0}-\overline{0A4}+\overline{8A0}-\overline{15}=110A+781$ . Thus $6A+1\equiv0\pmod{13}$ , implying $A\equiv2\pmod{13}$ so the answer is $\boxed{2}$ | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | As mentioned above, \[\binom{52}{10}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43} = 158A00A4AA0.\] We can divide both sides of $10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA0$ by 10 to obtain \[17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA,\] which means $A$ is simply the units digit of the left-hand side. This value is \[7 \cdot 3 \cdot 7 \cdot 7 \cdot 6 \cdot 1 \cdot 3 \equiv \boxed{2}.\] i_equal_tan_90 , revised by emerald_block | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | We note that: \[\frac{(52)(51)(50)(49)(48)(47)(46)(45)(44)(43)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)} = (13)(17)(7)(47)(46)(5)(22)(43).\] Let $K=(13)(17)(7)(47)(46)(5)(22)(43)$ . This will help us find the last two digits modulo $4$ and modulo $25$ .
It is obvious that $K \equiv 0 \pmod{4}$ . Also (although this not so obvious), \[K \equiv (13)(17)(7)(47)(46)(5)(22)(43)\] \[\equiv (13)(-8)(7)(-3)(-4)(5)(-3)(-7)\] \[\equiv (13)(-96)(21)(35)\] \[\equiv (13)(4)(-4)(10)\] \[\equiv (13)(-16)(10)\] \[\equiv (13)(9)(10)\] \[\equiv (117)(10)\] \[\equiv (-8)(10)\] \[\equiv 20 \pmod{25}.\] Therefore, $K \equiv 20 \mod 100$ . Thus $K=20$ , implying that $\boxed{2}$ | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | As in Solution 2, we see that
\[\binom{52}{10}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43,\]
which contains no factors of $3.$ Therefore, the sum of the digits must not be a multiple of $3.$ This sum is
\[1+5+8+A+0+0+A+4+A+A+0=18+4A.\]
It follows that $4A$ cannot be a multiple of $3,$ ruling out choices $(B)$ and $(D).$ Therefore, our possibilities are $A=2,4,$ and $7.$ Now, notice that $\binom{52}{10}$ is divisible by $7.$ Therefore, we can plug each possible value of $A$ into $158A00A4AA0$ and test for divisibility by $7.$ Conveniently, we see that the first value, $A=2,$ works. Thus, the answer is $\boxed{2}.$ (To make our argument more rigorous, we can also test divisibility by $7$ for $A=4$ and $7$ to show that these values do not work.) | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | The total number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
Using divisibility rules, we have that A is not a multiple of $3$ . Then, divide this equation by 10. This implies that the new number $158A00A4AA0$ is divisible by $2$ but not $4$ . This means that $A$ is either $2$ or $6$ . However, $6$ is a multiple of $3$ , meaning $A$ has to be $\boxed{2}$ | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | As stated in previous solutions, the number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
Canceling out common factors $(\dfrac{52}{2} = 26 \text{, } \dfrac{51}{3} = 17 \text{, } 5 \cdot 10 = 50 \text{, } \dfrac{49}{7} = 7 \text{, } 48 = 6 \cdot 8 \text{, } \dfrac{45}{9} = 5 \text{, } \dfrac{44}{11} = 4)$ , you get this - $\frac{\cancel{52}^{26}\cdot\cancel{51}^{17}\cdot\cancel{50}\cdot\cancel{49}^{7}\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}^5\cdot\cancel{44}^{4}\cdot43}{\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}$
When you multiply the remaining numbers, you get the product as $15820024220$ . From this product, we can then determine that $A$ is equal to $\boxed{2}$ | A | 2 |
80ed547a12beca4bf6dfdcd7502498bf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_19 | In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$ | Compute $\frac{52!}{10!42!} = 15820024220.$ Therefore our answer is $\boxed{2}.$ | A | 2 |
80d780837fb2c8c088bc8f294741bac0 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_20 | Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$ , together with its interior. For real $r\geq0$ , let $S(r)$ be the set of points in $3$ -dimensional space that lie within a distance $r$ of some point in $B$ . The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$ , where $a,$ $b,$ $c,$ and $d$ are positive real numbers. What is $\frac{bc}{ad}?$
$\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38$ | Split $S(r)$ into 4 regions:
1. The rectangular prism itself
2. The extensions of the faces of $B$
3. The quarter cylinders at each edge of $B$
4. The one-eighth spheres at each corner of $B$
Region 1: The volume of $B$ is $1 \cdot 3 \cdot 4 = 12$ , so $d=12$
