2022 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 1 Easy

Define x◊y to be \left| x-y\right| for all real numbers x and y. What's the value  of

\left( 1◊\left( 2◊3\right)\right)-\left( \left( 1◊2\right)◊3\right)?

  • A.

    -2

  • B.

    -1

  • C.

    0

  • D.

    1

  • E.

    2

Answer:A

Original Expression =(1◊1)-(1◊3)

=0-2

=2.

Link Problem
Problem 2 Easy

In rhombus ABCD, point P lies on segment \overline{AD} such that BP\bot AD, AP=3 and PD=2. What is the area of ABCD? (Note: The figure is not drawn to scale.)

  • A.

    3\sqrt{5}

  • B.

    10

  • C.

    6\sqrt{5}

  • D.

    20

  • E.

    25

Answer:D

A D=A P+P D=3+2=5 A B C D is a rhombus, so A B=A D=5. \triangle A P B is a 3-4-5 right triangle, hence B P=4. The area of the rhombus is base times height: b h=(A D)(B P)=5 \cdot 4=20.

Link Problem
Problem 3 Easy

A donkey suffers an attack of hiccups and the first hiccup happens at 4:00 one afternoon. Suppose that the donkey hiccups regularly every 5 seconds. At what time does the donkey's 700^{\text{th}} hiccup occur?

  • A.

    15 seconds after 4:58

  • B.

    20 seconds after 4:58

  • C.

    25 seconds after 4:58

  • D.

    30 seconds after 4:58

  • E.

    35 seconds after 4:58

Answer:A

The first hiccup at 4 pm needs to be counted. So the 700^{\text{th}} hiccup will occur after 5\times (700-1)=3495 seconds, or 58 minutes and 15 seconds, which will be 15 seconds after 4:58.

Link Problem
Problem 4 Easy

How many three-digit positive integers have an odd number of even digits?

  • A.

    150

  • B.

    250

  • C.

    350

  • D.

    450

  • E.

    550

Answer:D

There are 5\times 5\times 5\times 2 \text{ (odd/even/odd and odd/odd/even) }+4\times 5\times 5 \text{ (even/odd/odd)}=350 ways to let the number have one even digit, while 4 \times 5\times 5=100 ways to let the number have three even digits. So the answer is 350+100=450 such numbers.

Link Problem
Problem 5 Easy

What is the value of  \dfrac{\left( 1+\dfrac{1}{3}\right)\left( 1+\dfrac{1}{5}\right)\left( 1+\dfrac{1}{7}\right)}{\sqrt{\left( 1-\dfrac{1}{3^{2}}\right)\left( 1-\dfrac{1}{5^{2}}\right)\left( 1-\dfrac{1}{7^{2}}\right)}}?

  • A.

    \sqrt{3}

  • B.

    2

  • C.

    \sqrt{15}

  • D.

    4

  • E.

    \sqrt{105}

Answer:B

Take the difference of squares of the radicand of denominator, we can rewrite the expression in the form of \frac {a}{\sqrt{ab}}, where a=\frac 34 \times \frac 65 \times \frac 87 and b=\frac 23 \times \frac 45 \times \frac 67. Here b=\frac 14a. Substitute and simplify to get the result is 2.

Link Problem
Problem 6 Easy

How many of the first ten numbers of the sequence 121, 11211, 1112111,.\dots. are prime numbers?

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:A

Enumerate the terms and find the pattern:

\begin{aligned} 121 & =11 \cdot 11, \\ 11211 & =101 \cdot 111, \\ 1112111 & =1001 \cdot 1111, \end{aligned}, and so on. Therefore, no numbers in the sequence would be prime.

Link Problem
Problem 7 Easy

For how many values of the constant k will the polynomial x^{2}+kx+36 have two distinct integer roots?

  • A.

    6

  • B.

    8

  • C.

    9

  • D.

    14

  • E.

