2022 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 1 Easy

What is the value of

3+\frac{1}{3+\frac{1}{3+\frac{1}{3}}}?

  • A.

    \frac{31}{10}

  • B.

    \frac{49}{15}

  • C.

    \frac{33}{10}

  • D.

    \frac{109}{33}

  • E.

    \dfrac{15}{4}

Answer:D

3+\frac{1}{3+\frac{1}{3+\frac{1}{3}}}=3+\frac{1}{3+\frac{1}{\frac{10}{3}}}=3+\frac{1}{3+\frac{3}{10}}=3+\frac{1}{\frac{33}{10}}=3+\frac{10}{33}=\frac{109}{33}.

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Problem 2 Easy

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?

  • A.

    5

  • B.

    7

  • C.

    9

  • D.

    11

  • E.

    13

Answer:B

Assume the number of laps is x.

\frac{15}{57}=\frac{x}{27}

x=\frac{15 \times 27}{57} \approx 7.

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Problem 3 Easy

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?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:E

Assume x is the third number.

6x+(x+40)+x=96

x=7

(x+40)-6x=5

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Problem 4 Easy

In some countries, automobile fuel efficiency is measured in liters per 100 kilometers while other countries use miles per gallon. Suppose that 1 kilometer equals m miles, and 1 gallon equals l liters. Which of the following gives the fuel efficiency in liters per 100 kilometers for a car that gets x miles per gallon?

  • A.

    \frac{x}{100lm}

  • B.

    \dfrac{xlm}{100}

  • C.

    \frac{lm}{100x}

  • D.

    \frac{100}{xlm}

  • E.

    \frac{100lm}{x}

Answer:E

x \text{miles/gallon} = \frac x m \text{kilometers/} l \text{liters} = 1 \text{kilometer/} \frac {lm} {x} \text{liters} = 100 \text{kilometers/} \frac {100lm} {x} \text{liters}.

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Problem 5 Easy

Square ABCD has side length 1. Point P, Q, R, and S each lie on a side of ABCD such that APQCRS is an equilateral convex hexagon with side length s. What is s?

  • A.

    \frac{\sqrt{2}}{3}

  • B.

    \frac{1}{2}

  • C.

    2-\sqrt{2}

  • D.

    1-\frac{\sqrt{2}}{4}

  • E.

    \frac{2}{3}

Answer:C

Assume BP=x.

AP=PQ, PB=BQ,

\triangle {BPQ} is an isosceles right triangle.

\sqrt{2} x=1-x

(1+\sqrt{2})x=1

x=\frac{1}{1+\sqrt{2}}

AP=\frac{\sqrt{2}}{1+\sqrt{2}}=2-\sqrt{2}.

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Problem 6 Easy

Which expression is equal to \left| a-2-\sqrt{(a-1)^{2}}\right| for a\lt0?

  • A.

    3-2a

  • B.

    1-a

  • C.

    1

  • D.

    a+1

  • E.

    3

Answer:A

\because a<0

\therefore |a-2-\sqrt{(a-1)^2}|=|a-2-(1-a)|=|2a-3|=3-2a.

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Problem 7 Easy

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?

  • A.

    3

  • B.

    6

  • C.

    8

  • D.

    9

  • E.

    12

Answer:B

\because lcm(n,18)=180

\therefore 2^2 \cdot 5 | n and n | 2^2 \cdot 5 \cdot 3^2=180,

\therefore n=2^2 \cdot 5 \cdot m (m is a factor of 3^2).

\because gcd(n,45)=15,

\therefore 3 \cdot 5 | n and 3^2 \cdot 5 | n,

\therefore If you factorize n, there is only one factor 3,

\therefore n=2^2 \cdot 5 \cdot 3=60.

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Problem 8 Easy

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 positive values of X?

  • A.

    10

  • B.

    26

  • C.

    32

  • D.

    36

  • E.

    40

Answer:D

The possible values for the mean: 5, 7, x.

If mean =5, x=5\times6-(1+2+5+5+7)=10;

If mean =7, x=7\times6-(1+2+5+5+7)=22;

If mean =x, x=6x-20, then x=4.

Sum =10+22+4=36.

