problem_id
stringlengths 32
32
| link
stringlengths 75
84
| problem
stringlengths 14
5.33k
| solution
stringlengths 15
6.63k
| letter
stringclasses 5
values | answer
stringclasses 957
values |
|---|---|---|---|---|---|
873f32456f560f8353e47729bf0a87ad | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_24 | What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$ $0\le v\le1,$ and $0\le w\le1$
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$ | [asy] import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5); pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1); //draw(A--B--D--C--cycle); draw(A--B); label("1",midpoint(A--B),W); label("2",midpoint(D--B),N); draw(A--C,dashed); draw(B--D); draw(C--D, dashed); draw(A--AA); label("5",midpoint(A--AA),W); draw(B--BB,dashed); draw(C--CC,dashed); draw(D--DD); label("5",midpoint(D--DD),E); label("1",midpoint(CC--DD),E); label("2",midpoint(AA--CC),S); // Dotted vertices dot(A); dot(B); dot(C); dot(D); dot(AA); dot(BB); dot(CC); dot(DD); draw(AA--BB,dashed); draw(AA--CC); draw(BB--DD,dashed); draw(CC--DD); label("(0,0)",AA,W); label("(-3,4)",A,SW); label("(-1,5)",D,E); label("(2,1)",DD,NE); [/asy] Notice that we are given a parametric form of the region, and $w$ is used in both $x$ and $y$ . We first fix $u$ and $v$ to $0$ , and graph $(-3w,4w)$ from $0\le w\le1$ . When $w$ is $0$ , we have the point $(0,0)$ , and when $w$ is $1$ , we have the point $(-3,4)$ . We see that since this is a directly proportional function, we can just connect the dots like this:
[asy] import graph; Label f; size(5cm); unitsize(0.7cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); [/asy]
Now, when we vary $u$ from $0$ to $2$ , this line is translated to the right $2$ units:
[asy] import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((2,0)--(-1,4)); [/asy]
We know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$ , so we shade in the region:
[asy] import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((2,0)--(-1,4)); filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); [/asy]
We can also shift this quadrilateral one unit up, because of $v$ . Thus, this is our figure:
[asy] import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((2,0)--(-1,4)); filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); filldraw((0,1)--(-3,5)--(-1,5)--(2,1)--cycle, gray); draw((0,0)--(0,1),black+dashed); draw((2,0)--(2,1),black+dashed); draw((-3,4)--(-3,5),black+dashed); [/asy]
[asy] import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((1,0)--(-2,4)); filldraw((0,0)--(2,0)--(2,1)--(-1,5)--(-3,5)--(-3,4)--cycle, gray); [/asy]
The length of the boundary is simply $1+2+5+1+2+5$ $5$ can be obtained by Pythagorean theorem since we have side lengths $3$ and $4$ .). This equals $\boxed{16.}$ | E | 16. |
873f32456f560f8353e47729bf0a87ad | https://artofproblemsolving.com/wiki/index.php/2023_AMC_10B_Problems/Problem_24 | What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$ $0\le v\le1,$ and $0\le w\le1$
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$ | We can find the "boundary points" and work with our intuition to solve the problem. We set each of $u, v, w$ equal to $0, 1$ for a total of $8$ combinations in $u, v, w$ . We now test each one.
Case 1: $u = 0, v = 0, w = 0 \implies (0, 0)$
Case 2: $u = 0, v = 0, w = 1 \implies (-3, 4)$
Case 3: $u = 0, v = 1, w = 0 \implies (0, 1)$
Case 4: $u = 0, v = 1, w = 1 \implies (-3, 5)$
Case 5: $u = 1, v = 0, w = 0 \implies (2, 0)$
Case 6: $u = 1, v = 0, w = 1 \implies (-1, 4)$
Case 7: $u = 1, v = 1, w = 0 \implies (2, 1)$
Case 8: $u = 1, v = 1, w = 1 \implies (-1, 5)$
When graphed on a coordinate plane, the points appear as follows.
[asy] import graph; import geometry; Label f; size(5cm); unitsize(0.7cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); pair A = (0, 0); dot (A); pair B = (-3, 4); dot (B); pair C = (0, 1); dot (C); pair D = (-3, 5); dot (D); pair E = (2, 0); dot (E); pair F = (-1, 4); dot (F); pair G = (2, 1); dot (G); pair H = (-1, 5); dot (H); [/asy]
Notice how there are two distinct rectangles visible in the figure. This leads us to believe that the region tracks the motion of this region as it travels in space. To understand why this is true, we can imagine a fixed $w$ (as it is present in both the $x$ and $y$ coordinates). Then if we hold one of $u$ or $v$ fixed and let the other vary, we get a straight line parallel to the $x$ or $y$ axis respectively. If we let the other vary, we get the other type of straight line. Together, they form a rectangular region. In addition, $w$ serves as a diagonal translation, so if we now let $w$ vary, it traces out the motion of the rectangle. Keeping this in mind, we connect the dots.
[asy] import graph; import geometry; Label f; size(5cm); unitsize(0.7cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)--(-3,5)--(-1,5)--(2,1)--(2,0)--cycle, gray); [/asy]
Each of the diagonal sides have length $5$ by the distance formula on $(0,0)$ and $(-3,4)$ (the other diagonal side is congruent), so our total area is $2 + 1 + 5 + 2 + 1 + 5 = \boxed{16}$ | E | 16 |
790928f8d0ff228967807f6d0d6feb13 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_2 | Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$ | Mike's speed is $\frac{15}{57}=\frac{5}{19}$ laps per minute.
In the first $27$ minutes, he completed approximately $\frac{5}{19}\cdot27\approx\frac{1}{4}\cdot28=\boxed{7}$ laps. | B | 7 |
790928f8d0ff228967807f6d0d6feb13 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_2 | Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$ | Mike runs $1$ lap in $\frac{57}{15}=\frac{19}{5}$ minutes. So, in $27$ minutes, Mike ran about $\frac{27}{\frac{19}{5}} \approx \boxed{7}$ laps. | B | 7 |
790928f8d0ff228967807f6d0d6feb13 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_2 | Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$ | Mike's rate is \[\frac{15}{57}=\frac{x}{27},\] where $x$ is the number of laps he can complete in $27$ minutes.
If you cross multiply, $57x = 405$
So, $x = \frac{405}{57} \approx \boxed{7}$ | B | 7 |
790928f8d0ff228967807f6d0d6feb13 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_2 | Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$ | Note that $27$ minutes is a little bit less than half of $57$ minutes. Mike will therefore run a little bit less than $15/2=7.5$ laps, which is about $\boxed{7}$ | B | 7 |
790928f8d0ff228967807f6d0d6feb13 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_2 | Mike cycled $15$ laps in $57$ minutes. Assume he cycled at a constant speed throughout. Approximately how many laps did he complete in the first $27$ minutes?
$\textbf{(A) } 5 \qquad\textbf{(B) } 7 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 13$ | Note that $57$ minutes is almost equal to $1$ hour. Running $15$ laps in $1$ hour is running approximately $1$ lap every $4$ minutes. This means that in $27$ minutes, Mike will run approximately $\frac{27}{4}$ laps. This is very close to $\frac{28}{4} = \boxed{7}$ | B | 7 |
5cdbc783bde301c2bb337aac2c0ae46b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_3 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$
We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$
Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{5}.$ | E | 5 |
5cdbc783bde301c2bb337aac2c0ae46b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_3 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$
Therefore, we get \begin{align*} 6z+(z+40)+z&=96 \\ 8z+40&=96 \\ 8z&=56 \\ z&=7. \end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$
So, the answer is $|x-y|=|42-47|=\boxed{5}.$ | E | 5 |
5cdbc783bde301c2bb337aac2c0ae46b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_3 | The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{5}.\] [email protected], vvsss | E | 5 |
4e2135d395aa59ac6a38b40e9ce3e1fb | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_4 | In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles per gallon?
$\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}$ | The formula for fuel efficiency is \[\frac{\text{Distance}}{\text{Gas Consumption}}.\] Note that $1$ mile equals $\frac 1m$ kilometers. We have \[\frac{x\text{ miles}}{1\text{ gallon}} = \frac{\frac{x}{m}\text{ kilometers}}{l\text{ liters}} = \frac{1\text{ kilometer}}{\frac{lm}{x}\text{ liters}} = \frac{100\text{ kilometers}}{\frac{100lm}{x}\text{ liters}}.\] Therefore, the answer is $\boxed{100}.$ | E | 100 |
4e2135d395aa59ac6a38b40e9ce3e1fb | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_4 | In some countries, automobile fuel efficiency is measured in liters per $100$ kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals $m$ miles, and $1$ gallon equals $l$ liters. Which of the following gives the fuel efficiency in liters per $100$ kilometers for a car that gets $x$ miles per gallon?
$\textbf{(A) } \frac{x}{100lm} \qquad \textbf{(B) } \frac{xlm}{100} \qquad \textbf{(C) } \frac{lm}{100x} \qquad \textbf{(D) } \frac{100}{xlm} \qquad \textbf{(E) } \frac{100lm}{x}$ | Since it can be a bit odd to think of "liters per $100$ km", this statement's numerical value is equivalent to $100$ km per $1$ liter:
$1$ km requires $l$ liters, so the numerator is simply $l$ . Since $l$ liters is $1$ gallon, and $x$ miles is $1$ gallon, we have $1\text{ liter} = \frac{x}{l}$
Therefore, the requested expression is \[100\cdot\frac{m}{(\frac{x}{l})} = \boxed{100}.\] -Benedict T (countmath1) | E | 100 |
9afeb5638f92e71fc41bbbed7fdc7f7d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_7 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | Note that \begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*} Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:
Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{6}.$ | B | 6 |
9afeb5638f92e71fc41bbbed7fdc7f7d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_7 | The least common multiple of a positive integer $n$ and $18$ is $180$ , and the greatest common divisor of $n$ and $45$ is $15$ . What is the sum of the digits of $n$
$\textbf{(A) } 3 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 12$ | The options for $\text{lcm}(x, 18)=180$ are $20$ $60$ , and $180$ . The options for $\text{gcd}(y, 45)=15$ are $15$ $30$ $60$ $75$ , etc. We see that $60$ appears in both lists; therefore, $6+0=\boxed{6}$ | B | 6 |
ff96ebcbae6db7e51d9d4b757bd22a00 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_8 | A data set consists of $6$ (not distinct) positive integers: $1$ $7$ $5$ $2$ $5$ , and $X$ . The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$
$\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40$ | First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:
Case 1: the mean is $5$
$X = 5 \cdot 6 - 20 = 10$
Case 2: the mean is $7$
$X = 7 \cdot 6 - 20 = 22$
Case 3: the mean is $X$
$X= \frac{20+X}{6} \Rightarrow X=4$
Therefore, the answer is $10+22+4=\boxed{36}$ | D | 36 |
f0765456e01cdc45e64d4fa98795e027 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_9 | A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); [/asy]
$\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720$ | The top left rectangle can be $5$ possible colors. Then the bottom left region can only be $4$ possible colors, and the bottom middle can only be $3$ colors since it is next to the top left and bottom left. Similarly, we have $3$ choices for the top right and $3$ choices for the bottom right, which gives us a total of $5\cdot4\cdot3\cdot3\cdot3=\boxed{540}$ | D | 540 |
f0765456e01cdc45e64d4fa98795e027 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_9 | A rectangle is partitioned into $5$ regions as shown. Each region is to be painted a solid color - red, orange, yellow, blue, or green - so that regions that touch are painted different colors, and colors can be used more than once. How many different colorings are possible?