    16

Answer:B

By Vieta's Formulae, the product of two roots is 36=2^2\times 3^2, which contains 9 factors. Break down 36 to get 36=1\times 36=2\times 18=3\times 12=4\times 9=6\times 6, and the signs can be reversed. Since we want distinct roots, the 6\times 6 case will be neglected, so there are 4\times 2=8 choices of k.

Link Problem
Problem 8 Easy

Consider the following 100 sets of 10 elements each:

\left\{ 1,2,3,\dots,10,\right\}

\left\{ 11,12,13,.,20\right\}

\left\{ 21,22,23,..,30\right\}

:

\left\{ 991,992,993,\dots,1000\right\}.

How many of these sets contain exactly two multiples of 7?

  • A.

    40

  • B.

    42

  • C.

    43

  • D.

    49

  • E.

    50

Answer:B

Note that for each group, there is at least one multiple of 7 and at most two multiples of 7.

There are \left\lfloor \frac{1000}{7}\right\rfloor=142 multiples of 7 under 1000, while there are 100 groups. Therefore, there are 142-100=42 sets contain exactly two multiples of 7.

Link Problem
Problem 9 Easy

The sum \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!} can be expressed as a-\dfrac{1}{b!}  ,  where a and b are positive integers. What is a+b?

  • A.

    2020

  • B.

    2021

  • C.

    2022

  • D.

    2023

  • E.

    2024

Answer:D

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=2023.

Link Problem
Problem 10 Easy

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?

  • A.

    5

  • B.

    7

  • C.

    9

  • D.

    11

  • E.

    13

Answer:D

We have \text{mode}=\text{median}+2 and \text{median}=\text{mean}+2. If there are three modes, then \text{mode}=\text{median}, which is a contradiction. So there are two modes. Let the five numbers be a,b,c,d,d, respectively, with increasing order. Then d=c+2 and c=\frac{a+b+c+d+d}{5}+2.

4c=a+b+(c+2)\times 2+10

2c=a+b+14 \geqslant 8.5, so c_{min}=9 with d_{min}=9+2=11.

The five numbers will be 1,3,9,11,11.

Link Problem
Problem 11 Easy

All the high schools in a large school district are involved in a fundraiser selling T-shirts. Which of the choices below is logically equivalent to the statement "No school bigger than Euclid HS sold more T-shirts than Euclid HS"?

  • A.

    All schools smaller than Euclid HS sold fewer T-shirts than Euclid HS.

  • B.

    No school that sold more T-shirts than Euclid is bigger than Euclid HS.

  • C.

    All schools bigger than Euclid HS sold fewer T-shirts than Euclid HS.

  • D.

    All schools that sold fewer T-shirts than Euclid HS are samller than Euclid HS.

  • E.

    All schools smaller than Euclid HS sold more T-shirts than Euclid HS.

Answer:B

In mathematical logics, a contrapositive statement is equivalent to the original statement. We find answer choice B is such a statement.

Link Problem
Problem 12 Easy

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 \dfrac{1}{2}?

  • A.

    2

  • B.

    3

  • C.

    4

  • D.

    5

  • E.

    6

Answer:C

For each roll, the probability to have a sum of 7 is \frac 16.

To make the probability that the sum of the numbers face up on a roll equals 7 at least once is greater than \frac 12, we have P(\text{sum not 7})=1-\left(\frac 56 \right)^n<\frac 12, where n_{min}=4.

Link Problem
Problem 13 Easy

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?

  • A.

    8

  • B.

    10

  • C.

    11

  • D.

    13

  • E.

    16

Answer:E

Let the numbers be a and b, respectively. Then we have a-b=2 and a^3-b^3=31106. With manipulation, (a-b)^2=a^2-2ab+b^2=4 and a^2+ab+b^2=15553, so 3ab=15549 and ab=5183. Observe that a=73 and b=71. The next prime number is 79, so the answer is 7+9=16.