Region 2: This volume is equal to the surface area of $B$ times $r$ (these "extensions" are just more boxes). The volume is then $\text{SA} \cdot r=2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4)r=38r$ to get $c=38$
Region 3: We see that there are 12 quarter-cylinders, 4 of each type. We have 4 quarter-cylinders of height 1, 4 quarter-cylinders of height 3, 4 quarter-cylinders of height 4. Since 4 quarter-cylinders make a full cylinder, the total volume is $4 \cdot \dfrac{1\pi r^2}{4} + 4 \cdot \dfrac{3\pi r^2}{4} + 4 \cdot \dfrac{4 \pi r^2}{4}=8 \pi r^2$ . Therefore, $b=8\pi$
Region 4: There is an eighth-sphere of radius $r$ at each corner of $B$ . Since there are 8 corners, these eighth-spheres add up to 1 full sphere of radius $r$ . The volume of this sphere is then $\frac{4}{3}\pi \cdot r^3$ , so $a=\frac{4\pi}{3}$
Using these values, $\dfrac{bc}{ad}=\frac{(8\pi)(38)}{(4\pi/3)(12)} = \boxed{19}$ | B | 19 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | Completing the square, then difference of squares:
\begin{align*} 2^{202} + 202 &= (2^{101})^2 + 2\cdot 2^{101} + 1 - 2\cdot 2^{101} + 201\\ &= (2^{101} + 1)^2 - 2^{102} + 201\\ &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. \end{align*}
Thus, we see that the remainder is surely $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | Let $x=2^{50}$ . We are now looking for the remainder of $\frac{4x^4+202}{2x^2+2x+1}$
We could proceed with polynomial division, but the numerator looks awfully similar to the Sophie Germain Identity , which states that \[a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\]
Let's use the identity, with $a=1$ and $b=x$ , so we have
\[1+4x^4=(1+2x^2+2x)(1+2x^2-2x)\]
Rearranging, we can see that this is exactly what we need:
\[\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1\]
So \[\frac{4x^4+202}{2x^2+2x+1} = \frac{4x^4+1}{2x^2+2x+1} +\frac{201}{2x^2+2x+1}\]
Since the first half divides cleanly as shown earlier, the remainder must be $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | We let \[x = 2^{50}\] and \[2^{202} + 202 = 4x^{4} + 202\] .
Next we write \[2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1\] .
We know that \[4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)\] by the Sophie Germain identity so to find \[4x^{4} + 202,\] we find that \[4x^{4} + 202 = 4x^{4} + 201 + 1\] which shows that the remainder is $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | We let $x=2^{50.5}$ . That means $2^{202}+202=x^{4}+202$ and $2^{101}+2^{51}+1=x^{2}+x\sqrt{2}+1$ . Then, we simply do polynomial division, and find that the remainder is $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | Let $n=2^{101}+2^{51}+1$ . Then, mod $n$
$2^{202}+202 \equiv (-2^{51}-1)^2 + 202$
$\equiv 2^{102}+2^{52}+203$
$= 2(n-1)+203 \equiv 201 \pmod{n}$
Thus, the remainder is $\boxed{201}$ | D | 201 |
4a07eea4da15190b18dbbea5a0680860 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_22 | What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$
$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$ | We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:
\[\frac{2^{202}+202}{2^{101}+2^{51}+1}\] \[= \frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[= 2^{101} - \frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\] \[=2^{101} - \frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}\] \[=2^{101} - 2^{51} + \frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}\] \[= 2^{101} - 2^{51} + \frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}\] \[= 2^{101} - 2^{51} + 1 + \frac{201}{2^{101} + 2^{51} + 1}.\]
Clearly, $201 < 2^{201} + 2^{51} + 1$ , hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is $\boxed{201}$ | D | 201 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = 400$ and $n = 2500$ , the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation, which we abbreviate as $L=R$
For $n = 330$ $L=19$ but $18^2 < 330 < 19^2$ so $R=18$
For $n = 400$ $L=20$ and $R=20$
For $n = 470$ $L=21$ $R=21$
For $n = 540$ $L=22$ but $540 > 23^2$ so $R=23$
Now we move to $n = 2500$
For $n = 2430$ $L=49$ and $49^2 < 2430 < 50^2$ so $R=49$
For $n = 2360$ $L=48$ and $48^2 < 2360 < 49^2$ so $R=48$
For $n = 2290$ $L=47$ and $47^2 < 2360 < 48^2$ so $R=47$
For $n = 2220$ $L=46$ but $47^2 < 2220$ so $R=47$
For $n = 2500$ $L=50$ and $R=50$
For $n = 2570$ $L=51$ but $2570 < 51^2$ so $R=50$
Therefore we have 6 total solutions, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{6}$ | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | This is my first solution here, so please forgive me for any errors.
