problem_id stringlengths 32 32 | link stringlengths 75 84 | problem stringlengths 14 5.33k | solution stringlengths 15 6.63k | letter stringclasses 5
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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... | 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 \imp... | 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 \tex... | 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 \tex... | 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+4... | 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 \tex... | In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{5}.\] vladimir.shelomovskii@gmail.com, 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 ... | 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{ kilom... | 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 ... | 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 requeste... | 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 \te... | 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)-... | 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 ... | 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)-... | 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 an... | 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 o... | [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("... | 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 re... | 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 ... | 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 state... | 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 state... | 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. Th... | 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 state... | 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 ... | 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 \tex... | 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 $... | 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 \tex... | 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 ... | 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 \tex... | 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... | 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 $AD?$
$\textbf{(A) } 8 \qquad \tex... | [asy] size(300); pair A, B, C, P, XX, D; B = (0,0); P = (2,0); C = (5,0); A=(0,4.47214); D = A + (10,0); draw(A--B--C--cycle, linewidth(1)); dot("$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,... | 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 w... | 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... | 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 $a+b+c?$
$\t... | 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}... | 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 $a+b+c?$
$\t... | 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-... | 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 $a+b+c?$
$\t... | 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 formu... | 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}... | 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... | 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}... | 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: \[... | 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}... | 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... | 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}... | 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... | 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}... | 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}... | 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 mul... | 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{\unde... | 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... | 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... | 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)... | 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... | 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 $x$ -axi... | 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... | 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(... | 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 \... | 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... | 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 \... | 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... | 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 \... | 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 wa... | 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?
$\te... | 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{a... | 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?
$\te... | 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^... | 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); ... | 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... | 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); ... | (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 (... | 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); ... | 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 m... | 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); ... | 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 dif... | 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); ... | 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 p... | 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... | 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 $... | 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... | 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 pas... | 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... | 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... | 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... | 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... | 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... | 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 th... | 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 $... | 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 counterclockw... | 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 $... | 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 $... | 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... | 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 $... | 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 $... | 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 t... | 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 $... | 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
... | 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 $\fra... | 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 c... | 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 $\fra... | 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 \pm... | 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 $\fra... | 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... | 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))... | [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 +... | 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))... | [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, ... | 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 ... | 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 las... | 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)\lef... | 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\cdo... | 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, ... | 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*} Therefor... | 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 th... | 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... | 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 $\... | 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 ev... | 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 $9$ $1... | 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)... | 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)... | 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 ... | 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)... | 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... | 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(\... | 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... | 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 degr... | 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)}\ ... | 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 $a+b$ mus... | 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)}\ ... | 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,... | 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 $n$ time... | 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 th... | 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 th... | 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 \q... | 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'... | E | 16 |
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