Link Problem
Problem 14 Easy

Suppose that S is a subset of \left\{ 1,2,3\cdots ..,25\right\} such that the sum of any two (not necessarily distinct) elements of S is never an element of S. What is the maximum number of element S may contain?

  • A.

    12

  • B.

    13

  • C.

    14

  • D.

    15

  • E.

    16

Answer:B

The sum of two elements must be at least 26 to avoid overlap with the original set. By this fact, 1 can only pair with 25, 2 can pair with 24 and 25, \cdots, 13 can pair with all numbers greater than itself up to 25. Therefore, the set \{13,14,\cdots, 25\} will be the desired set with 13 elements.

Link Problem
Problem 15 Medium

Let S_{n} be the sum of the first n term of an arithmetic sequence that has a common difference of 2. The quotient \dfrac{S_{3n}}{S_{n}} does not depend on n. What is S_{20}?

  • A.

    340

  • B.

    360

  • C.

    380

  • D.

    400

  • E.

    420

Answer:D

\begin{aligned} \frac{S_{3 n}}{S_n} & =\frac{[a+a+2 \times(3 n-1)] \times 3 n}{[a+a+2 \times(n-1)] \times n} \\ & =\frac{3[2 a+2(3 n-1)]}{2 a+2(n-1)} \\ & =\frac{18 n+6 a-6}{2 n+2 a-2}\end{aligned}.

Since the quotient \dfrac{S_{3n}}{S_{n}} does not depend on n, a=1, and S_{20}=1+3+5+\cdots+39=400.

Link Problem
Problem 16 Easy

The diagram below shows a rectangle with side lengths 4 and 8 and a square with side length 5. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?

  • A.

    15\dfrac{1}{8}

  • B.

    15\dfrac{3}{8}

  • C.

    15\dfrac{1}{2}

  • D.

    15\dfrac{5}{8}

  • E.

    15\dfrac{7}{8}

Answer:D

We first notice that the side length of the square is 5 by Pythagorean Theorem. Then, by Triple Perpendicular Model, the four right triangles are similar to each other. A_{\triangle A B C}: A_{\triangle C D E}: A_{\triangle E F G}: \text{top triangle}=1: 1: \frac{1}{16}: \frac {25}{16}. The area of the top right triangle is therefore \frac{25}{16}\times 6=\frac {75}{8}. Hence the shaded area is 25 - \frac{75}{8}=\frac{125}{8}=15\frac{5}{8}.

Link Problem
Problem 17 Medium

One of the following numbers is not divisible by any prime number less than 10. Which is it?

  • A.

    2^{606}-1

  • B.

    2^{606}+1

  • C.

    2^{607}-1

  • D.

    2^{607}+1

  • E.

    2^{607}+3^{607}

Answer:C

A. 2^{606} \equiv 4^{303} \equiv 1(\bmod 3) \Rightarrow 3 \mid 2^{606}-1.

B. 2^{606} \equiv 4^{203} \equiv -1(\bmod 5) \Rightarrow 5 \mid 2^{606}+1.

D. 2^{607} \equiv 4^{303} \times 2 \equiv 2(\bmod 3) \Rightarrow 3 \mid 2^{607}+1.

E. 2^{607}+3^{607} \equiv 2^{607}+(-2)^{607} \equiv 0(\bmod 5) \Rightarrow 5\mid 2^{607}+3^{607}.

So the answer is C.

Link Problem
Problem 18 Easy

Consider systems of three linear equations with unknowns x, y, and z

a_{1}x+b_{1}y+c_{1}z=0

a_{2}x+b_{2}y+c_{2}z=0

a_{3}x+b_{3}y+c_{3}z=0

where each of the coefficients is either 0 or 1 and the system has a solution other than x=y=z=0. For example, one such system is \left\{ 1x+1y+0z=0,\ 0x+1y+1z=0,0x+0y+0z=0\right\} with a nonzero solution of  \left\{ x,y,z=\left\{ 1,-1,1\right\}\right\}. How many such systems are there?(The equations in a system need not be distinct, and two systems containing the same equations in a different order are considered different.)