We are given that \[\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor\]
$\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$ . As $1000\equiv 20\pmod{70}$ , this means that $n\equiv 50\pmod{70}$ , so we can write $n=70k+50$ for $k\in\mathbb{Z}$
Therefore, \[\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor\]
Also, we can say that $\sqrt{70k+50}-1 < k+15$ and $k+15\leq\sqrt{70k+50}$
Squaring the second inequality, we get $k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35$
Similarly solving the first inequality gives us $k < 19-\sqrt{155}$ or $k > 19+\sqrt{155}$
$\sqrt{155}$ is larger than $12$ and smaller than $13$ , so instead, we can say $k\leq 6$ or $k\geq 32$
Combining this with $5\leq k\leq 35$ , we get $k=5,6,32,33,34,35$ are all solutions for $k$ that give a valid solution for $n$ , meaning that our answer is $\boxed{6}$ .
-Solution By Qqqwerw | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We start with the given equation \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor\] From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$ . This means that \[\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}\] Solving each inequality separately gives us two inequalities: \[n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50\] \[n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}\] Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence, the answer is $2+4 = \boxed{6}$ | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$ ; let $n=70j-20$ . The given equation becomes \[j+14 = \lfloor \sqrt{70j-20} \rfloor\]
Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$ , we can take $x=\sqrt{70j-20}$ with $\lfloor x \rfloor =j+14$ to get \[j+14 \leq \sqrt{70j-20} < j+15\] We can square the inequality to get \[196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}\] The left inequality simplifies to $(j-36)(j-6) \leq 0$ , which yields \[6 \le j \le 36.\] The right inequality simplifies to $(j-20)^2 - 155 > 0$ , which yields \[j < 20 - \sqrt{155} < 8 \quad \text{or} \quad j > 20 + \sqrt{155} > 32\]
Solving $j < 8$ , and $6 \le j \le 36$ , we get $6 \le j < 8$ , for $2$ values $j\in \{6, 7\}$
Solving $j >32$ , and $6 \le j \le 36$ , we get $32 < j \le 36$ , for $4$ values $k\in \{33, \ldots , 36\}$
Thus, our answer is $2 + 4 = \boxed{6}$ | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$ . Set $k = \lfloor x \rfloor \ge 0$ ; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$ , we get \[k^2 \le 70k - 1000 < k^2 + 2k + 1.\] The left inequality simplifies to $(k-20)(k-50) \le 0$ , which yields \[20 \le k \le 50.\] The right inequality simplifies to $(k-34)^2 > 155$ , which yields \[k < 34 - \sqrt{155} < 22 \quad \text{or} \quad k > 34 + \sqrt{155} > 46\] Solving $k < 22$ , and $20 \le k \le 50$ , we get $20 \le k < 22$ , for $2$ values $k\in \{20, 21\}$
Solving $k >46$ , and $20 \le k \le 50$ , we get $46 < k \le 50$ , for $4$ values $k\in \{47, \ldots , 50\}$
Thus, our answer is $2 + 4 = \boxed{6}$ | C | 6 |
84e5dd35acc8075c78483b52f8a820db | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_24 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$ , so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into
\begin{align*}
\frac{k^2 + 1000}{70} &= k \\
k^2 - 70k + 1000 &= 0 \\
(k-20)(k-50) &= 0,
\end{align*}
so $k = 20$ or $k= 50$ , so $n = 400, 2500.$
If $n$ is not square, then we can say that, for a positive integer $k$ , we have
\begin{align*}
k^2 < &n < (k+1)^2 \\
k^2 + 1000 < &n + 1000 = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\
k^2 + 1000 < &70k < (k+1)^2 + 1000.
\end{align*}
To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities $k^2 + 1000 < 70k$ and $70k < (k+1)^2 + 1000$ . To solve the first one, we have
\begin{align*}
k^2 - 70k + 1000 &< 0 \\
(k-20)(k-50) &< 0\\
\end{align*} $k\in (20, 50),$ because the portion of the parabola between its two roots will be negative.
The second inequality yields
\begin{align*}
70k &< k^2 + 2k + 1 + 1000 \\
0 &< k^2 -68k + 1001.
\end{align*}
This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are $34 + \sqrt{155}>46$ and $34-\sqrt{155}<22$ (they are roughly equal, but this is to ensure that we do not miss any solutions).