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Problem 9 Easy

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?

  • A.

    120

  • B.

    270

  • C.

    360

  • D.

    540

  • E.

    720

Answer:D

D is special as it touches all the other rectangles.

Step 1: Paint D first, we have 5 choices;

Step 2: Paint C, we have 4 choices;

Step 3: Paint E, we have 5 choices (it's different from D);

Step 4: Paint A and B at last, we have 3 choices for each.

Total number of ways =5\times4\times3\times3\times3=540.

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Problem 10 Easy

Daniel finds a rectangular index card and measures its diagonal to be 8 centimeters. Daniel then cuts out equal squares of side 1cm at two opposite cormers of the index card and measures the distance between the two closest vertices of these squares to be 4\sqrt{2}  centimeters, as shown below. What is the area of the original index card?

  • A.

    14

  • B.

    10\sqrt{2}

  • C.

    16

  • D.

    12\sqrt{2}

  • E.

    18

Answer:E

Assume the length of the index is x and the width is y.

x^2+y^2=8^2=64

(x-2)^2+(y-2)^2=(4\sqrt{2})^2=32

x^2-4x+4+y^2-4y+4=32

(x^2+y^2)-4(x+y)=24

4(x+y)=40

x+y=10

(x+y)^2=x^2+y^2+2xy=100

2xy=100-64=36

Area =xy=18

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Problem 11 Easy

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?

  • A.

    5

  • B.

    6

  • C.

    7

  • D.

    8

  • E.

    9

Answer:C

2^m \cdot \left(\frac {1} {4096} \right)^\frac 1 2=2 \cdot \left(\frac {1} {4096} \right)^\frac 1 m

2^m \cdot \left(2^{-12} \right)^\frac1 2=2 \cdot \left(2^{-12} \right)^\frac1 m

2^m \cdot 2^{-6}=2 \cdot 2^{-\frac{12}{m}}

m-6=1-\frac{12}{m}

m+\frac{12}{m}=7

m^2-7m+12=0

Using Vieta's formula, sum of roots =7.

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Problem 12 Easy

On Halloween 31 children walked into the principal's office asking for candy. They can be classified into three types: Some always lie; some always tell the truth; and some alternately lie and tell the truth. The alternaters arbitrarily choose their first response, either a lie or the truth, but each subsequent statement has the opposite truth value from its predecessor. The principal asked everyone the same three questions in this order.

"Are you a truth-teller?" The principal gave a piece of candy to each of the 22 children who answered yes.

"Are you an alternater?" The principal gave a piece of candy to each of the 15 children who answered yes.

"Are you a liar?" The principal gave a piece of candy to each of the 9 children who answered yes.

How many pieces of candy in all did the principal give to the children who always tell the truth?

  • A.

    7

  • B.

    12

  • C.

    21

  • D.

    27

  • E.

    31

Answer:A

Answer (A): Let T denote the number of truth-tellers, A_{\text{o}} the number of alternators who tell the truth on odd-numbered questions, A_{\text{e}} the number of alternators who tell the truth on evennumbered questions, and L the number of liars. These four variables must satisfy the following equations.

\begin{aligned} \text{Total:} 31 & =T+L+A_{\text{e}}+A_{\text{o}} \\ \text{Question 1:}22 & =T+L+A_{\text{e}} \\ \text{Question 2:}15 & =L+A_{\text{e}} \\ \text{Question 3:}9 & =A_{\text{e}} \end{aligned}

Subtract the third equation from the second to find that T=7. Because they received candy only in response to the first question, they received 7 pieces of candy. Note: It may be further deduced from the system of equations that A_{\text{e}}=9, L=6, and A_{\text{o}}=9.

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Problem 13 Easy

Let  △ABC  be a scalene triangle. Point P lies on \overline{BC} so that \overline{AP} bisects  ∠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?

  • A.

    8

  • B.

    9

  • C.

    10

  • D.

    11

  • E.

    12

Answer:C

\because AP \perp BD

\therefore AM=AB

\therefore \frac{AB}{AC}=\frac{BP}{PC}=\frac 23

\therefore \frac{MC}{AM}=\frac{AC-AM}{AM}=\frac 12

\because AD \parallel BC

\therefore \triangle {AMD} \sim \triangle {CMB}

\therefore \frac{AD}{BC}=\frac{AM}{MC}=2

\therefore AD=10

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Problem 14 Easy

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?