[asy] size(5.5cm); draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((2,0)--(8,0)--(8,2)--(2,2)--cycle); draw((8,0)--(12,0)--(12,2)--(8,2)--cycle); draw((0,2)--(6,2)--(6,4)--(0,4)--cycle); draw((6,2)--(12,2)--(12,4)--(6,4)--cycle); [/asy]
$\textbf{(A) }120\qquad\textbf{(B) }270\qquad\textbf{(C) }360\qquad\textbf{(D) }540\qquad\textbf{(E) }720$ | Case 1: All the rectangles are different colors. It would be $5! = 120$ choices.
Case 2: Two rectangles that are the same color. Grouping these two rectangles as one gives us $5\cdot4\cdot3\cdot2 = 120$ . But, you need to multiply this number by three because the same-colored rectangles can be chosen at the top left and bottom right, the top right and bottom left, or the bottom right and bottom left, which gives us a grand total of $360$
Case 3: We have two sets of rectangles chosen from these choices (top right & bottom left, top left & bottom right) that have the same color. However, the choice of the bottom left and bottom right does not work for this case, as the second pair would be chosen from two touching rectangles. Again, grouping the same-colored rectangles gives us $5\cdot4\cdot3 = 60$
Therefore, we have $120 + 360 + 60 = \boxed{540}$ | D | 540 |
3b39d7e0d510dec63756450f5a41acd5 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_10 | Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters.
Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters, as shown below. What is the area of the original index card? [asy] // Diagram by MRENTHUSIASM, edited by Djmathman size(200); defaultpen(linewidth(0.6)); draw((489.5,-213) -- (225.5,-213) -- (225.5,-185) -- (199.5,-185) -- (198.5,-62) -- (457.5,-62) -- (457.5,-93) -- (489.5,-93) -- cycle); draw((206.29,-70.89) -- (480.21,-207.11), linetype ("6 6"),Arrows(size=4,arrowhead=HookHead)); draw((237.85,-182.24) -- (448.65,-95.76),linetype ("6 6"),Arrows(size=4,arrowhead=HookHead)); label("$1$",(450,-80)); label("$1$",(475,-106)); label("$8$",(300,-103)); label("$4\sqrt 2$",(300,-173)); [/asy] $\textbf{(A) } 14 \qquad \textbf{(B) } 10\sqrt{2} \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 12\sqrt{2} \qquad \textbf{(E) } 18$ | [asy] /* Edited by MRENTHUSIASM */ size(250); real x, y; x = 6; y = 3; draw((0,0)--(x,0)); draw((0,0)--(0,y)); draw((0,y)--(x,y)); draw((x,0)--(x,y)); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((x-0.5,y)--(x-0.5,y-0.5)--(x,y-0.5)); draw((0.5,0.5)--(x-0.5,y-0.5),dashed,Arrows()); draw((x,0)--(0,y),dashed,Arrows()); label("$1$",(x-0.5,y-0.25),W); label("$1$",(x-0.25,y-0.5),S); label("$8$",midpoint((0.5,y-0.5)--(x/2,y/2)),(0,2.5)); label("$4\sqrt{2}$",midpoint((0.5,0.5)--(x/2,y/2)),S); label("$A$",(0,0),SW); label("$E$",(0,0.5),W); label("$F$",(0.5,0),S); label("$I$",(0.5,0.5),N); label("$D$",(x,y),NE); label("$G$",(x-0.5,y),N); label("$H$",(x,y-0.5),E); label("$J$",(x-0.5,y-0.5),S); Label L1 = Label("$w$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$\ell$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw((0,-1)--(x,-1), L=L1, arrow=Arrows(),bar=Bars(15)); draw((x+1,0)--(x+1,y), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Label the bottom left corner of the larger rectangle (without the square cut out) as $A$ and the top right as $D$ $w$ is the width of the rectangle and $\ell$ is the length. Now we have vertices $E, F, G, H$ as vertices of the irregular octagon created by cutting out the squares. Let $I, J$ be the two closest vertices formed by the squares.
The distance between the two closest vertices of the squares is thus $IJ=\left(4\sqrt{2}\right).$ Substituting, we get
\[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\] Using the fact that the diagonal of the rectangle is $8,$ we get \[w^2+\ell^2 = 64.\] Subtracting the first equation from the second equation, we get \[4w+4\ell=40 \implies w+\ell = 10.\] Squaring yields \[w^2 + 2w\ell + \ell^2 = 100.\] Subtracting the second equation from this, we get $2w\ell = 36,$ and thus area of the original rectangle is $w\ell = \boxed{18}.$ | E | 18 |
6e59cea6c9ffc86137767010cb42812f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_11 | Ted mistakenly wrote $2^m\cdot\sqrt{\frac{1}{4096}}$ as $2\cdot\sqrt[m]{\frac{1}{4096}}.$ What is the sum of all real numbers $m$ for which these two expressions have the same value?
$\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$ | We are given that \[2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.\] Converting everything into powers of $2,$ we have \begin{align*} 2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \\ 2^{m-6} &= 2^{1-\frac{12}{m}} \\ m-6 &= 1-\frac{12}{m}. \end{align*} We multiply both sides by $m$ , then rearrange as \[m^2-7m+12=0.\] By Vieta's Formulas, the sum of such values of $m$ is $\boxed{7}.$ | C | 7 |
6e59cea6c9ffc86137767010cb42812f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_11 | Ted mistakenly wrote $2^m\cdot\sqrt{\frac{1}{4096}}$ as $2\cdot\sqrt[m]{\frac{1}{4096}}.$ What is the sum of all real numbers $m$ for which these two expressions have the same value?
$\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$ | We can rewrite the equation using fractional exponents and take logarithms of both sides: \[\log_2{(2^{m}\cdot4096^{-1/2}}) = \log_2{(2\cdot4096^{-1/m})}.\] We can then use the additive properties of logarithms to split them up: \[\log_2{(2^{m})} + \log_2{(4096^{-1/2})} = \log_2{2} + \log_2{(4096^{-1/m})}.\] Using the power rule, the fact that $4096 = 2^{12},$ and bringing the exponents down, we get \begin{align*} m - 6 &= 1 - \frac{12}{m} \\ m + \frac{12}{m} &= 7 \\ m^{2} + 12 &= 7m \\ m^{2} - 7m + 12 &= 0 \\ (m-3)(m-4) &= 0, \end{align*} from which $m = 3$ or $m = 4$ . Therefore, the answer is $3+4 = \boxed{7}.$ | C | 7 |
6e59cea6c9ffc86137767010cb42812f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_11 | Ted mistakenly wrote $2^m\cdot\sqrt{\frac{1}{4096}}$ as $2\cdot\sqrt[m]{\frac{1}{4096}}.$ What is the sum of all real numbers $m$ for which these two expressions have the same value?
$\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$ | Since surd roots are conventionally positive integers, assume $m$ is an integer, so $m$ can only be $1$ $2$ $3$ $4$ $6$ , and $12$ $\sqrt{\frac{1}{4096}}=\frac{1}{64}$ . Testing out $m$ , we see that only $3$ and $4$ work. Hence, $3+4=\boxed{7}$ | C | 7 |
5555ebfd218fffae74e29ec6a1ad6f81 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_12 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$ | Note that:
Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that \begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*} Subtracting the second equation from the first, we have $T=22-15=\boxed{7}.$ | A | 7 |
5555ebfd218fffae74e29ec6a1ad6f81 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_12 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$ | Consider when the principal asks "Are you a liar?": The truth tellers truthfully say no, and the liars lie and say no. This leaves only alternaters who lie on this question to answer yes. Thus, all $9$ children that answered yes are alternaters that falsely answer Questions 1 and 3, and truthfully answer Question 2. The rest of the alternaters, however many there are, have the opposite behavior.
Consider the second question, "Are you an alternater?": The truth tellers again answer no, the liars falsely answer yes, and alternaters that truthfully answer also say yes. From the previous part, we know that $9$ alternaters truthfully answer here. Because only liars and $9$ alternaters answer yes, we can deduce that there are $15-9=6$ liars.
Consider the first question, "Are you a truth teller?": Truth tellers say yes, liars also say yes, and alternaters that lie on this question also say yes. From the first part, we know that $9$ alternaters lie here. From the previous part, we know that there are $6$ liars. Because only the number of truth tellers is unknown here, we can deduce that there are $22-9-6=7$ truth tellers.
The final question is how many pieces of candy did the principal give to truth tellers. Because truth tellers answer yes on only the first question, we know that all $7$ of them said yes once, resulting in $\boxed{7}$ pieces of candy. | A | 7 |
5555ebfd218fffae74e29ec6a1ad6f81 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_12 | On Halloween $31$ children walked into the principal's office asking for candy. They
can be classified into three types: Some always lie; some always tell the truth; and
some alternately lie and tell the truth. The alternaters arbitrarily choose their first
response, either a lie or the truth, but each subsequent statement has the opposite
truth value from its predecessor. The principal asked everyone the same three
questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who
answered yes.
How many pieces of candy in all did the principal give to the children who always
tell the truth?
$\textbf{(A) } 7 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 31$ | Note that we have $4$ types of people:
Given this information, it is reasonable to ignore the fourth type, because they will never answer yes to any question. Hence, we only consider people of type $1, 2,$ and $3.$
The principal's first question implies that \[T + L + A = 22.\] The second question implies that \[L + A = 15.\] The third question implies that \[A = 9.\]
Solving, we find that $T = 7,$ so $\boxed{7}$ is our answer. We can also note that $T = 7, L = 6, A = 9,$ and there are $9$ alternators who answer no to every question. | A | 7 |
3bb15d2c554e600eda5e886b9a27c5bd | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | Suppose that $\overline{BD}$ intersects $\overline{AP}$ and $\overline{AC}$ at $X$ and $Y,$ respectively. By Angle-Side-Angle, we conclude that $\triangle ABX\cong\triangle AYX.$
Let $AB=AY=2x.$ By the Angle Bisector Theorem, we have $AC=3x,$ or $YC=x.$
By alternate interior angles, we get $\angle YAD=\angle YCB$ and $\angle YDA=\angle YBC.$ Note that $\triangle ADY \sim \triangle CBY$ by the Angle-Angle Similarity, with the ratio of similitude $\frac{AY}{CY}=2.$ It follows that $AD=2CB=2(BP+PC)=\boxed{10}.$ | C | 10 |
3bb15d2c554e600eda5e886b9a27c5bd | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | Let the intersection of $AC$ and $BD$ be $M$ , and the intersection of $AP$ and $BD$ be $N$ . Draw a line from $M$ to $BC$ , and label the point of intersection $O$
By adding this extra line, we now have many pairs of similar triangles. We have $\triangle BPN \sim \triangle BOM$ , with a ratio of $2$ , so $BO = 4$ and $OC = 1$ . We also have $\triangle COM \sim \triangle CAP$ with ratio $3$ . Additionally, $\triangle BPN \sim \triangle ADN$ (with an unknown ratio). It is also true that $\triangle BAN \cong \triangle MAN$
Suppose the area of $\triangle COM$ is $x$ . Then, $[\triangle CAP] = 9x$ . Because $\triangle CAP$ and $\triangle BAP$ share the same height and have a base ratio of $3:2$ $[\triangle BAP] = 6x$ . Because $\triangle BOM$ and $\triangle COM$ share the same height and have a base ratio of $4:1$ $[\triangle BOM] = 4x$ $[\triangle BPN] = x$ , and thus $[OMNP] = 4x - x = 3x$ . Thus, $[\triangle MAN] = [\triangle BAN] = 5x$
Finally, we have $\frac{[\triangle BAN]}{[\triangle BPN]} = \frac{5x}{x} = 5$ , and because these triangles share the same height $\frac{AN}{PN} = 5$ . Notice that these side lengths are corresponding side lengths of the similar triangles $BPN$ and $ADN$ . This means that $AD = 5\cdot BP = \boxed{10}$ | C | 10 |
3bb15d2c554e600eda5e886b9a27c5bd | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | Let point $B$ be the origin, with $C$ being on the positive $x$ -axis and $A$ being in the first quadrant.