Notation wise, we need all integers $k$ such that
\[k \in \left(20, 50\right) \cap \left(-\infty,34 - \sqrt{155} \right)\] or \[k \in \left(20, 50\right) \cap \left(34 + \sqrt{155}, \infty \right).\]
For the first one, since our uppoer bound is a little less than $22$ , the $k$ that works is $21$ . For the second, our lower bound is a little more than $46$ , so the $k$ that work are $47, 48,$ and $49$
$\boxed{6}$ total solutions for $n$ , since each value of $k$ corresponds to exactly one value of $n$ | C | 6 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Note that $96 = 2^5 \cdot 3$ . Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$ , we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$ : We see that there is $1$ way, merely $96$
$k=2$ : This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$ 's leave us with $5$ ways and symmetry doubles us so we have $10$
$k=3$ : We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of $2$ in a manner that is $3-0-0$ $2-1-0$ or $1-1-1$ . A $3-0-0$ split has $6 + 3 = 9$ ways of happening ( $24-2-2$ and symmetry; $2-3-16$ and symmetry), a $2-1-0$ split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a $1-1-1$ split has $3$ ways of happening ( $6-4-4$ and symmetry) so in this case we have $9+18+3=30$ ways.
$k=4$ : We have $3, 2, 2, 2$ as our baseline, and for the two other $2$ 's, we have a $2-0-0-0$ or $1-1-0-0$ split. The former grants us $4+12=16$ ways ( $12-2-2-2$ and symmetry and $3-8-2-2$ and symmetry) and the latter grants us also $12+12=24$ ways ( $6-4-2-2$ and symmetry and $3-4-4-2$ and symmetry) for a total of $16+24=40$ ways.
$k=5$ : We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.
$k=6$ : We have $3, 2, 2, 2, 2, 2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways.
Thus, adding, we have $1+10+30+40+25+6=\boxed{112}$ | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | As before, note that $96=2^5\cdot3$ , and we need to consider 6 different cases, one for each possible value of $k$ , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization needs to contain one factor that is itself a multiple of $3$ . There are $k$ to choose from. That leaves $k-1$ slots left to fill, each of which must contain at least one factor of $2$ . Once we have filled in a $2$ to each of the remaining slots, we're left with $5-(k-1)=6-k$ twos.
Consider the remaining $6-k$ factors of $2$ left to assign to the $k$ factors. Using stars and bars, the number of ways to do this is: \[{{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}\] This makes $k{5\choose{6-k}}$ possibilities for each k.
To obtain the total number of factorizations, add all possible values for k: \[\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{112}.\] | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Begin by examining $f_1$ $f_1$ can take on any value that is a factor of $96$ except $1$ . For each choice of $f_1$ , the resulting $f_2...f_k$ must have a product of $96/f_1$ . This means the number of ways the rest $f_a$ $1<a<=k$ can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)$ , since the product of $f_2 \cdot f_3... \cdot f_k=x$ is counted as one valid product if and only if $f_1 \cdot x=96$ , the product $x$ has the properties that factors are greater than $1$ , and differently ordered products are counted separately.
For example, say the first factor is $2$ . Then, the remaining numbers must multiply to $48$ , so the number of ways the product can be written beginning with $2$ is $D(48)$ . To add up all the number of solutions for every possible starting factor, $D(96/f_1)$ must be calculated and summed for all possible $f_1$ , except $96$ and $1$ , since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$ . This means
$D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$
Instead of calculating D for the larger factors first, reduce $D(48)$ $D(32)$ , and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation. Following the recursive definition $D(n)=($ sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to
$D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)$ $(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.$
$D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$ , so the sum further simplifies to
$D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5$ , after combining terms. From quick casework,
$D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1$ and $D(2)=1$ . Substituting these values into the expression above,
$D(96)=3 \cdot 8+4 \cdot 8+5 \cdot 4+4 \cdot 3+5 \cdot 2+4 \cdot 1+5 \cdot 1+5=\boxed{112}$ | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Note that $96 = 3 \cdot 2^5$ , and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$
First, calculate $D(32)$ . Since $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ , all you have to do was find the number of ways to divide the $2$ 's into groups, such that each group has at least one $2$ . By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$ , is not needed for the remaining calculations.)
Then, to get $D(96)$ , in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five.
The resulting number of possible sequences is $3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{112}$ . ~ emerald_block | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Consider the arrangement of the prime factors of 96 in a line $(2,2, 2, 2, 2, 3)$ . An arrangement of factors can be created by placing "dividers" to group primes. For example, $(2, 2, |, 2, 2, 2, |, 3)$ is equivalent to the arrangement $4 \cdot 8 \cdot 3$ . Because there are $6$ ways to order the prime factors, and $2^5$ ways to place dividers, this gives us an initial $6 \cdot 2^5$ ways to arrange divisors.
However, through this method, we overcount cases where $3$ is combined with another factor. For example, the arrangement $4 \cdot 6 \cdot 4$ can be written as $(2, 2, |, 2, 3, |, 2, 2)$ or $(2, 2, |, 3, 2, |, 2, 2)$ . Precisely, we double count any case with $6$ as a factor, triple count any case with $12$ , quadruple count any case with $24$ , etc.