  • A.

    302

  • B.

    338

  • C.

    340

  • D.

    343

  • E.

    344

Answer:B

The sufficient and necessary condition for a homogeneous linear system of equations to have a non-zero solution is that the rank of its coefficient matrix is less than the number of unknowns. In simple terms, it means that among these three equations, at least one must be "redundant" and cannot provide additional constraints to solve the equation. It is an identity that always holds true. Then we can discuss the following cases:

1^\circ One of the equations has coefficients that are all zeros. For each ordered triple of coefficients in the equations P_i=(a_i,b_i,c_i),i=1,2,3, there are a total of 2^3=8 choices. So for this case, we have 8^3-(8-1)^3=169 choices.

2^\circ None of the equations has coefficients that are all zeros and one of the equations can be obtained by linearly combining the other two equations (without loss of generality, we consider only addition).

1' Among the three equations, there are two equations that are identical. For this case, we have 7^3-7\cdot 6\cdot 5=133 choices.

2' In the equation that is a "sum", only two coefficients are 1 and the other one is 0. For example, for the ordered triples of coefficients in the equations we have (1,0,0)+(0,1,0)=(1,1,0). For this case, we have _3C_1\cdot _3P_3=18 choices.

3' In the equation that is a "sum", all of the three coefficients are 1. For example, for the ordered triples of coefficients in the equations we have (1,1,0)+(0,0,1)=(1,1,1). For this case, we have _3C_1\cdot _3P_3=18 choices.

So in total, we have 169+133+18+18=338 choices.

Link Problem
Problem 19 Hard

Each square in a 5\times5 grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

-Any filled square with two or three filled neighbors remains filled.

-Any empty square with exactly three filled neighbors becomes a filled square.

-All other squares remain empty or become empty.

A sample transformation is shown in the figure below.

\to

Suppose the 5\times5  grid has a border of empty squares surrounding a 3\times3  subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

\to

  • A.

    14

  • B.

    18

  • C.

    22

  • D.

    26

  • E.

    30

Answer:C

This is actually a very famous mathematical game called "Conway's Game of Life", which is the most well-known two-dimensional cellular automaton. Its inventor, Professor John Horton Conway, passed away in 2020, and the question writer may have been paying tribute to him in this way.

After discussing the background, let's get back to the problem itself. We will discuss two cases:

1^\circ The first case is when the central square is initially empty. Therefore, among the eight squares surrounding it, there must be exactly three that are filled.

1' Its three filled neighbours are all in the corner.

​

For this case, we have 4 initial configurations.

2' Two of its three filled neighbours are in the corner.

​

As shown in the diagram, we have 4+8=12 initial configurations.

3' One of its three filled neighbours is in the corner.

​

As shown in the diagram, we have 4 initial configurations.

4' None of its three filled neighbours is in the corner. At this point, there must be a filled square with two filled neighbours so that does not hold.

2^\circ The second case is when the central square is initially filled. Therefore, among the eight squares surrounding it, there must be two or three that are filled.

If one of it is not in the corner, then no matter how the other square is placed, there will always be a new filled point generated.

So all points must be in the corners, and at this point, there can only be two filled points and they must be placed in opposite corners. We have 2 initial configurations.

So in total, we have 4+12+4+2=22 initial configurations.

Link Problem
Problem 20 Hard

Let ABCD be a rhombus with ∠ADC=46^{\circ} . Let E be the midpoint of \overline{CD}, and let F be the point on \overline{BE} such that \overline{AF} is perpendicular to \overline{BE}. What is the degree measure of ∠BFC?

  • A.

    110

  • B.

    111

  • C.

    112

  • D.

    113

  • E.

    114

Answer:D

​

Refer to the figure. AF \perp BE and BO \perp AC. Denote the intersection between AC and BD as H. Here OH is the orthocenter of \triangle ABG.