By the Angle Bisector Theorem, $AB:AC = 2:3$ . Thus, assume that $AB = 4$ , and $AC = 6$
Let the perpendicular from $A$ to $BC$ be $AM$
Using Heron's formula, \[[ABC] = \sqrt{\frac{15}{2}\left(\frac{15}{2}-4\right)\left(\frac{15}{2}-5\right)\left(\frac{15}{2}-6\right)} = \frac{15}{4}\sqrt{7}.\]
Hence, \[AM = \frac{\frac{15}{4}\sqrt{7}}{\frac{5}{2}} = \frac{3}{2}\sqrt{7}.\]
Next, we have \[BM^2 + AM^2 = AB^2\] \[\therefore BM = \sqrt{16 - \frac{63}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \textrm{ and } PM = \frac{3}{2}.\]
The slope of line $AP$ is thus \[\frac{-\frac{3}{2}\sqrt{7}}{\frac{3}{2}} = -\sqrt{7}.\]
Therefore, since the slopes of perpendicular lines have a product of $-1$ , the slope of line $BD$ is $\frac{1}{\sqrt{7}}$ . This means that we can solve for the coordinates of $D$
\[y = \frac{3}{2}\sqrt{7}\] \[y = \frac{1}{\sqrt{7}}x\] \[\frac{1}{\sqrt{7}}x = \frac{3}{2}\sqrt{7}\] \[x = \frac{7 \cdot 3}{2} = \frac{21}{2}\] \[D = \left(\frac{21}{2}, \frac{3}{2}\sqrt{7}\right).\]
We also know that the coordinates of $A$ are $\left(\frac{1}{2}, \frac{3}{2}\sqrt{7}\right)$ , because $BM = \frac{1}{2}$ and $AM = \frac{3}{2}\sqrt{7}$
Since the $y$ -coordinates of $A$ and $D$ are the same, and their $x$ -coordinates differ by $10$ , the distance between them is $10$ . Our answer is $\boxed{10}.$ | C | 10 |
3bb15d2c554e600eda5e886b9a27c5bd | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_13 | Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | [asy] size(300); pair A, B, C, P, XX, D; B = (0,0); P = (2,0); C = (5,0); A=(0,4.47214); D = A + (10,0); draw(A--B--C--cycle, linewidth(1)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, E); dot("$P$", P, S); dot("$D$", D, E); markscalefactor = 0.1; draw(anglemark(B,A,P)); markscalefactor = 0.12; draw(anglemark(P,A,C)); draw(P--A--D--B, linewidth(1)); XX = intersectionpoints(A--P,B--D)[0]; dot("$X$", XX, dir(150)); markscalefactor = 0.03; draw(rightanglemark(A,B,C)); draw(rightanglemark(D,XX,P)); [/asy] Since there is only one possible value of $AD$ , we assume $\angle{B}=90^{\circ}$ . By the angle bisector theorem, $\frac{AB}{AC}=\frac{2}{3}$ , so $AB=2\sqrt{5}$ and $AC=3\sqrt{5}$ . Now observe that $\angle{BAD}=90^{\circ}$ . Let the intersection of $BD$ and $AP$ be $X$ . Then $\angle{ABD}=90^{\circ}-\angle{BAX}=\angle{APB}$ . Consequently, \[\bigtriangleup DAB \sim \bigtriangleup ABP\] and therefore $\frac{DA}{AB} = \frac{AB}{BP}$ , so $AD=\boxed{10}$ , and we're done! | C | 10 |
32b78ca89eab905f8b4bb9861baab0f1 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_14 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$
Note that $6$ can pair with either $12$ or $13.$ From here, we consider casework:
Together, the answer is $72+72=\boxed{144}.$ | E | 144 |
32b78ca89eab905f8b4bb9861baab0f1 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_14 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | As said in Solution 1, clearly, the integers from $8$ through $14$ must be in different pairs.
We know that $8$ or $9$ can pair with any integer from $1$ to $4$ $10$ or $11$ can pair with any integer from $1$ to $5$ , and $12$ or $13$ can pair with any integer from $1$ to $6$ . Thus, $8$ will have $4$ choices to pair with, $9$ will then have $3$ choices to pair with ( $9$ cannot pair with the same number as the one $8$ pairs with). $10$ cannot pair with the numbers $8$ and $9$ has paired with but can also now pair with $5$ , so there are $3$ choices. $11$ cannot pair with $8$ 's, $9$ 's, or $10$ 's paired numbers, so there will be $2$ choices for $11$ $12$ can pair with an integer from $1$ to $5$ that hasn't been paired with already, or it can pair with $6$ $13$ will only have one choice left, and $7$ must pair with $14$
So, the answer is $4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{144}.$ | E | 144 |
32b78ca89eab905f8b4bb9861baab0f1 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_14 | How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$ | The integers $x \in \{8, \ldots , 14 \}$ must each be the larger elements of a distinct pair.
Assign partners in decreasing order for $x \in \{7, \dots, 1\}$
Note that $7$ must pair with $14$ $\mathbf{1} \textbf{ choice}$
For $5 \leq x \leq 7$ , the choices are $\{2x, \dots, 14\} - \{ \text{previous choices}\}$ . As $x$ decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding $\mathbf{3!} \textbf{ combined choices}$
After assigning a partner to $5$ , there are no invalid pairings for yet-unpaired numbers, so there are $\mathbf{4!} \textbf{ ways}$ to choose partners for $\{1,2,3,4\}$
The answer is $3! \cdot 4! = \boxed{144}$ | E | 144 |
b966ac6a585433afcbaaac722be7f8d9 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_15 | Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$
$\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997$ | Opposite angles of every cyclic quadrilateral are supplementary, so \[\angle B + \angle D = 180^{\circ}.\] We claim that $AC=25.$ We can prove it by contradiction:
By the Inscribed Angle Theorem, we conclude that $\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\frac{AC}{2}=\frac{25}{2}.$
The area of the requested region is \[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\] Therefore, the answer is $a+b+c=\boxed{1565}.$ | D | 1565 |
b966ac6a585433afcbaaac722be7f8d9 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_15 | Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$
$\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997$ | When we look at the side lengths of the quadrilateral we see $7$ and $24,$ which screams out $25$ because of Pythagorean triplets. As a result, we can draw a line through points $A$ and $C$ to make a diameter of $25.$ See Solution 1 for a rigorous proof.
This can also be shown using the Law of Cosines: Since $7^2+24^2-2\cdot7\cdot24\cdot\cos B=15^2+20^2-2\cdot15\cdot20\cdot\cos D$ and $\cos B + \cos D = 0,$ it follows that $\cos B = \cos D = 0.$
Since the diameter is $25,$ we can see the area of the circle is just $\frac{625\pi}{4}$ from the formula of the area of the circle with just a diameter.
Then we can use Brahmagupta Formula $\sqrt{(s - a)(s - b)(s - c)(s - d)}$ where $a,b,c,d$ are side lengths, and $s$ is semi-perimeter to find the area of the quadrilateral.
If we just plug the values in, we get $\sqrt{54756}=234.$ So now the area of the region we are trying to find is $\frac{625\pi}{4} - 234 = \frac{625\pi-936}{4}.$
Therefore, the answer is $a+b+c=\boxed{1565}.$ | D | 1565 |
b966ac6a585433afcbaaac722be7f8d9 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_15 | Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$
$\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997$ | We can guess that this quadrilateral is actually made of two right triangles: $\triangle CDA$ has a $3 \text{-} 4 \text{-} 5$ ratio in the side lengths, and $\triangle ABC$ is a $7 \text{-} 24 \text{-} 25$ triangle.
(See Solution 1 for a proof.)
Next, we can choose one of these triangles and use the circumradius formula to find the radius. Let's choose the $15-20-25$ triangle. The area of the triangle is equal to the product of the side lengths divided by $4$ times the circumradius. Therefore, $150 = \frac{15\cdot20\cdot25}{4r}$ . Solving this simple algebraic equation gives us $r = \frac{25}{2}$
Plugging in the values, we have $\frac{25}{2}^2\cdot\pi - \left(\frac{15\cdot20}{2}+\frac{7\cdot24}{2}\right) = \frac{625\cdot\pi}{4} - 234$ . Rewriting this gives us $\frac{625\pi-936}{4}$
Therefore, adding these values gets us $\boxed{1565}.$ | D | 1565 |
83705160375d5ad382c2620a56a43698 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_16 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48$ | Let $a$ $b$ $c$ be the three roots of the polynomial. The lengthened prism's volume is \[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$ $b$ $c$ satisfies: \begin{alignat*}{8} a+b+c &= -\frac{B}{A} &&= \frac{39}{10}, \\ ab+ac+bc &= \hspace{2mm}\frac{C}{A} &&= \frac{29}{10}, \\ abc &= -\frac{D}{A} &&= \frac{6}{10}. \end{alignat*} We can substitute these into the expression, obtaining \[V = \frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right) + 8 = \boxed{30}.\] | D | 30 |
83705160375d5ad382c2620a56a43698 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_16 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48$ | From the answer choices, we can assume the roots are rational numbers, and therefore this polynomial should be easily factorable.
The coefficients of $x$ must multiply to $10$ , so these coefficients must be $5,2,1$ or $10,1,$ in some order. We can try one at a time, and therefore write the factored form as follows: \[(5x-p)(2x-q)(x-r).\] Note that $p, q, r$ have to multiply to $6$ , so they must be either $3,2,1$ or $6,1,1$ in some order. Again, we can try one at a time in different positions and see if they multiply correctly.
We try $(5x-2)(2x-1)(x-3)$ and multiply the $x-$ terms, and sure enough they add up to $29$ . You can try to add up the $x^2$ terms and they add up to $-39$ . Therefore the roots are $\frac{2}{5}$ $\frac{1}{2}$ and $3$ . Now if you add $2$ to each root, you get the volume is $\frac{12}{5} \cdot \frac{5}{2} \cdot 5 = 6 \cdot 5 = 30 = \boxed{30}$ | D | 30 |
83705160375d5ad382c2620a56a43698 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_16 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48$ | We can find the roots of the cubic using the Rational Root Theorem, which tells us that the rational roots of the cubic must be in the form $\frac{p}{q}$ , where $p$ is a factor of the constant $(-6)$ and $q$ is a factor of the leading coefficient $(10)$ . Therefore, $p$ is $\pm (1, 2, 3, 6)$ and q is $\pm (1, 2, 5, 10).$
Doing Synthetic Division, we find that $3$ is a root of the cubic: \[\begin{array}{c|rrrr}&10&-39&29&-6\\3&&30&-27&6\\\hline\\&10&-9&2&0\\\end{array}.\]
Then, we have a quadratic $10x^2-9x+2.$ Using the Quadratic Formula, we can find the other two roots: \[x=\frac{9 \pm \sqrt{(-9)^2-4(10)(2)}}{2 \cdot 10},\] which simplifies to $x=\frac{1}{2}, \frac{2}{5}.$
To find the new volume, we add $2$ to each of the roots we found: \[(3+2)\cdot\left(\frac{1}{2}+2\right)\cdot\left(\frac{2}{5}+2\right).\] Simplifying, we find that the new volume is $\boxed{30}.$ | D | 30 |
83705160375d5ad382c2620a56a43698 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_16 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48$ | Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ , and let $a, b, c$ be the roots of $P(x)$ . The roots of $P(x-2)$ are then $a + 2, b + 2, c + 2,$ so the product of the roots of $P(x-2)$ is the area of the desired rectangular prism.