  • A.

    108

  • B.

    120

  • C.

    126

  • D.

    132

  • E.

    144

Answer:E

Integers from 8 to 14 have to be separated into differentt groups, because none of them could be the 'lesser number' in the pair to be the larger number. The lesser number candidates are 1,2,3,4,5,6,7.

7 \rightarrow 14

6 \rightarrow 12, 13 (let's go with 13)

5 \rightarrow 10, 11, 12 (let's go with 12)

4 \rightarrow 8, 9, 10, 11 (let's go with 11)

3 \rightarrow 8, 9, 10 (let's go with 10)

2 \rightarrow 8, 9 (let's go with 9)

1 \rightarrow 8

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Problem 15 Easy

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?

  • A.

    260

  • B.

    855

  • C.

    1235

  • D.

    1565

  • E.

    1997

Answer:D

7^2+24^2=25^2=15^2+20^2

\angle{B} + \angle {D}= 180 ^{\circ}

If \angle{B}>90^{\circ}, AC>75

\because \angle{B}+ \angle{D}=180^{\circ}

\therefore \angle D<90^{\circ},AC<25, so there is a contradiction.

If \angle{B}<90^{\circ}, we also have a contradiction.

\therefore \angle B=\angle D = 90^{\circ},

\therefore AC is the diameter of the circle.

\therefore A_{ABCD}=A_{ABC}+A_{ACD}=\frac 12 \times 7 \times 24+\frac 12 \times 15 \times 20=234,

A_{circle}=\pi \cdot \left(\frac{25}{2} \right)^2=\frac{625\pi}{4}.

\therefore \frac{a\pi-b}{c}=\frac{625\pi}{4}-234=\frac{625\pi-936}{4}

\therefore a+b+c=625+936+4=1565.

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Problem 16 Easy

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?

  • A.

    \frac{24}{5}

  • B.

    \dfrac{42}{5}

  • C.

    \frac{81}{5}

  • D.

    30

  • E.

    48

Answer:D

Assume the roots of the polynomial are a,b,c.

According to Vieta's formula,

\begin{cases}a+b+c=\frac{39}{10} \\ ab+bc+ca=\frac{29}{10} \\ abc=\frac 35 \end{cases}

(a+2)(b+2)(c+2)=abc+2(ab+bc+ca)+4(a+b+c)+8=30.

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Problem 17 Easy

How many three-digit positive integers \underline{a} \underline{b} \underline{c} are there whose nonzero digits a, b, and c satisfy

0.\overline{\underline{a}\ \underline{b}\ \underline{c}}=\frac{1}{3}(0.\overline{a}+0.\overline{b}+0.\overline{c})?

(The bar indicates repetition, thus  0.\overline{\underline{a}\ \underline{b}\ \underline{c}} is the infinite repeating decimal 0.\underline{a} \underline{b} \underline{c} \underline{a} \underline{b} \underline{c} \cdot \cdot \cdot)

  • A.

    9

  • B.

    10

  • C.

    11

  • D.

    13

  • E.

    14

Answer:D

0.\overline{\underline{a}\ \underline{b}\ \underline{c}} is a recuesive decimal, so as 0.\overline{a}, 0.\overline{b} and 0.\overline{c}.

\frac{100a+10b+c}{999}=\frac{1}{3}\cdot \left(\frac a9+\frac b9+\frac c9 \right)

\frac{100a+10b+c}{999}=\frac{a+b+c}{27}

100a+10b+c=37a+37b+37c

63a=27b+36c

3b+4c=7a

\therefore b\equiv c( \text{mod} 7).

Thus, either b=c or (b,c)=(1,8),(2,9),(9,2). Each of these will have an unique solution, yielding the 13 intergers: 111,222,333,444,555,666,777,888,999,518,629,481,592.

Therefore, the answer is 13.

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Problem 18 Easy

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}\ ,..,T_{n} returns the point (1,0) back to itself?

  • A.