Now, consider all cases where $3$ must be grouped with at least one $2$ . This can be expressed in the same "line" format as $(2, 2, 2, 2, 6)$ , where dividers can again be placed to group divisors. In this case, there are $5$ ways to order divisors, and $2^4$ ways to place dividers, so we have an $5 \cdot 2^4$ possible sequences for this case. Notice that in this format, we double count cases where $12$ is a factor, we triple count cases where $24$ is a factor, etc. Precisely, for any case counted $n$ times in the first step, it is counted $n - 1$ times in this step. Thus, if we subtract, we count each case exactly once.
So, we get:
$6 \cdot 2^5 - 5 \cdot 2^4 = \boxed{112}$ . ~ hdai1122 | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$ . By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$ 's, or multiply it to one of the $g_i$ 's, a total of $2m+1$ ways. Hence the answer to the problem is
\[\sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{112}.\] | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Note that $96 = 2^5 \cdot 3$ $D(n)$ depends on dividing $2^5$ into different terms, which is the integer partition of $5$
Divide $96$ into $1$ term:
There is only one way. $\underline{\textbf{1}}$
Divide $96$ into $2$ terms:
$2 + 8 = \underline{\textbf{10}}$
Divide $96$ into $3$ terms:
$12 + 18 = \underline{\textbf{30}}$
Divide $96$ into $4$ terms:
$24 + 16 = \underline{\textbf{40}}$
Divide $96$ into $5$ terms:
When dividing $5$ into $5$ parts there are $2$ cases.
$20 + 5 = \underline{\textbf{25}}$
Divide $96$ into $6$ terms:
$5 = 1 + 1 + 1 + 1 + 1$ . The number of arrangements of $(2, 2, 2, 2, 2, 3)$ is $\frac{6!}{5!} = \underline{\textbf{6}}$
$1 + 10 + 30 + 40 + 25 + 6 = \boxed{112}$ | A | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Ignore the $3$ first and first count $2^{x_1+x_2+...+x^n}=32$ which $x_1+x_2+...+x_n=5$ . This implies that $n$ is less than or equal to $5$ . Now, we can see that $3$ can lie between the two $x_i, x_{i+1}$ , or contribute to one of them. This gives $2k+1$ if $x_1+...+x_k=5$ . Now, just sum up gives $\binom{5-1}{5-1}\cdot 11+\binom{5-1}{4-1}\cdot 9+\binom{5-1}{3-1}\cdot 7+\binom{5-1}{2-1}\cdot 5+\binom{5-1}{1-1}\cdot 3=\boxed{112}$ | null | 112 |
faab445e43dddbc89304325efb42144f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_25 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Consider how we partition the factors of $96 = 2^5\cdot 3$ . For each $k$ , there are two cases. Either we can put the $2$ s into $k$ nonzero parts, so that the $3$ shares a partition with some $2$ s, which can be done in $k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}$ ways, or we can put the $2$ s into $k-1$ nonzero parts and put the $3$ in its own partition, which can be done in $k\binom{5-k+k-1}{k-1} = k\binom{4}{k-1}$ ways. Summing over all $k$ , we have \[\sum_{k=1}^6 \binom{4}{k-2} + \sum_{k=1}^6 \binom{4}{k-1}.\]
But $\sum_{k=1}^6 \binom{4}{k-2} = 2\binom{4}{0}+3\binom{4}{1} + \dots + 6\binom{4}{4} = 4 \cdot \sum_{i=0}^4 \binom{4}{i} = 4 \cdot 2^4 = 64$ . Similarly, $\sum_{k=1}^6 \binom{4}{k-1} = 3\cdot 2^4 = 48$ . So our answer is $64+48 = \boxed{112}$ | A | 112 |
91e01dfe831ce598f96a687a425bc3b7 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_1 | What is the value of \[2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?\]
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | $2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9$
$= 1+1 = \boxed{2}$ | C | 2 |
2f2814b5af9841b0952522d9c21d94b4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_3 | Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,
\[A-1 = 5(B-1)\] and \[A = B^2.\]
Substituting the second equation into the first gives us
\[B^2-1 = 5(B-1).\]
By using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$
The answer is $16-4 = 12 \implies \boxed{12}$ | D | 12 |
2f2814b5af9841b0952522d9c21d94b4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_3 | Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | Simple guess and check works. Start with all the square numbers - $1$ $4$ $9$ $16$ $25$ $36$ , etc. (probably stop at around $100$ since at that point it wouldn't make sense). If Ana is $9$ , then Bonita is $3$ , so in the previous year, Ana's age was $4$ times greater than Bonita's. If Ana is $16$ , then Bonita is $4$ , and Ana's age was $5$ times greater than Bonita's in the previous year, as required. The difference in the ages is $16 - 4 = 12. \boxed{12}$ | D | 12 |
2f2814b5af9841b0952522d9c21d94b4 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_3 | Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | The second sentence of the problem says that Ana's age was once $5$ times Bonita's age. Therefore, the difference of the ages $n$ must be divisible by $4.$ The only answer choice which is divisible by $4$ is $12 \rightarrow \boxed{12}.$ | D | 12 |
fc30ba5fbc0078b42ae28e42306bab6c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_4 | A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?