$P(x-2)$ has leading coefficient $10$ and constant term $P(0-2) = P(-2) = 10(-2)^3 - 39(-2)^2 + 29(-2) - 6 = -300$
Thus, by Vieta's Formulas, the product of the roots of $P(x-2)$ is $\frac{-(-300)}{10} = \boxed{30}$ | D | 30 |
83705160375d5ad382c2620a56a43698 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_16 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48$ | Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ . This can be factored m as $P(x) = 10(x-a)(x-b)(x-c)$ , where $a$ $b$ , and $c$ are the roots of $P(x)$ . We want $V = (a+2)(b+2)(c+2)$
"Luckily" $P(-2) = 10(-2-a)(-2-b)(-2-c) = -10V$ $P(-2) = -300$ , giving $V = \boxed{30}$ | D | 30 |
83705160375d5ad382c2620a56a43698 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_16 | The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
$\textbf{(A) } \frac{24}{5} \qquad \textbf{(B) } \frac{42}{5} \qquad \textbf{(C) } \frac{81}{5} \qquad \textbf{(D) } 30 \qquad \textbf{(E) } 48$ | By Vieta's, we can see that $ABC = \frac{6}{10}$ .
Using this, we can see that if each side $ABC$ is the same length, the length is between $0.8$ $0.512$ ) and $0.9$ $0.729$ ). Adding $2$ to these numbers would give us three numbers that are close to $3$ . Rounding up, we will just assume they are all three.
If we multiply all of them, it gives us $27$ .
The closest answer choice is $\boxed{30},$ as all of the other choices are far from this number (the second closest answer choice being $11$ away). | D | 30 |
bdf9142ed2bed6d886b651ccb048f0f3 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_17 | How many three-digit positive integers $\underline{a} \ \underline{b} \ \underline{c}$ are there whose nonzero digits $a,b,$ and $c$ satisfy \[0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?\] (The bar indicates repetition, thus $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ is the infinite repeating decimal $0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots$
$\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14$ | We rewrite the given equation, then rearrange: \begin{align*} \frac{100a+10b+c}{999} &= \frac13\left(\frac a9 + \frac b9 + \frac c9\right) \\ 100a+10b+c &= 37a + 37b + 37c \\ 63a &= 27b+36c \\ 7a &= 3b+4c. \end{align*} Now, this problem is equivalent to counting the ordered triples $(a,b,c)$ that satisfies the equation.
Clearly, the $9$ ordered triples $(a,b,c)=(1,1,1),(2,2,2),\ldots,(9,9,9)$ are solutions to this equation.
The expression $3b+4c$ has the same value when:
We find $4$ more solutions from the $9$ solutions above: $(a,b,c)=(4,8,1),(5,1,8),(5,9,2),(6,2,9).$ Note that all solutions are symmetric about $(a,b,c)=(5,5,5).$
Together, we have $9+4=\boxed{13}$ ordered triples $(a,b,c).$ | D | 13 |
6582b63a0e193a0d75b9cbf8276d9b9d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_18 | Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive
integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point $(1,0)$ back to itself?
$\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721$ | Let $P=(r,\theta)$ be a point in polar coordinates, where $\theta$ is in degrees.
Rotating $P$ by $k^{\circ}$ counterclockwise around the origin gives the transformation $(r,\theta)\rightarrow(r,\theta+k^{\circ}).$ Reflecting $P$ across the $y$ -axis gives the transformation $(r,\theta)\rightarrow(r,180^{\circ}-\theta).$ Note that \begin{align*} T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\ T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}). \end{align*} We start with $(1,0^{\circ})$ in polar coordinates. For the sequence of transformations $T_1, T_2, T_3, \cdots, T_k,$ it follows that
The least such positive integer $k$ is $180.$ Therefore, the least such positive integer $n$ is $2k-1=\boxed{359}.$ | A | 359 |
6582b63a0e193a0d75b9cbf8276d9b9d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_18 | Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive
integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point $(1,0)$ back to itself?
$\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721$ | Note that since we're reflecting across the $y$ -axis, if the point ever makes it to $(-1,0)$ then it will flip back to the original point. Note that after $T_1$ the point will be $1$ degree clockwise from the negative $x$ -axis. Applying $T_2$ will rotate it to be $1$ degree counterclockwise from the negative $x$ -axis, and then flip it so that it is $1$ degree clockwise from the positive $x$ -axis. Therefore, after every $2$ transformations, the point rotates $1$ degree clockwise. To rotate it so that it will rotate $179$ degrees clockwise will require $179 \cdot 2 = 358$ transformations. Then finally on the last transformation, it will rotate on to $(-1,0)$ and then flip back to its original position. Therefore, the answer is $358+1 = 359 = \boxed{359}$ | A | 359 |
6582b63a0e193a0d75b9cbf8276d9b9d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_18 | Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$ -axis. What is the least positive
integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point $(1,0)$ back to itself?
$\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721$ | In degrees:
Starting with $n=0$ , the sequence goes ${0}\rightarrow {179}\rightarrow {359}\rightarrow {178}\rightarrow {358}\rightarrow {177}\rightarrow {357}\rightarrow\cdots.$
We see that it takes $2$ steps to downgrade the point by $1^{\circ}$ . Since the $1$ st point in the sequence is ${179}$ , the answer is $1+2(179)=\boxed{359}.$ | A | 359 |
595c2d26bf849a1cdbd0e4f3cdc8b51a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_19 | Define $L_n$ as the least common multiple of all the integers from $1$ to $n$ inclusive. There is a unique integer $h$ such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}\] What is the remainder when $h$ is divided by $17$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Notice that $L_{17}$ contains the highest power of every prime below $17$ since higher primes cannot divide $L_{17}$ . Thus, $L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$
When writing the sum under a common fraction, we multiply the denominators by $L_{17}$ divided by each denominator. However, since $L_{17}$ is a multiple of $17$ , all terms will be a multiple of $17$ until we divide out $17$ , and the only term that will do this is $\frac{1}{17}$ . Thus, the remainder of all other terms when divided by $17$ will be $0$ , so the problem is essentially asking us what the remainder of $\frac{L_{17}}{17} = L_{16}$ divided by $17$ is. This is equivalent to finding the remainder of $16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ divided by $17$
We use modular arithmetic to simplify our answer:
This is congruent to $-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}$
Evaluating, we get: \begin{align*} (-1) \cdot 9 \cdot 35 \cdot 11 \cdot 13 &\equiv (-1) \cdot 9 \cdot 1 \cdot 11 \cdot 13 \pmod{17} \\ &\equiv 9 \cdot 11 \cdot (-13) \pmod{17} \\ &\equiv 9 \cdot 11 \cdot 4\pmod{17} \\ &\equiv 2 \cdot 11 \pmod{17} \\ &\equiv 5\pmod{17} \end{align*} Therefore the remainder is $\boxed{5}$ | C | 5 |
595c2d26bf849a1cdbd0e4f3cdc8b51a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_19 | Define $L_n$ as the least common multiple of all the integers from $1$ to $n$ inclusive. There is a unique integer $h$ such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}\] What is the remainder when $h$ is divided by $17$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | As in solution 1, we express the LHS as a sum under one common denominator. We note that \[\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{17} = \frac{\frac{17!}{1}}{17!} + \frac{\frac{17!}{2}}{17!} + \frac{\frac{17!}{3}}{17!} + \dots + \frac{\frac{17!}{17}}{17!}\]
Now, we have $h = L_{17}\left(\frac{\frac{17!}{1} + \frac{17!}{2} + \frac{17!}{3} + \dots + \frac{17!}{17}}{17!}\right)$ . We'd like to find $h \pmod{17},$ so we can evaluate our expression $\pmod{17}.$ Since $\frac{\frac{17!}{1}}{17!}, \frac{\frac{17!}{2}}{17!}, \dots, \frac{\frac{17!}{16}}{17!}$ don't have a factor of $17$ in their denominators, and since $L_{17}$ is a multiple of $17,$ multiplying each of those terms and adding them will get a multiple of $17.$ $\pmod{17}$ , that result is $0.$ Thus, we only need to consider $L_{17}\cdot \frac{\frac{17!}{17}}{17!} = \frac{L_{17}}{17} \pmod{17}.$ Proceed with solution $1$ to get $\boxed{5}$ | C | 5 |
595c2d26bf849a1cdbd0e4f3cdc8b51a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_19 | Define $L_n$ as the least common multiple of all the integers from $1$ to $n$ inclusive. There is a unique integer $h$ such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}\] What is the remainder when $h$ is divided by $17$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | Using Wolstenholmes' Theorem, we can rewrite $1 + \frac{1}{2} \dots + \frac{1}{16}$ as $\frac{17^2 n}{(17 - 1)!} = \frac{17^2 n}{16!}$ (for some $n \in \mathbb{Z}$ ). Adding the $\frac{1}{17}$ to $\frac{17^2 n}{16!}$ , we get $\frac{17^3 n + 16!}{17!}$
Now we have $\frac{17^3 n + 16!}{17!} = \frac{h}{L_{17}}$ and we want $h \pmod{17}$ . We find that $\frac{L_{17}(17^3 n + 16!)}{17!} = \frac{L_{16}(17^3 n + 16!)}{16!} = h$ . Taking $\pmod{17}$ and multiplying, we get $L_{16}(17^3 n + 16!) \equiv 16! \cdot h \pmod{17}$
Applying Wilson's Theorem on $16!$ and reducing, we simplify the congruence to $L_{16}(0 - 1) \equiv -L_{16} \equiv -h \pmod{17}$ . Now we proceed with Solution 1 and find that $L_{16} \equiv 5 \pmod{17}$ , so our answer is $\boxed{5}$ | C | 5 |
5fc9c01db26031361d56fde4feb2a8e4 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_20 | A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57$ $60$ , and $91$ . What is the fourth term of this sequence?
$\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206$ | Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,br^2,br^3.$
We are given that \begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91, \end{align*} and we wish to find $a+3d+br^3.$
Subtracting the first equation from the second and the second equation from the third, we get \begin{align*} d+b(r-1)&=3, \\ d+br(r-1)&=31. \end{align*} Subtract these results, we get \[b(r-1)^2=28.\]
Note that either $b=28$ or $b=7.$ We proceed with casework:
Therefore, The answer is $a+3d+br^3=17+189=\boxed{206}.$ | E | 206 |
5fc9c01db26031361d56fde4feb2a8e4 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_20 | A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57$ $60$ , and $91$ . What is the fourth term of this sequence?
$\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206$ | Start similarly to Solution 1 and deduce the three equations \begin{align*} a+b&=57, \\ a+d+br&=60, \\ a+2d+br^2&=91. \end{align*} Then, add the last two equations and take away the first equation to get $a+3d+br^2+br-b=94$ We can solve for this in terms of what we want: $a+3d=-br^2-br+b+94$
We're looking for $a+3d+br^3$ . We can substitute our value of $a+3d$ in here to get \[br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94.\] Since our sequence only has positive integers we can now check by the answer choices. For each answer choice, we can subtract $94$ and factor it to see if it has a perfect square factor and at least one other factor and those should differ by $2$ \begin{alignat*}{8} \textbf{(A)} \ 190-94&=96&&=2^5\cdot3, \\ \textbf{(B)} \ 194-94&=100&&=2^2\cdot5^2, \\ \textbf{(C)} \ 198-94&=104&&=2^3\cdot13, \\ \textbf{(D)} \ 202-94&=108&&=2^2\cdot3^3, \\ \textbf{(E)} \ 206-94&=112&&=2^4\cdot7. \end{alignat*} From this, the only possible answer choices are $\textbf{(A)}$ and $\textbf{(E)}$ , where $r=3$ . To solve for $b$ , we look back to the given equations above.