\angle A F O=\angle A B O=23^{\circ}

\angle O F E=90^{\circ}-\angle AFO=90^{\circ}-23^{\circ}=67^{\circ}.

∵ \angle ACD=67^{\circ},

∴ OFCE are concyclic,

∴ \angle CFE=\angle COE=90^{\circ}-23^{\circ}=67^{\circ},

∴ \angle BFC=180^{\circ}-67^{\circ}=113^{\circ}.

Link Problem
Problem 21 Hard

Let P(x) be a polynomial with rational coefficients such that when P(x) is divided by the polynomial x^{2}+x+1, the remainder is x+2, and when P(x) is divided by the polynomial x^{2}+1  the remainder is 2x+1. There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?

  • A.

    10

  • B.

    13

  • C.

    19

  • D.

    20

  • E.

    23

Answer:E

p(x) \div \left(x^2+x+1\right) R \ (x+2);

p(x) \div \left(x^2+1\right) R \ (2 x+1).

Let the remainder of p(x) \div\left(x^2+1\right)\left(x^2+x+1\right) be \theta(x)=A x^3+B x^2+cx+d.

Then \left\{\begin{array}{l}\theta(x)=\left(x^2+x+1\right)(m x+n)+(x+2) \\ \theta(x)=\left(x^2+1\right)(k x+b)+(2 x+1)\end{array}\right..

Simplify to get

\left\{\begin{array}{l}\theta(x)=m x^3+(n+m) x^2+(m+n-1) x+n+2 \\ \theta(x)=k x^3+b x^2+(k+2) x+b+1\end{array}\right.

So \left\{\begin{array}{l}m=k \\ m+n=b \\ m+n+1=k+2 \\ n+2=b+1\end{array}\right.

and \left\{\begin{array}{l}k=1 \\ m=1 \\ n=1 \\ b=2\end{array}\right..

\theta(x)=x^3+2 x^2+3 x+3,

Answer is 1^2+2^2+3^2+3^2=23.

Link Problem
Problem 22 Hard

Let S be the set of circles in the coordinate plane that are tangent to each of the three circles with equations x^2+y^2=4, x^2+y^2=64 and \left( x-5\right)^{2}+y^{2}=3. What is sum of the areas of all circles in S.

  • A.

    48\pi

  • B.

    68\pi

  • C.

    96\pi

  • D.

    102\pi

  • E.

    136\pi

Answer:E

There are two circles that are externally tangent to the two small circles and internally tangent to the large circle. Here r=3 and A=18\pi. (blue part)

There are two circles that are internally tangent to the concentric circles and externally tangent to the other circle. Here r=5 and A=50\pi. (red part)

There are two circles that are internally tangent to the big and non-center circles, but externally tangent to the small center circle. Here r=3 and A=18\pi. (green part)

There are two circles that are internally tangent to all three circles. Here r=5 and A=50\pi. (brown part)

The sum of area of all circles is 136\pi.

Link Problem
Problem 23 Hard

Ant Amelia starts on the number line at 0 and crawls in the following manner. For n=1,2,3, Amelia chooses a time duration t_{n} and an increment x_{n} independently and uniformly at random from the interval (0,1). During the nth step of the process, Amelia moves x_{n} units in the positive direction, using up t_{n} minutes. If the total elapsed time has exceeded 1 minute during the n^{\text{th}} step, she stops at the end of that step; otherwise, she continues with the next step, taking at most 3 steps in all. What is the probability that Amelia's position when she stops will be greater than 1?

  • A.

    \dfrac{1}{3}

  • B.

    \dfrac{1}{2}

  • C.

    \dfrac{2}{3}

  • D.

    \dfrac{3}{4}

  • E.