    359

  • B.

    360

  • C.

    719

  • D.

    720

  • E.

    721

Answer:A

Let the angle between 'the point', 'the origin' and the positive x-axis be \alpha.

T_{x}: \alpha \rightarrow \alpha + x \rightarrow 180-(\alpha+x)

T_x+T_y: 180-(\alpha +x) \rightarrow 180-(\alpha +x)+y \rightarrow \alpha+x-y

Initially: 0^{\circ} after T_1, T_2 \dots T_n.

When n=2k, the angle:

(1-2)+(2-3)+\dots [(2k+1)-2k]=-k.

Back to the original position: k=360^{\circ}, n=720.

When n=2k+1, the angle:

(1-2)+(2-3)+\dots [(2k+1)-2k]+T_{2k+1}=-k+T_{2k+1}=180-(-k+2k+1)=179-k

When k=179, n=359.

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Problem 19 Easy

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?

  • A.

    1

  • B.

    3

  • C.

    5

  • D.

    7

  • E.

    9

Answer:C

Rearrange:

h=\frac{L_{17}}{1}+\frac{L_{17}}{2}+\frac{L_{17}}{3}+\frac{L_{17}}{4}+\dots+\frac{L_{17}}{17}

Every term above is divisible by 17.

h \equiv \frac{L_{17}}{17}=L_{16} (\text{mod} 17)

L_{16}=16 \cdot 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5

h \equiv 5 (\text{mod} 17)

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Problem 20 Easy

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?

  • A.

    190

  • B.

    194

  • C.

    198

  • D.

    202

  • E.

    206

Answer:E

Arithematic sequence: a, a+d,a+2d,a+3d.

Geometric sequence: x,xr,xr^2,xr^3.

\begin{cases}a+x=57 \\ a+d+xr=60 \\ a+2d+xr^2=91 \end{cases}

Then, \begin{cases} d+(r-1)x=3\\ d+(r^2-r)x=31 \end{cases}

So, (r^2-2r+1)x=28

(r-1)^2x=28

(r-1)\sqrt{x}=2\sqrt{7}

Since x is an integer, x has factor of 7.

Let x=7m^2 (m is an interger), q=\frac2m+1.

\because a+x=57, ax>0

\therefore m=1 or 2.

When m=1, q=2+1=3.

Arithematic sequence: 7,21,63,189.

Geometric sequence: 50,39,28,17.

189+17=206.

When m=2, q=1+1=2.

Arithematic sequence: 28,56,112.

Geometric sequence: 22,4,-16 (contradiction).

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Problem 21 Easy

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?

  • A.

    6

  • B.

    7

  • C.

    5+2\sqrt{2}

  • D.

    8

  • E.

    9

Answer:B

Create an tetraledron, each side of the  tetraledron gives an equilateral triangle.

A_{ABCD}=3\times3=9, A_{octagon}=8-4\times\frac12 \times\frac12=7.

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Problem 22 Easy

Suppose that 13 cards numbered 1, 2, 3,\cdots,13 are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards 1, 2, 3 are picked up on the first pass, 4 and 5 on the second pass, 6 on the third pass 7, 8, 9, 10 on the fourth pass, and 11, 12, 13 on the fifth pass. For how many of the 13! possible orderings of the cards will the 13 cards be picked up in exactly two passes?

​

  • A.

    4082

  • B.

    4095

  • C.

    4096

  • D.

    8178

  • E.

    8191

Answer:D

Picking up in exactly two passes is equivalent to cutting 13 cards in half.

For example, [1,2,3,4,5] [6,7,8,9,10,11,12,13].

When we rearrange, we need to make sure [1,2,3,4,5],[6,7,8,9,10,11,12,13] are arranged in increasing order, but we cannot include the one pass (1,2,3,4,5,6,7,8,9,10,11,12,13).

[1] [2,3,4,5,6,7,8,9,10,11,12,13]: 13C1-1

(Note: there are 13 possible spot for [1], but we need to subtract the situation when [1] is at the first place.)

[1,2] [3,4,5,6,7,8,9,10,11,12,13]: 13C2-1

\dots \dots \dots \dots

[1,2,3,4,5,6,7,8,9,10,11,12] [13]: 13C12-1

In total: (13C1+13C2+\dots+13C12)-12=2^{13}-14=8178.