We are looking for $a+3d+27b$ . If $\textbf{(A)}$ were the answer, then we know that $a$ would have to be divisible by $3$ and $b$ would equal $6$ . Looking at our second equation, if this were the case, then $d$ would also have to be divisible by $3$ . However, this contradicts the third equation, as all variables are divisible by $3$ , but their sum isn't. So, $\boxed{206}$ is our answer. | E | 206 |
9617e81145ebd3092d9cb47a53600959 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? [asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy] $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | We extend line segments $\ell,m,$ and $n$ to their point of concurrency, as shown below: [asy] /* Made by AoPS; edited by MRENTHUSIASM */ import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; draw(surface(uVsl[0]--uVsr[0]--uVsl[1]--uVsr[1]--uVsl[2]--uVsr[2]--uVsl[3]--uVsr[3]--cycle),yellow); for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); draw(uVsl[1]--uVsr[0],red+2bp); draw(uVsr[3]--uVsl[4],red+2bp); draw(lVs[0]--mVs[0],red+2bp); draw(uVsr[0]--uVsr[0]+uVsr[0]-uVsl[1],red+dashed+2bp); draw(uVsl[4]--uVsl[4]+uVsl[4]-uVsr[3],red+dashed+2bp); draw(mVs[0]--mVs[0]+mVs[0]-lVs[0],red+dashed+2bp); label("$\ell$",midpoint(lVs[0]--mVs[0]),(1,2,0),red); label("$m$",midpoint(uVsr[3]--uVsl[4]),(1,0,-2),red); label("$n$",midpoint(uVsl[1]--uVsr[0]),(-1,0,-2),red); [/asy] We claim that lines $\ell,m,$ and $n$ are concurrent: In the lateral faces of the bowl, we know that lines $\ell$ and $m$ must intersect, and lines $\ell$ and $n$ must intersect. In addition, line $\ell$ intersects the top plane of the bowl at exactly one point. Since lines $m$ and $n$ are both in the top plane of the bowl, we conclude that lines $\ell,m,$ and $n$ are concurrent.
In the lateral faces of the bowl, the dashed red line segments create equilateral triangles. So, the dashed red line segments all have length $1.$ In the top plane of the bowl, we know that $\overleftrightarrow{m}\perp\overleftrightarrow{n}.$ So, the dashed red line segments create an isosceles triangle with leg-length $1.$
Note that octagon has four pairs of parallel sides, and the successive side-lengths are $1,\sqrt2,1,\sqrt2,1,\sqrt2,1,\sqrt2,$ as shown below: [asy] /* Made by AoPS; edited by MRENTHUSIASM */ size(225); real r = 1/3; draw((0,r)--(0,0)--(r,0)^^(1-r,0)--(1,0)--(1,r)^^(1,1-r)--(1,1)--(1-r,1)^^(r,1)--(0,1)--(0,1-r),red+2bp+dashed); fill((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,yellow); draw((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,black+2bp); label("$1$",(0.5,0),S); label("$1$",(1,0.5),E); label("$1$",(0.5,1),N); label("$1$",(0,0.5),W); label("$\sqrt2$",(1-r/2,r/2),NW); label("$\sqrt2$",(1-r/2,1-r/2),SW); label("$\sqrt2$",(r/2,1-r/2),SE); label("$\sqrt2$",(r/2,r/2),NE); label("$1$",(1-r/2,0),S); label("$1$",(1,r/2),E); label("$1$",(1,1-r/2),E); label("$1$",(1-r/2,1),N); label("$1$",(r/2,1),N); label("$1$",(0,1-r/2),W); label("$1$",(0,r/2),W); label("$1$",(r/2,0),S); [/asy] The area of the octagon is \[3^2-4\cdot\left(\frac12\cdot1^2\right)=\boxed{7}.\] | B | 7 |
9617e81145ebd3092d9cb47a53600959 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? [asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy] $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | (This is an alternate way of analyzing the red extension line segments drawn in Solution 1.)
The perimeter of the square bottom of the bowl is $4$ .
Halfway up the bowl, the boundary is still a square, with perimeter $4$ times the hexagon circumradius, aka $4 \times 2 = 8$ times the hexagon (also square) side length (1), an increase of $4$
Extending the bottom half of the bowl to twice its height (full height of the bowl) would increase the perimeter by the same amount again, forming a square with perimeter $4 + (8-4)\times 2= 12$ . Thus the top octagon is cut out of a square of side length $12\div4=3$ and thus area $9$
The difference between the above-constructed square and the octagon is four right triangles, and (by rotational and reflection symmetry), each is isosceles with equal-length perpendicular bases of length $(3-1)/2 = 1$ , and thus having area $\frac12$ . Therefore the area of the octagon is $9-(4\times 1/2) = \boxed{7}$ | B | 7 |
9617e81145ebd3092d9cb47a53600959 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? [asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy] $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | Note that the octagon is equiangular by symmetry, but it is not equilateral. $4$ of its sides are shared with the hexagon's sides, so each of those sides have side length $1$ . However, the other $4$ sides are touching the triangles, so we wish to find the length of these sides.
Notice that when two adjacent hexagons meet at a side, their planes make the same dihedral angle at the bottom-most point of intersection and at the top-most point of intersection by symmetry. Therefore, the triangle that is wedged between the two hexagons has the same angle as the square at the bottom wedged between the hexagons. Thus, the triangle is a $45-45-90$ isosceles triangle.
This conclusion can also be reached by cutting the bottom square across a diagonal and noticing that each resulting triangle is congruent to each triangle wedged between the hexagons by symmetry.
Furthermore, notice that if you take a copy of this bowl and invert it and place it on top of this bowl, you will get a polyhedron with faces of hexagons and squares, a truncated octahedron, and therefore this triangle has a $90^\circ$ -angle:
Now that we have come to this conclusion, by simple Pythagorean theorem, we have that the other $4$ sides of the octagon are $\sqrt{2}$
We can draw a square around the octagon so that the area of the octagon is the area of the square minus each corner triangle. The hypotenuse of these corner triangles are $1$ and they are $45-45-90$ triangles because the octagon is equiangular, so each has dimensions $\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},1$ .
The side length of the square is $\sqrt{2}$ for the larger sides of the octagon, and adding $2$ of $\frac{\sqrt{2}}{2}$ for each width of the triangle. Therefore, the area of the square is: \[\left(\sqrt{2} + 2 \cdot \frac{\sqrt{2}}{2}\right)^2 \implies \left(2\sqrt{2}\right)^2 = 8\] The area of each triangle is $\frac{1}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{4}$ and there are $4$ of them, so we subtract $1$ from the area of the square. The area of the octagon is thus $\boxed{7}$ | B | 7 |
9617e81145ebd3092d9cb47a53600959 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? [asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy] $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | Denote a square $ABCD, AB = 1,\pi$ is the plane $ABC,$ regular hexagons $ABFKSE, BCHMLF, CDGPNH, ADGQRE,$ triangles $FKL, ESR, GPQ, HMN.$
The main diagonal of each regular hexagon $EF = 2 \implies EFGH$ is square with side $2$ parallel to $\pi.$
The area of this square $[EFGH] = 4 \implies [EFGH] - [ABCD] = 3.$
The difference $3$ is the area of the projection of $4$ half of hexagons on the plane $\pi.$
So the area of the projections of another $4$ half of hexagons is $3.$
It is evident (may be not only for me) that projections of the coincide sides of hexagons are along diagonals of $ABCD$ (for example $A, E, C,$ and $H$ are collinear.)
So the projections on $\pi$ of the coincide sides of hexagons and triangles are perpendicular to this lines $(SE \perp AE).$
Therefore in plane $\pi$ projections of points $S, E,$ and $R$ are collinear and plane of $\triangle ESR$ is perpendicular to $\pi.$
We get $1 +3 +3 = \boxed{7}.$ | B | 7 |
9617e81145ebd3092d9cb47a53600959 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_21 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$ . The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl? [asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy] $\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | Through observation, we can reasonably assume that each of the triangles on this shape is a right triangle. Since each side length of the hexagons is $1$ , the hypotenuse of the triangles would be $\sqrt2$ . Now we know the side lengths of the octagon whose area we are solving for. The octagon can be broken into nine pieces. We have four triangles whose side lengths are 1, and their hypotenuse is a side whose length is $\sqrt2$ . Next, we have $5$ $1$ by $1$ squares. The triangles each have an area of $\frac{1}{2}$ , and the squares each have an area of $1$
Then, we add these up, so we get $\boxed{7}.$ | B | 7 |
97d1015f54330aea361b80ccdef0237d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$ | For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass.
Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards.
For each value of $k,$ there are $\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement [asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("1", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("2", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("3", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("4", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("5", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("6", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("7", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("8", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("9", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("10", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("11", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("12", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("13", (37,1.5)); [/asy] in which the first pass consists of all $13$ cards.
Therefore, the answer is \[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{8178}.\] | D | 8178 |
97d1015f54330aea361b80ccdef0237d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$ | Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$ .
Also note that if $m$ , which is the card that is numbered one more than $n$ , is placed before $n$ , then $m$ will not be picked up on the first pass since cards are picked up in order. Therefore we desire $m$ to be placed before $n$ to create a second pass, and that after the first pass, the numbers $m$ through $13$ are lined up in order from least to greatest.
To construct this, $n$ cannot go in the $n$ th position because all cards $1$ to $n-1$ will have to precede it and there will be no room for $m$ . Therefore $n$ must be in slots $n+1$ to $13$ .
Let's do casework on which slot $n$ goes into to get a general idea for how the problem works.
Case 1: With $n$ in spot $n+1$ , there are $n$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve these slots for the $n-1$ cards is $\binom{n}{n-1}$ . Then there is only $1$ way to order these cards (since we want them in increasing order). Then card $m$ goes into whatever slot is remaining, and the $13-m$ cards are ordered in increasing order after slot $n+1$ , giving only $1$ way. Therefore in this case there are $\binom{n}{n-1}$ possibilities.
Case 2: With $n$ in spot $n+2$ , there are $n+1$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve slots for these cards are $\binom{n+1}{n-1}$ . Then there is one way to order these cards. Then cards $m$ and $m+1$ must go in the remaining two slots, and there is only one way to order them since they must be in increasing order. Finally, cards $m+2$ to $13$ will be ordered in increasing order after slot $n+1$ , which yields $1$ way. Therefore, this case has $\binom{n+1}{n-1}$ possibilities.
I think we can see a general pattern now. With $n$ in slot $x$ , there are $x-1$ slots to distribute to the previous $n-1$ cards, which can be done in $\binom{x-1}{n-1}$ ways. Then the remaining cards fill in in just $1$ way. Since the cases of $n$ start in slot $n+1$ and end in slot $13$ , this sum amounts to: \[\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\] for any $n$
Hmmm ... where have we seen this before?