$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$ | We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee.
Namely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any one color. Drawing one more ball guarantees that we will get $15$ balls of one color — either red, green, or yellow. Thus, the answer is $75 + 1 = \boxed{76}$ | B | 76 |
c6ba51c90cefa9e1c985fd8b02333488 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_5 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | We might at first think that the answer would be $9$ , because $1+2+3 \dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$ . Thus, the answer is, intuitively, $\boxed{90}$ integers. | D | 90 |
c6ba51c90cefa9e1c985fd8b02333488 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_5 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\frac{45}{\frac12}=\boxed{90}$ | D | 90 |
c730c698b920a0d85cc8dfeb4dfed712 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_6 | For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$ | This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer is $\boxed{3}$ | C | 3 |
c730c698b920a0d85cc8dfeb4dfed712 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_6 | For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$ | We can use a process of elimination. Going down the list, we can see a square obviously works. A rectangle that is not a square works as well. Both rhombi and parallelograms don't have a point that is equidistant, but isosceles trapezoids do have such a point, so the answer is $\boxed{3}$ | C | 3 |
c730c698b920a0d85cc8dfeb4dfed712 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_6 | For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$ | The perpendicular bisector of a line segment is the locus of all points that are equidistant from the endpoints. The question then boils down to finding the shapes where the perpendicular bisectors of the sides all intersect at a point. This is true for a square, rectangle, and isosceles trapezoid, so the answer is $\boxed{3}$ | C | 3 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ .
Now we find the intersection points between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2}$ , whose area is $\boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem , we can directly find that the area is $\boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at $(4, 6)$ and $(6, 4)$ . Then apply Heron's Formula: the semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$ , so the area reduces nicely to a difference of squares, making it $\boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . We can now draw the bounding square with vertices $(2, 2)$ $(2, 6)$ $(6, 6)$ and $(6, 2)$ , and deduce that the triangle's area is $16-4-2-4=\boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\frac62 + 4 - 1 = \boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45\] Therefore, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$ , so the area is $\frac12 bc \sin A = \frac12 \cdot 2\sqrt5 \cdot 2\sqrt5 \cdot \frac35 = \boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \[\frac12\begin{Vmatrix} 2&2&1\\ 4&6&1\\ 6&4&1\\ \end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 = \boxed{6}\] | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . Then vectors $\overrightarrow{AB} = \langle 2, 4 \rangle$ and $\overrightarrow{AC} = \langle 4, 2 \rangle$ . The area of the triangle is half the magnitude of the cross product of these two vectors. \[\frac12\begin{Vmatrix} i&j&k\\ 2&4&0\\ 4&2&0\\ \end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{6}\] | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . By the Pythagorean theorem, this is an isosceles triangle with base $2\sqrt2$ and equal length $2\sqrt5$ . The area of an isosceles triangle with base $b$ and equal length $l$ is $\frac{b\sqrt{4l^2-b^2}}{4}$ . Plugging in $b = 2\sqrt2$ and $l = 2\sqrt5$ \[\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{6}\] | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45\] Therefore, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$ . By the extended Law of Sines, \[2R = \frac{a}{\sin A} = \frac{2\sqrt2}{\frac35} = \frac{10\sqrt2}{3}\] \[R = \frac{5\sqrt2}{3}\] Then the area is $\frac{abc}{4R} = \frac{2\sqrt2 \cdot 2\sqrt5^2}{4 \cdot \frac{5\sqrt2}{3}} = \boxed{6}$ | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | The area of a triangle formed by three lines, \[a_1x + a_2y + a_3 = 0\] \[b_1x + b_2y + b_3 = 0\] \[c_1x + c_2y + c_3 = 0\] is the absolute value of \[\frac12 \cdot \frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \cdot \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\\ \end{vmatrix}^2\] Plugging in the three lines, \[-x + 2y - 2 = 0\] \[-2x + y + 2 = 0\] \[x + y - 10 = 0\] the area is the absolute value of \[\frac12 \cdot \frac{1}{(-2-1)(-1-2)(-1+4)} \cdot \begin{vmatrix} -1&2&-2\\ -2&1&2\\ 1&1&-10\\ \end{vmatrix}^2 = \frac12 \cdot \frac{1}{27} \cdot 18^2 = \boxed{6}\] Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991. | C | 6 |