    \dfrac{5}{6}

Answer:C

Let x and y be random variables that are independently and uniformly distributed in the interval (0,1). Note that P(x+y \leq 1)=\frac{\frac{1}{2} \cdot 1^2}{1^2}=\frac{1}{2}, as shown below:

Let x, y, and z be random variables that are independently and uniformly distributed in the interval (0,1). Note that P(x+y+z \leq 1)=\frac{\frac{1}{3} \cdot\left(\frac{1}{2} \cdot 1^2\right) \cdot 1}{1^3}=\frac{1}{6}, as shown below:

We have two cases: 1. Amelia takes exactly 2 steps. We need x_1+x_2>1 and t_1+t_2>1. So, the probability is P\left(x_1+x_2>1\right) \cdot P\left(t_1+t_2>1\right)=\left(1-\frac{1}{2}\right) \cdot\left(1-\frac{1}{2}\right)=\frac{1}{4}.

2. Amelia takes exactly 3 steps. We need x_1+x_2+x_3>1 and t_1+t_2 \leq 1. So, the probability is P\left(x_1+x_2+x_3>1\right) \cdot P\left(t_1+t_2 \leq 1\right)=\left(1-\frac{1}{6}\right) \cdot \frac{1}{2}=\frac{5}{12} . Together, the answer is \frac{1}{4}+\frac{5}{12}= (C) \frac{2}{3}.

Link Problem
Problem 24 Hard

Consider functions f that satisfy |f(x)-f(y)|\leqslant\dfrac{1}{2}|x-y| for all real numbers x and y. Of all such functions that also satisfy the equation f(300)=f(900), what is the greatest possible value of

f\left( f(800)\right)-f(f(400))?

  • A.

    25

  • B.

    50

  • C.

    100

  • D.

    150

  • E.

    200

Answer:B

|f(800)-f(900) |\leqslant \frac{1}{2} |800-900|=50;

|f(400)-f(300)| \leqslant \frac{1}{2} |400-300|=50.

|f(800)-f(400)| = |f(800-f(900)+f(300)-f(400)|

\leqslant |f(800)-f(900)|+|f(300)-f(400)|

\leqslant 50+50 =100.

\begin{aligned} f[f(800)]-f[f(400)] & \leqslant \frac{1}{2}|f(800)-f(400)| \\ & \leqslant \frac{1}{2} \times 100 \\ & =40\end{aligned}

Link Problem
Problem 25 Hard

Let  x_{0},\ x_{1},\ x_{2}\cdots .  be a sequence of numbers, where each x_{k}  is either 0 or 1. For each positive integer n, define

S_{n}=\sum _{k=0}^{n-1}x_{k}2^{k}

Suppose  7S_{n}=1 (mod 2^{n}) for all n\geqslant1. What is the value of the sum

x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}?

  • A.

    6

  • B.

    7

  • C.

    12

  • D.

    14

  • E.

    15

Answer:A

\begin{aligned} & 7 S_1=3 \times 2^1+1 \Rightarrow x_0=1 \\ & 7 S_2=5 \times 2^2+1 \Rightarrow x_1=1 \\ & 7 S_3=6 \times 2^3+1 \Rightarrow x_2=1 \\ & 7 S_4=3 \times 2^4+1 \Rightarrow x_3=0 \\ & 7 S_5=5 \times 2^5+1 \Rightarrow x_4=1 \\ & 7 S_6=6 \times 2^6+1 \Rightarrow x_5=1 \\ & 7 S_7=3 \times 2+1 \Rightarrow x_6=0\end{aligned}

\cdots

With the patterns, we notice that x_{3n}=0 and x_{3n+1}=x_{3n+2}=1. Therefore, x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=0=2+4+0=6.

Link Problem
Related Paper
2023 AMC 10 B 2022 AMC 10 A 2021 AMC 10 A Fall 2021 AMC 10 B Fall
Table of Contents
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
  • 11
  • 12
  • 13
  • 14
  • 15
  • 16
  • 17
  • 18
  • 19
  • 20
  • 21
  • 22
  • 23
  • 24
  • 25
Think Academy Powered by ChatGPT