We use wishful thinking to add a term of $\binom{n-1}{n-1}$ \[\binom{n-1}{n-1}+\binom{n}{n-1}+\binom{n+1}{n-1}+\binom{n+2}{n-1} + \cdots + \binom{12}{n-1}\]
This is just the hockey stick identity! Applying it, this expression is equal to $\binom{13}{n}$ . However, we added an extra term, so subtracting it off, the total number of ways to order the $13$ cards for any $n$ is \[\binom{13}{n}-1\]
Finally, to calculate the total for all $n$ , we sum from $n=0$ to $13$ . This yields us:
\[\sum_{n=0}^{13} \binom{13}{n}-1 \implies \sum_{n=0}^{13} \binom{13}{n} - \sum_{n=0}^{13} 1\] \[\implies 2^{13} - 14 = 8192 - 14 = 8178 = \boxed{8178}.\] | D | 8178 |
97d1015f54330aea361b80ccdef0237d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$ | To solve this problem, we can use recursion on $n$ . Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$
If $a_1 = 1$ , the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in the case of $n-1$ numbers. So there are $A_{n-1}$ arrangements.
If $a_1 = 2$ , then we need to choose 1 position from position 2 to $n-1$ to put 1, and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{1}$ such arrangements.
If $a_1 = k$ , then we need to choose $k-1$ positions from position 2 to $n-1$ to put $1, 2,\cdots k-1$ , and all remaining numbers must be arranged in increasing order, so there are $\binom{n-1}{k-1}$ such arrangements.
So we can write \[A_n = A_{n-1} + \binom{n-1}{1} + \binom{n-1}{2} + \cdots + \binom{n-1}{n-1}\] which can be simplified to \[A_n = A_{n-1} + 2^{n-1} - 1\] We can solve this recursive sequence by summing up $n-1$ lines of the recursive formula \[A_n - A_{n-1} = 2^{n-1} - 1\] \[A_{n-1} - A_{n-2} = 2^{n-2} - 1\] \[\cdots\] \[A_2 - A_{1} = 2^{1} - 1\] to get \[A_n - A_1 = \sum_{k=1}^{n-1} (2^k - 1) = 2^n - 2 - (n-1) = 2^n - n - 1\] since $A_1=0$ , we have \[A_n = 2^n - n - 1\] and $A_{13} = 2^{13} - 14 = \boxed{8178}$ | D | 8178 |
97d1015f54330aea361b80ccdef0237d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$ | When we have $3$ cards arranged in a row, after listing out all possible arrangements, we see that we have $4$ ones: $(1, 3, 2), (2, 1, 3), (2, 3, 1),$ and $(3, 1, 2)$ . When we have $4$ cards, we find $11$ possible arrangements: $(1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3, 4, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 4, 1, 2),$ and $(4, 1, 2, 3).$ Hence, we recognize the pattern that for $n$ cards, we have $2^n - n - 1$ valid arrangements, so our answer is $2^{13} - 13 - 1 = \boxed{8178}.$ ~eibc | D | 8178 |
97d1015f54330aea361b80ccdef0237d | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_22 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("7", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("11", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("8", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("6", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("4", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("5", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("9", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("12", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("1", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("13", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("10", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("2", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("3", (37,1.5)); [/asy] $\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$ | Notice that for each card "position", we can choose for it to be picked up on the first or second pass, for a total of $2^{13}$ options. However, if all of the cards selected to be picked up first are before all of the cards to be picked up second, then this means that the list is in consecutive ascending order (and thus all cards will be picked up on the first pass instead). This can happen in 14 ways, so our answer is $2^{13}-14=\boxed{8178}$ | D | 8178 |
2b185240be5f2762ce05eaec3b1b2489 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$ | For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$ , we can add $1$ to each $c_i$ (mod $n+1$ ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings.
Plugging in $n = 5$ gives $\boxed{1296}$ such strings. | E | 1296 |
2b185240be5f2762ce05eaec3b1b2489 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$ | Note that a valid string must have at least one $0.$
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
Together, the answer is $1+75+500+600+120=\boxed{1296}.$ | E | 1296 |
2b185240be5f2762ce05eaec3b1b2489 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$ | Denote by $N \left( p, q \right)$ the number of $p$ -digit strings formed by using numbers $0, 1, \cdots, q$ , where for each $j \in \{1,2, \cdots , q\}$ , at least $j$ of the digits are less than $j$
We have the following recursive equation: \[N \left( p, q \right) = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1\] and the boundary condition $N \left( p, 0 \right) = 1$ for any $p \geq 0$
By solving this recursive equation, for $q = 1$ and $p \geq q$ , we get \begin{align*} N \left( p , 1 \right) & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\ & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\ & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\ & = 2^p - 1 . \end{align*}
For $q = 2$ and $p \geq q$ , we get \begin{align*} N \left( p , 2 \right) & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\ & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right) - p \\ & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\ & = 3^p - 2^p - p . \end{align*}
For $q = 3$ and $p \geq q$ , we get \begin{align*} N \left( p , 3 \right) & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\ & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} \left( p - i \right) \right) - \frac{3}{2} p \left( p - 1 \right) \\ & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1} - \frac{3}{2} p \left( p - 1 \right) \\ & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) . \end{align*}
For $q = 4$ and $p = 5$ , we get \begin{align*} N \left( 5 , 4 \right) & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\ & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\ & = \boxed{1296} | E | 1296 |
2b185240be5f2762ce05eaec3b1b2489 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$ | The number of strings is $(n+1)^{(n-1)}$ as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose $\boxed{1296}$ | E | 1296 |
2b185240be5f2762ce05eaec3b1b2489 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$ | Solution 4 tried to observe the answer modulo $5$ to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set $\{a_1, a_2, a_3, a_4, a_5\}$ , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by $5$ . Therefore to evaluate the answer modulo $5$ , we only need to look at sets where each element has a multiplicity of at least $2$ (i.e. appears twice or more in the set).
These sets are of the form $\{a,a,b,b,b\}$ and $\{a,a,a,a,a\}$ . The first set can be permuted in $\binom{5}{2,3}=10 \equiv 0\pmod{5}$ , and the second set can be permuted one way, and the only set of the form $\{a,a,a,a,a\}$ is $\{0,0,0,0,0\}$ . Therefore the answer is congruent to $1\pmod{5}$ and you CAN choose $\boxed{1296}$ | E | 1296 |
2b185240be5f2762ce05eaec3b1b2489 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_24 | How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$ | We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string.
00000 - 1
00001 - 5
00002 - 5
00003 - 5
00004 - 5
00011 - 10
00012 - 20
00013 - 20
00014 - 20
00022 - 10
00023 - 20
00024 - 20
00033 - 10
00034 - 20
00111 - 10
00112 - 30
00113 - 30
00114 - 30
00122 - 30
00123 - 60
00124 - 60
00133 - 30
00134 - 60
00222 - 10
00223 - 30
00224 - 30
00233 - 30
00234 - 60
01111 - 5
01112 - 20
01113 - 20
01114 - 20
01122 - 30
01123 - 60
01124 - 60
01133 - 30
01134 - 60
01222 - 20
01223 - 60
01224 - 60
01233 - 60
01234 - 120
Finally, adding all of these up, we obtain $1\cdot 1+5\cdot 5+5\cdot 10+10\cdot 20+10\cdot 30 + 10\cdot 60 + 1\cdot 120 = \boxed{1296}$ . (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5) | E | 1296 |
8f35d89f5841bc48a6455dde93baef53 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_25 | Let $R$ $S$ , and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$ . The top two vertices of $T$ are in $R \cup S$ , and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale). [asy] size(8cm); label(scale(.8)*"$y$", (0,60), N); label(scale(.8)*"$x$", (60,0), E); filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white); label(scale(1.3)*"$R$", (55/2,55/2)); filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white); label(scale(1.3)*"$S$",(-14,14)); filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white); label(scale(1.3)*"$T$",(3.5,25/2)); draw((0,-10)--(0,60),EndArrow()); draw((-34,0)--(60,0),EndArrow()); [/asy] The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$ . What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$
$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$ | Let $r$ be the number of lattice points on the side length of square $R$ $s$ be the number of lattice points on the side length of square $S$ , and $t$ be the number of lattice points on the side length of square $T$ . Note that the actual lengths of the side lengths are the number of lattice points minus $1$ , so we can work in terms of $r, s, t$ and subtract $3$ to get the actual answer at the end. Furthermore, note that the number of lattice points inside a rectangular region is equal to the number of lattice points in its width times the number of lattice points along its length.
Using this fact, the number of lattice points in $R$ is $r^2$ , the number of lattice points in $S$ is $s^2$ , and the number of lattice points in $T$ is $t^2$
Now, by the first condition, we have \[r^2=\frac{9}{4}\cdot s^2 \implies r = \frac{3}{2}s \quad \quad \quad \quad \quad (1)\]
The second condition, the number of lattice points contained in $T$ is a fourth of the number of lattice points contained in $R \cup S$ . The number of lattice points in $R \cup S$ is equal to the sum of the lattice points in their individually bounded regions, but the lattice points along the y-axis for the full length of square $S$ is shared by both of them, so we need to subtract that out.
In all, this condition yields us $t^2 = \frac{1}{4}\cdot(r^2 + s^2 - s )\implies t^2 = \frac{1}{4}\cdot\left(\frac{9}{4}\cdot s^2 + s^2 - s \right)$ $\implies t^2=\frac{1}{4}\cdot\frac{13s^2-4s}{4} \implies 16t^2= s(13s-4)$
Note from $(1)$ that $s$ is a multiple of $2$ . We can write $s=2j$ and substitute: $16t^2=2j(26j-4) \implies 4t^2=j(13j-2)$ .
Note that $j$ must be divisible by two for the product to be divisible by 4. Thus we make another substitution, $j=2k$ \[4t^2=2k(26k-2) \implies t^2 = k(13k-1) \quad \quad \quad \quad \quad (2)\]
Finally we look at the last condition; that the fraction of the lattice points inside $S$ that are inside $S \cap T$ is $27$ times the fraction of lattice points inside $r$ that are inside $R \cap T$
Let $x$ be the number of lattice points along the bottom of the rectangle formed by $S \cap T$ , and $y$ be the number of lattice points along the bottom of the the rectangle formed by $R \cap T$
Therefore, the number of lattice points in $S\cap T$ is $xt$ and the number of lattice points in $R \cap T$ is $yt$
Thus by this condition, $\frac{xt}{s^2} = 27 \cdot \frac{yt}{r^2} \implies \frac{x}{s^2} = 27 \cdot \frac{y}{\frac{9}{4}\cdot s^2} \implies x= 12y$
Finally, notice that $t=x+y-1=12y+y-1$ (subtracting overlap), and so we have \[t=13y-1 \quad \quad \quad \quad \quad (3)\]
Now notice that by $(3)$ $t\equiv -1 \pmod{13}\implies t^2 \equiv 1 \pmod{13}$
However, by $(2)$ $t^2 \equiv k \cdot -1 \pmod{13}$ . Therefore, $-k \equiv 1 \pmod{13} \implies k \equiv -1 \pmod{13}$
Also, by $(2)$ , we know $k$ must be a perfect square since $k$ is relatively prime to $13k-1$ (Euclids algorithm) and the two must multiply to a perfect square. Hence we know two conditions on $k$ , and we can now guess and check to find the smallest that satisfies both.