db9ab8a6f647f4e20488b6bfc5ca541b | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_7 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that our triangle is isosceles with legs of $2\sqrt5$ and base $2\sqrt2$ . Then, the semi - perimeter of our triangle is, \[\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.\] Applying Heron's formula, we find that the area of this triangle is equivalent to \[\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqrt5-\sqrt2)}{(2)}} = \sqrt{{(20-2)}{(2)}} = \boxed{6}.\] | C | 6 |
ddba2142409c4131a9e16a1178b1c8f2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_9 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$ | The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ , and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$ , which means they would be able to cancel with the factors in $n!$ . Thus, the sum of $n$ positive integers would be a divisor of $n!$ when $n+1$ is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, $n+1$ must instead be prime. The greatest three-digit integer that is prime is $997$ , so we subtract $1$ to get $n=\boxed{996}$ | B | 996 |
ddba2142409c4131a9e16a1178b1c8f2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_9 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$ | As in Solution 1, we deduce that $n+1$ must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of $n$ . Choices $A$ $C$ , and $E$ don't work because $n+1$ is even, and all even numbers are divisible by two, which makes choices $A$ $C$ , and $E$ composite and not prime. Choice $D$ also does not work since $999$ is divisible by $9$ , which means it's a composite number and not prime. Thus, the correct answer must be $\boxed{996}$ | B | 996 |
ddba2142409c4131a9e16a1178b1c8f2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_9 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is $\underline{not}$ a divisor of the product of the first $n$ positive integers?
$\textbf{(A) } 995 \qquad\textbf{(B) } 996 \qquad\textbf{(C) } 997 \qquad\textbf{(D) } 998 \qquad\textbf{(E) } 999$ | The sum of the first $n$ positive integers is $\frac{(n)(n+1)}{2}$ and the product of the positive integers upto $n$ is $n!$ . The quotient of the two is -
$\frac{(2)(n!)}{(n)(n+1)}$
which simplifies to $\frac{(2)((n-1)!)}{n+1}$ . Thus, $n+1$ must be odd for the remainder to not be 0 (as $2$ will multiply with some number in $n!$ , cancelling out $n+1$ if it is even, which leaves us with the answer choices $n = 996$ and $n = 998$ . Notice that $n + 1$ must also be prime as otherwise there will be a factor of $n + 1$ in $2$ $n!$ somewhere. So either $997$ or $999$ must be prime - $999$ is obviously not prime as it is divisible by 9, so our answer should be $n$ where $n + 1 = 997$ , and so our answer is $n = 996$ or $\boxed{996}$ | B | 996 |
37a52accb87db88639f07386c5011bd6 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_10 | A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$ | The number of tiles the bug visits is equal to $1$ plus the number of times it crosses a horizontal or vertical line. As it must cross $16$ horizontal lines and $9$ vertical lines, it must be that the bug visits a total of $16+9+1 = \boxed{26}$ squares. | C | 26 |
37a52accb87db88639f07386c5011bd6 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_10 | A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$ | We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ feet sections.
[asy] unitsize(20); for(int i =0; i<= 7; ++i) { for(int j =0; j<= 4; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle); } for(int k =0; k<= 4; ++k) { draw((8,k)--(8.5,k)--(8.5,k+1)--(8,k+1)--cycle); } } draw((0,5)--(8.5,0)--cycle); [/asy]
Though it may appear that the line we drew comes very close to several points, we know that since $10$ and $17$ are relatively prime (numbers where the only positive integer that divides both of them is 1, a.k.a. numbers with a gcd of 1), the line will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count the number of squares the line passes through using the diagram, we get $13$ squares. We can then multiply this by 2 to find out the total number of squares the bug passes through on the rectangular floor giving us a total of $2 \cdot 13 = \boxed{26}$ | C | 26 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | Prime factorizing $201^9$ , we get $3^9\cdot67^9$ .
A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get perfect square are $0, 2, 4, 6, 8$ for both $3$ and $67$ . This yields $5\cdot5 = 25$ perfect squares.
Perfect cubes must have multiples of $3$ for each of their prime factors' exponents, so we have either $0, 3, 6$ , or $9$ for both $3$ and $67$ , which yields $4\cdot4 = 16$ perfect cubes, for a total of $25+16 = 41$
Subtracting the overcounted powers of $6$ $3^0\cdot67^0$ $3^0\cdot67^6$ $3^6\cdot67^0$ , and $3^6\cdot67^6$ ), we get $41-4 = \boxed{37}$ | C | 37 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | Observe that $201 = 67 \cdot 3$ . Now divide into cases:
Case 1 : The factor is $3^n$ . Then we can have $n = 2$ $3$ $4$ $6$ $8$ , or $9$
Case 2 : The factor is $67^n$ . This is the same as Case 1.