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Problem 23 Easy

Isosceles trapezoid ABCD has parallel sides \overline{AD} and \overline{BC}, with BC\lt AD and AB=CD.

There is a point P in the plane such that PA=1,\ PB=2,\ PC=3, and  PD=4. What is \frac{BC}{AD} ?

  • A.

    \dfrac{1}{4}

  • B.

    \dfrac{1}{3}

  • C.

    \dfrac{1}{2}

  • D.

    \dfrac{2}{3}

  • E.

    \dfrac{3}{4}

Answer:B

\begin{cases}m^2+n^2=1 \\ n^2+(m-a)^2=16 \\ (m-b)^2+(n-t)^2=2^2 \\ (m-c)^2+(n-t)^2=3^2 \end{cases}

So \begin{cases}a^2-2ma=15 \\ c^2-b^2+2(b-c)m=5 \end{cases}

Since ABCD is equilateral triangle, so a=c+b.

(c+b)(c-b)+2(b-c)m=\frac15

(c-b)(a-2m)=5

a(a-2m)=15

Thus, \frac{c-b}{a}=\frac 13.

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Problem 24 Easy

How many strings of length 5 formed from the digits 0, 1, 2, 3, 4 are there such that for each j\in\left\{ 1,2,3,4\right\}, at least j of the digits are less than j? (For example, 02214 satisfies the condition because it contains at least 1 digit less than 1, at least 2 digits less than 2, at least 3 digits less than 3, and at least 4 digits less than 4. The string 23404 does not satisfy the condition because it does not contain at least 2 digits less than 2.)

  • A.

    500

  • B.

    625

  • C.

    1089

  • D.

    1199

  • E.

    1296

Answer:E

Suppose a<b<c<d<e. Then let's do the casework.

① If 5 numbers are all distinct, 5!=120.

② If one number is counted twice, 10\times \frac{5!}{2}=600.

③ If two numbers are counted twice, 10\times \frac{5!}{2 \cdot 2}=300.

④ If one number is counted 3 times, 10\times \frac{5!}{3!}=200.

⑤If one number is counted 3 times and the others are counted twice, 5\times \frac{5!}{2!\cdot3!}=50.

⑥ If one number is counted 4 times, 5 \times 5=25.

⑦ If one number is counted 5 times, only 1 possible way.

In total, 120+600+300+200+50+25+1=1296.

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Problem 25 Easy

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 as many lattice points as does S. The top two vertices of T are in R\cup S, and T contains \dfrac{1}{4} of the lattice points contained in R\cup S. See the figure (not drawn to scale).

The fraction of lattice points in S that are in S\cap T is 27 times the fraction of lattice points in R that are in R\cap T. What is the minimum possible value of the edge length of R plus the edge length of S plus the edge length of T?

  • A.

    336

  • B.

    337

  • C.

    338

  • D.

    339

  • E.

    340

Answer:B

① \frac94 (a+1)^2=(b+1)^2

② S\cup R= (a+1)^2+(b+1)^2-(a+1)

T=\frac14 S\cup R=\frac14[(a+1)^2+(b+1)^2-(a+1)]

=\frac14[(a+1)^2+\frac94(a+1)^2-(a+1)]

=\frac{1}{16}(a+1)(13a+9)

=(c+1)^2

③ T\cap S=(x+1)(c+1)

T\cap R=(y+1)(c+1)

\frac{(x+1)(c+1)}{(b+1)^2}=27 \times \frac{(y+1)(c+1)}{(a+1)^2}

Then x+1=27 \times \frac 49 (y+1)=12(y+1)

c=x+y=13y+11

y=1\cdot 2\cdot3\cdot5\cdot5\cdot6\dots

When y=6, c=89.

The equation: \frac{1}{16}(a+1)(13a+9)=(c+1)^2) has integer solution.

a=99, b=149, so (a+b+c)_{min}=89+99+149=337.

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2023 AMC 10 A 2023 AMC 10 B 2022 AMC 10 B 2021 AMC 10 A Fall
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