We check $k=12$ first since its one less than a multiple of $13$ , but this does not work. Next, we have $k=25$ which works because $25$ is a perfect square. Thus we have found the smallest $k$ , and therefore the smallest $r, s, t$
Now we just work backwards: $j= 2k = 50$ and $s=2j=100$ . Then $r=\frac{3}{2}\cdot 100 = 150$ . Finally, from $(2)$ $t^2=25(13\cdot25-1) \implies t^2 = 25 \cdot 324 \implies t=5\cdot 18=90$
Finally, the sum of each square’s side lengths is $r+s+t-3=340-3=337=\boxed{337}$ | B | 337 |
8f35d89f5841bc48a6455dde93baef53 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_25 | Let $R$ $S$ , and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$ . The top two vertices of $T$ are in $R \cup S$ , and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale). [asy] size(8cm); label(scale(.8)*"$y$", (0,60), N); label(scale(.8)*"$x$", (60,0), E); filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white); label(scale(1.3)*"$R$", (55/2,55/2)); filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white); label(scale(1.3)*"$S$",(-14,14)); filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white); label(scale(1.3)*"$T$",(3.5,25/2)); draw((0,-10)--(0,60),EndArrow()); draw((-34,0)--(60,0),EndArrow()); [/asy] The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$ . What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$
$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$ | Notice that each answer choice has a different residue mod $13$ . Therefore, we can just find the residue of $r+s+t-3$ mod $13$ and find the unique answer choice that fits, without actually finding $r, s, t$
From Solution 1, we have $16t^2 = s(13s-4)$ from the second condition. From the third condition, $t\equiv -1 \pmod{13} \implies t^2 \equiv 1 \pmod{13}$ .
Substituting, we get $16 \cdot 1 \equiv s \cdot -4 \pmod{13}$ . Therefore, $s \equiv -4 \pmod{13}$ .
From the first condition, we have $r=\frac{3}{2} \cdot s$ , so $r \equiv -6 \pmod{13}$
Therefore $r+s+t \equiv -6-4-1 \equiv -11 \equiv 2 \pmod {13}$
We want to find $r+s+t-3$ , so our answer will have a remainder of $-1$ when divided by $13$
We divide $340$ by $13$ and find that the remainder is $2$ . Therefore the answer that will give us a remainder of $-1$ will be $340-3=337=\boxed{337}$ | B | 337 |
8f35d89f5841bc48a6455dde93baef53 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_25 | Let $R$ $S$ , and $T$ be squares that have vertices at lattice points (i.e., points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the $x$ -axis. The left edge of $R$ and the right edge of $S$ are on the $y$ -axis, and $R$ contains $\frac{9}{4}$ as many lattice points as does $S$ . The top two vertices of $T$ are in $R \cup S$ , and $T$ contains $\frac{1}{4}$ of the lattice points contained in $R \cup S.$ See the figure (not drawn to scale). [asy] size(8cm); label(scale(.8)*"$y$", (0,60), N); label(scale(.8)*"$x$", (60,0), E); filldraw((0,0)--(55,0)--(55,55)--(0,55)--cycle, yellow+orange+white+white); label(scale(1.3)*"$R$", (55/2,55/2)); filldraw((0,0)--(0,28)--(-28,28)--(-28,0)--cycle, green+white+white); label(scale(1.3)*"$S$",(-14,14)); filldraw((-10,0)--(15,0)--(15,25)--(-10,25)--cycle, red+white+white); label(scale(1.3)*"$T$",(3.5,25/2)); draw((0,-10)--(0,60),EndArrow()); draw((-34,0)--(60,0),EndArrow()); [/asy] The fraction of lattice points in $S$ that are in $S \cap T$ is $27$ times the fraction of lattice points in $R$ that are in $R \cap T$ . What is the minimum possible value of the edge length of $R$ plus the edge length of $S$ plus the edge length of $T$
$\textbf{(A) }336\qquad\textbf{(B) }337\qquad\textbf{(C) }338\qquad\textbf{(D) }339\qquad\textbf{(E) }340$ | Solution: Let $r$ $s$ $t$ be the edge length of square $R$ $S$ , and $T$ respectively. Then we have \[(r+1)^2=\dfrac{9}{4}(s+1)^2\ \ \ \ \ (t+1)^2=\dfrac{1}{4}((s+1)^2+(r+1)^2-(s+1))\] Therefore \[r=\dfrac{3s+1}{2}\ \ \ \ \ t=\dfrac{1}{4}\sqrt{(s+1)(13s+9)}-1\] Therefore \[r+s+t=\dfrac{3s+1}{2}+s+\dfrac{1}{4}\sqrt{(s+1)(13s+9)}-1\] \[\approx\dfrac{5}{2}s+\dfrac{\sqrt{13}}{4}s-\dfrac{1}{2}\approx 3.4\cdot s\]
Given that average of the answer choices is around $340$ , therefore $s\approx 100$ . Since $t$ is an integer, therefore $(s+1)(13s+9)$ must be a perfect square divisible by 16. Plugging in $s=99$ $t=89$ and $r=149$ . Therefore $r+s+t=99+89+149=337$ . So the answer is $\boxed{337}$ | B | 337 |
58db2fa1acb8dcbd4b773846a1324ce3 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_1 | Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$ | We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{2} ~MRENTHUSIASM | A | 2 |
58db2fa1acb8dcbd4b773846a1324ce3 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_1 | Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$ | Observe that the $\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\boxed{2}.$ | A | 2 |
a44fec80677fee2bd201cc78636a5cac | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_2 | In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy]
$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$ | [asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]
\[\textbf{Figure redrawn to scale.}\]
$AD = AP + PD = 3 + 2 = 5$
$ABCD$ is a rhombus, so $AB = AD = 5$
$\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$
The area of the rhombus is base times height: $bh = (AD)(BP) = 5 \cdot 4 = \boxed{20}$ | D | 20 |
a44fec80677fee2bd201cc78636a5cac | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_2 | In rhombus $ABCD$ , point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$ $AP = 3$ , and $PD = 2$ . What is the area of $ABCD$ ? (Note: The figure is not drawn to scale.)
[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy]
$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$ | [asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(D--B); draw(B--P); draw(rightanglemark(B,P,D)); [/asy]
The diagram is from as Solution 1, but a line is constructed at $BD$
When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that $\angle ABD \cong \angle BDC$ , by the Alternate Interior Angles Theorem.
By SAS Congruence, we get $\triangle ABD \cong \triangle BDC$
Since $AP=3$ and $AB=5$ , we know that $BP=4$ because $\triangle APB$ is a 3-4-5 right triangle, as stated in Solution 1.
The area of $\triangle ABD$ would be $10$ , since the area of the triangle is $\frac{bh}{2}$
Since we know that $\triangle ABD \cong \triangle BDC$ and that $ABCD=\triangle ABD + \triangle BDC$ , so we can double the area of $\triangle ADB$ to get $10 \cdot 2 = \boxed{20}$ | D | 20 |
7d9d1d12e5ce66f075241d2765dfd453 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_3 | How many three-digit positive integers have an odd number of even digits?
$\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550$ | We use simple case work to solve this problem.
Case 1: even, even, even = $4 \cdot 5 \cdot 5 = 100$
Case 2: even, odd, odd = $4 \cdot 5 \cdot 5 = 100$
Case 3: odd, even, odd = $5 \cdot 5 \cdot 5 = 125$
Case 4: odd, odd, even = $5 \cdot 5 \cdot 5 = 125$
Simply sum up the cases to get your answer. $100 + 100 + 125 + 125 = \boxed{450}$ | D | 450 |
7d9d1d12e5ce66f075241d2765dfd453 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_3 | How many three-digit positive integers have an odd number of even digits?
$\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550$ | We will show that the answer is $450$ by proving a bijection between the three digit integers that have an even number of even digits and the three digit integers that have an odd number of even digits. For every even number with an odd number of even digits, we increment the number's last digit by $1$ , unless the last digit is $9$ , in which case it becomes $0$ . It is very easy to show that every number with an even number of even digits is mapped to every number with an odd number of even digits, and vice versa. Thus, the answer is half the number of three digit numbers, or $\boxed{450}$ | D | 450 |
5151e0ddfb1f63638e5dd0731ef8a736 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_5 | What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\] $\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}$ | We apply the difference of squares to the denominator, and then regroup factors: \begin{align*} \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}} &= \frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}\cdot\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}}{\sqrt{\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)}} \\ &= \frac{\sqrt{\frac43\cdot\frac65\cdot\frac87}}{\sqrt{\frac23\cdot\frac45\cdot\frac67}} \\ &= \frac{\sqrt{4\cdot6\cdot8}}{\sqrt{2\cdot4\cdot6}} \\ &= \frac{\sqrt8}{\sqrt2} \\ &= \boxed{2} ~MRENTHUSIASM | B | 2 |
5151e0ddfb1f63638e5dd0731ef8a736 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_5 | What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\] $\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}$ | Since these numbers are fairly small, we can use brute force as follows: \[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} =\frac{\frac{4}{3}\cdot\frac{6}{5}\cdot\frac{8}{7}}{\sqrt{\frac{8}{9}\cdot\frac{24}{25}\cdot\frac{48}{49}}} =\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} =\frac{\frac{64}{35}}{\frac{96}{105}}=\frac{64}{35}\cdot\frac{105}{96}=\boxed{2}.\] ~not_slay | B | 2 |
5151e0ddfb1f63638e5dd0731ef8a736 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_5 | What is the value of \[\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?\] $\textbf{(A)}\ \sqrt3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \sqrt{15} \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ \sqrt{105}$ | This solution starts precisely like the one above. We simplify to get the following:
\[\frac{(1+\frac{1}{3})(1+\frac{1}{5})(1+\frac{1}{7})}{\sqrt{(1-\frac{1}{3^2})(1-\frac{1}{5^2})(1-\frac{1}{7^2})}} = \frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}}\]
But now, we can get a nice simplification as shown: \[\frac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{(2^3)(2^3\cdot3^1)(2^4\cdot3^1)}{(3^2)(5^2)(7^2)}}} = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\sqrt{\frac{2^{10} \cdot 3^{2}}{3^2\cdot 5^2\cdot 7^2}}} = \dfrac{\frac{4\cdot6\cdot8}{3\cdot5\cdot7}}{\frac{2^5 \cdot 3}{3\cdot5\cdot 7}} =\dfrac{4\cdot6\cdot8}{3\cdot5\cdot7} \hspace{0.05 in} \cdot \hspace{0.05 in}\dfrac{3\cdot5\cdot 7}{2^5 \cdot 3} =\dfrac{2^6\cdot 3}{2^5\cdot 3} = \boxed{2}.\] | B | 2 |
c312f7a3641fae100f31df4b73d468db | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | The $n$ th term of this sequence is \[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\] It follows that the terms are \begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence. | A | 0 |
c312f7a3641fae100f31df4b73d468db | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | Denote this sequence as $a_{n}$ , then we can find that \begin{align*} a_{1} &= 121 = 10^2 + 2\cdot10 + 1 = (10^2 + 10) + (10 + 1), \\ a_{2} &= 11211 = (10^4 + 10^3 + 10^2) + (10^2 + 10 + 1), \\ a_{3} &= 1112111 = (10^6 + 10^5 + 10^4 + 10^3) + (10^3 + 10^2 + 10 + 1), \\ & \ \vdots \end{align*} So, we can induct that the general term is \begin{align*} a_n &= (10^{2n} + 10^{2n-1} + \ldots + 10^{n+1} + 10^n) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= 10^n\cdot(10^n + 10^{n-1} + \ldots +10 + 1) + (10^n + 10^{n-1} + \ldots +10 + 1) \\ &= \left(10^n+1\right)\sum_{k=0}^{n}10^k. \end{align*} Therefore, there are $\boxed{0}$ prime numbers in this sequence. | A | 0 |
c312f7a3641fae100f31df4b73d468db | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | Observe how \begin{align*} 121 &= 110 + 11, \\ 11211 &= 11100 + 111, \\ 1112111 &= 1111000 + 1111, \\ & \ \vdots \end{align*} all take the form of \[\underbrace{111\ldots}_{n+1}\underbrace{00\ldots}_{n} + \underbrace{111\ldots}_{n+1} = \underbrace{111\ldots}_{n+1}(10^{n} + 1).\] Factoring each of the sums, we have \[11(10+1), 111(100+1), 1111(1000+1), \ldots\] respectively. With each number factored, there are $\boxed{0}$ primes in the set. | A | 0 |
c312f7a3641fae100f31df4b73d468db | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_6 | How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3 \qquad \textbf{(E) }4$ | Note that $121$ is divisible by $11$ and $11211$ is divisible by $3$ . Because this is Problem 6 of the AMC 10, we assume we do not need to check two-digit prime divisibility or use obscure theorems. Therefore, the answer is $\boxed{0}.$ | A | 0 |
03b882a2e3ed85af172d72f2dc4d0e4a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_7 | For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$ | Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formulas , we have $p+q=-k$ and $pq=36.$
It follows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\] Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{8}$ values of $k,$ namely $\pm37,\pm20,\pm15,\pm13.$ | B | 8 |
03b882a2e3ed85af172d72f2dc4d0e4a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_7 | For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$ | Note that $k$ must be an integer. Using the quadratic formula $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $4$ divides $144$ evenly, $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.
Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is even, $k+n$ and $k-n$ must both be even. Assuming that $k$ is positive, we get $5$ possible values of $k+n$ , namely $2, 4, 8, 6, 12$ , which will give distinct positive values of $k$ , but $k+n=12$ gives $k+n=k-n$ and $n=0$ , giving $2$ identical integer roots. Therefore, there are $4$ distinct positive values of $k.$ Multiplying that by $2$ to take the negative values into account, we get $4\cdot2=\boxed{8}$ values of $k$ | B | 8 |
03b882a2e3ed85af172d72f2dc4d0e4a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_7 | For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$ | Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than $0$ to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle, where $k$ is the hypotenuse, and $12$ is a leg. There are four ways we could have this: a $9$ $12$ $15$ triangle, a $12$ $16$ $20$ triangle, a $5$ $12$ $13$ triangle, and a $12$ $35$ $37$ triangle.
Multiply by $2$ to account for negative $k$ values (since $k$ is being squared), and our answer is $\boxed{8}$ | B | 8 |
03b882a2e3ed85af172d72f2dc4d0e4a | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_7 | For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$ | Since $36 = 2^2\cdot3^2$ , that means there are $(2+1)(2+1) = 9$ possible factors of $36$ . Since $6 \cdot 6$ violates the distinct root condition, subtract $1$ from $9$ to get $8$ . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get $\boxed{8}$ | B | 8 |
7451a3c1e6a863fb225731e9f419262f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_8 | Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$ | We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{42}.$ | B | 42 |
7451a3c1e6a863fb225731e9f419262f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_8 | Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$ | We find a pattern. \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets.
We see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\cdot3= \boxed{42}$ | B | 42 |
7451a3c1e6a863fb225731e9f419262f | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_8 | Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$ | Each set contains exactly $1$ or $2$ multiples of $7$
There are $\dfrac{1000}{10}=100$ total sets and $\left\lfloor\dfrac{1000}{7}\right\rfloor = 142$ multiples of $7$
Thus, there are $142-100=\boxed{42}$ sets with $2$ multiples of $7$ | B | 42 |
ecbc373b6bc3c29ff231abaabf128d97 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9 | The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$ | Note that $\frac{n}{(n+1)!} = \frac{1}{n!} - \frac{1}{(n+1)!}$ , and therefore this sum is a telescoping sum, which is equivalent to $1 - \frac{1}{2022!}$ . Our answer is $1 + 2022 = \boxed{2023}$ | D | 2023 |
ecbc373b6bc3c29ff231abaabf128d97 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9 | The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$ | We add $\frac{1}{2022!}$ to the original expression \[\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\right)+\frac{1}{2022!}=\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2020}{2021!}\right)+\frac{1}{2021!}.\] This sum clearly telescopes, thus we end up with $\left(\frac{1}{2!}+\frac{2}{3!}\right)+\frac{1}{3!}=\frac{2}{2!}=1$ . Thus the original expression is equal to $1-\frac{1}{2022!}$ , and $1+2022=\boxed{2023}$ | D | 2023 |
ecbc373b6bc3c29ff231abaabf128d97 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9 | The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$ | By looking for a pattern, we see that $\tfrac{1}{2!} = 1 - \tfrac{1}{2!}$ and $\tfrac{1}{2!} + \tfrac{2}{3!} = \tfrac{5}{6} = 1 - \tfrac{1}{3!}$ , so we can conclude by engineer's induction that the sum in the problem is equal to $1 - \tfrac{1}{2022!}$ , for an answer of $\boxed{2023}$ , completing the proof.
~eibc | D | 2023 |
ecbc373b6bc3c29ff231abaabf128d97 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9 | The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$ | Let $x=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}.$
Note that \begin{align*} \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=x+1-\frac{1}{2022!}\\ \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)&=1-\frac{1}{2022!}. \end{align*} Therefore, the answer is $1+2022=\boxed{2023}.$ | D | 2023 |
ecbc373b6bc3c29ff231abaabf128d97 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9 | The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$ | Because the fractions get smaller, it is obvious that the answer is less than $1$ , so we can safely assume that $a=1$ (this can also be guessed by intuition using similar math problems). Looking at the answer choices, $2018<b<2024$ . Because the last term consists of $2022!$ (and the year is $2022$ ) we can guess that $b=2022$ . Adding them yields $1+2022=\boxed{2023}$ | D | 2023 |
ecbc373b6bc3c29ff231abaabf128d97 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_9 | The sum \[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\] can be expressed as $a-\frac{1}{b!}$ , where $a$ and $b$ are positive integers. What is $a+b$
$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$ | Knowing that the answer will be in the form $a-\frac{1}{b!}$ , we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite $\frac{n-1}{n!}$ as $\frac{A}{(n-1)!}-\frac{B}{n}$ . Setting these equal and multiplying by $n!$ , we get $n-1=An-B(n-1)!$ . Since $An$ is the only term with $n$ with degree $1$ , we can conclude that $A=1$ . This means that $B=\frac{1}{(n-1)!}$ . Substituting, we find that $\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$ . This sum clearly telescopes and we obtain $1-\frac{1}{2022!}$ . This means that our desired answer is $1+2022=\boxed{2023}.$ | D | 2023 |
6afdb68f75cdc1fe70757a783bbb3eac | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_10 | Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$ | Let $M$ be the median. It follows that the two largest integers are both $M+2.$
Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{11}.$ | D | 11 |
6afdb68f75cdc1fe70757a783bbb3eac | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_10 | Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13$ | We can also easily test all the answer choices. (This strategy is generally good to use for multiple-choice questions if you don't have a concrete method to proceed with!)
For answer choice $\textbf{(A)},$ the mode is $5,$ the median is $3,$ and the arithmetic mean is $1.$ However, we can quickly see this doesn't work, as there are five integers, and they can't have an arithmetic mean of $1$ while having a mode of $5.$
Trying answer choice $\textbf{(B)},$ the mode is $7,$ the median is $5,$ and the arithmetic mean is $3.$ From the arithmetic mean, we know that all the numbers have to sum to $15.$ We know three of the numbers: $\underline{\hspace{3mm}},\underline{\hspace{3mm}},5,7,7.$ This exceeds the sum of $15.$
Now we try answer choice $\textbf{(C)}.$ The mode is $9,$ the median is $7,$ and the arithmetic mean is $5.$ From the arithmetic mean, we know that the list sums to $25.$ Three of the numbers are $\underline{\hspace{3mm}},\underline{\hspace{3mm}},7,9,9,$ which is exactly $25.$ However, our list needs positive integers, so this won't work.
Since we were really close on answer choice $\textbf{(C)},$ we can intuitively feel that the answer is probably going to be $\textbf{(D)}.$ We can confirm this by creating a list that satisfies the problem and choose $\textbf{(D)}: 1,3,9,11,11.$
So, our answer is $\boxed{11}.$ | D | 11 |
5d6cb404898e2a6949a495659e0fd61b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_12 | A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | Rolling a pair of fair $6$ -sided dice, the probability of getting a sum of $7$ is $\frac16:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\frac16=\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\left(\frac56\right)^n.$
Therefore, we have $1-\left(\frac56\right)^n>\frac12,$ or \[\left(\frac56\right)^n<\frac12.\] Since $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ the least integer $n$ satisfying the inequality is $\boxed{4}.$ | C | 4 |
5d6cb404898e2a6949a495659e0fd61b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_12 | A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | Let's try the answer choices. We can quickly find that when we roll $3$ dice, either the first and second sum to $7$ , the first and third sum to $7$ , or the second and third sum to $7$ . There are $6$ ways for the first and second dice to sum to $7$ $6$ ways for the first and third to sum to $7$ , and $6$ ways for the second and third dice to sum to $7$ . However, we overcounted (but not by much) so we can assume that the answer is $\boxed{4}$ | C | 4 |
5d6cb404898e2a6949a495659e0fd61b | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_12 | A pair of fair $6$ -sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | We can start by figuring out what the probability is for each die to add up to $7$ if there is only $1$ roll. We can quickly see that the probability is $\frac16$ , as there are $6$ ways to make $7$ from $2$ numbers on a die, and there are a total of $36$ ways to add $2$ numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for $n$ rolls would be $\frac16$ $n$ . The smallest number that satisfies this is $\boxed{4}$ | C | 4 |
5ad4bf9836198b7b75cbec11530e9400 | https://artofproblemsolving.com/wiki/index.php/2022_AMC_10B_Problems/Problem_13 | The positive difference between a pair of primes is equal to $2$ , and the positive difference between the cubes of the two primes is $31106$ . What is the sum of the digits of the least prime that is greater than those two primes?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 11 \qquad\textbf{(D)}\ 13 \qquad\textbf{(E)}\ 16$ | Let the two primes be $a$ and $b$ . We would have $a-b=2$ and $a^{3}-b^{3}=31106$ . Using difference of cubes, we would have $(a-b)(a^{2}+ab+b^{2})=31106$ . Since we know $a-b$ is equal to $2$ $(a-b)(a^{2}+ab+b^{2})$ would become $2(a^{2}+ab+b^{2})=31106$ . Simplifying more, we would get $a^{2}+ab+b^{2}=15553$
Now let's introduce another variable. Instead of using $a$ and $b$ , we can express the primes as $x+2$ and $x$ where $a$ is $x+2$ and b is $x$ . Plugging $x$ and $x+2$ in, we would have $(x+2)^{2}+x(x+2)+x^{2}$ . When we expand the parenthesis, it would become $x^{2}+4x+4+x^{2}+2x+x^{2}$ . Then we combine like terms to get $3x^{2}+6x+4$ which equals $15553$ . Then we subtract 4 from both sides to get $3x^{2}+6x=15549$ . Since all three numbers are divisible by 3, we can divide by 3 to get $x^{2}+2x=5183$
Notice how if we add 1 to both sides, the left side would become a perfect square trinomial: $x^{2}+2x+1=5184$ which is $(x+1)^{2}=5184$ . Since $2$ is too small to be a valid number, the two primes must be odd, therefore $x+1$ is the number in the middle of them. Conveniently enough, $5184=72^{2}$ so the two numbers are $71$ and $73$ . The next prime number is $79$ , and $7+9=16$ so the answer is $\boxed{16}$ | E | 16 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.