Case 3 : The factor is some combination of $3$ s and $67$ s.
This would be easy if we could just have any combination, as that would simply give $6 \cdot 6$ . However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for $n$
$n = 2$ is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.
$n = 3$ is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of $3$
$n = 4$ is a "square".
$n = 6$ is interesting, since it's both a "square" and a "cube". Don't count this as either because this would double-count, so we will count this in another case.
$n = 8$ is a "square"
$n = 9$ is a "cube".
Now let's consider subcases:
Subcase 1 : The squares are with each other.
Since we have $3$ square terms, and they would pair with $3$ other square terms, we get $3 \cdot 3 = 9$ possibilities.
Subcase 2 : The cubes are with each other.
Since we have $2$ cube terms, and they would pair with $2$ other cube terms, we get $2 \cdot 2 = 4$ possibilities.
Subcase 3 : A number pairs with $n=6$
Since any number can pair with $n=6$ (as it gives both a square and a cube), there would be $6$ possibilities. Remember however that there can be two different bases ( $3$ and $67$ ), and they would produce different results. Thus, there are in fact $6 \cdot 2 = 12$ possibilities.
Finally, summing the cases gives $6+6+9+4+12 = \boxed{37}$ | C | 37 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | We first prime factorize $201^9 = 3^9 \cdot 67^9$ . Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of $3$ in the exponent. The largest square for $3$ can be $3^8$ , so their must be $\dfrac {8}{2} = 4$ ways. The largest cube is $3^9$ , so there must be $\dfrac{9}{3} = 3$ . Minus one $3^6$ due to overlapping and we get $4 + 3 -1 = 6$ ways for $3$ to be a cube/square. We can see that this same thing happens for $67^9$ due to the same exponent. Adding $0$ as a case, we have our answer; $6 \cdot 6 + 1 = \boxed{37}$ | C | 37 |
c5ad213bba44545028b1fdacc24ae607 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_11 | How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)? $\textbf{(A) }32 \qquad \textbf{(B) }36 \qquad \textbf{(C) }37 \qquad \textbf{(D) }39 \qquad \textbf{(E) }41$ | Notice that $201=3 \cdot 67$ . We factorize $201^9$ to get $3^9 \cdot 67^9$ . We then list perfect squares and cubes. $3^2$ $3^4$ $3^6$ $3^8$ $3^3$ $3^6$ $3^9$ $67^2$ $67^4$ $67^6$ $67^8$ $67^3$ $67^6$ $67^9$ . Notice that the powers of $6$ overlap. We must not forget $1$ though. Of course, all of these factors already work. This gives us $15-2=3$ . Next, we count the perfect squares. Since there are $4$ options we have $4 \cdot 4=16$ . We do the same for the perfect cubes except with 3 options this time, and we have $3 \cdot 3=9$ . However, we accidentally overcounted $3^6 \cdot 67^6$ . We add our answers and subtract $1$ to get $13+16+9-1 = \boxed{37}$ | C | 37 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$ | [asy] unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));[/asy]
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$ . We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$
\[\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}\]
\[\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}\]
Then, we take triangle $BFC$ , and find $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{110}.$ | D | 110 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$ | Alternatively, we could have used similar triangles. We start similarly to Solution 1.
Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Therefore, \[\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.\]
So, $\triangle BEF \sim BDA$ by AA Similarity, since $\angle EBF = \angle DBA$ and $\angle BEC = 90^{\circ} = \angle BDA$ .
Thus, we know \[\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.\]
Finally, we deduce \[\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{110}.\] | D | 110 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$ | Through the property of angles formed by intersecting chords, we find that \[m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}\]
Through the Outside Angles Theorem, we find that \[m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}\]
Adding the two equations gives us \[m\angle BFC + m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB\]
Since $\overarc{BC}$ is the diameter, $m\overarc{BC}=180^{\circ}$ , and because $\triangle ABC$ is isosceles and $m\angle ACB=40^{\circ}$ , we have $m\angle CAB=70^{\circ}$ . Thus \[m\angle BFC=180^{\circ}-70^{\circ}=\boxed{110}\] | D | 110 |
514e7909abcc9a9bd591269b5aa67039 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$ | Notice that if $\angle BEC = 90^{\circ}$ , then $\angle BCE$ and $\angle ACE$ must be $20^{\circ}$ . Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that $\angle EBD \cong \angle ECD = 20^{\circ}$ . Thus $\angle CBF = 70 - 20 = 50^{\circ}$ , and so $\angle BFC = 180 - 20 - 50 = 110^{\circ}$ , which is $\boxed{110}$ | D | 110 |
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