## AMC 8 2023

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 1 point for each correct answer. There is no penalty for wrong answers.
- No aids are permitted other than plain scratch paper, writing utensils, ruler, and erasers. In particular, graph paper, compass, protractor, calculators, computers, smartwatches, and smartphones are not permitted.
- Figures are not necessarily drawn to scale.
- You will have
**40 minutes**working time to complete the test.

What is the value of $(8 \times 4 + 2) - (8 + 4 \times 2)$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 24$

$\textbf{D}$

By the order of operations, we have\[(8 \times 4 + 2) - (8 + 4 \times 2) = (32+2) - (8+8) = 34 - 16 = 18\]

A square piece of paper is folded twice into four equal quarters, as shown below, then cut along the dashed line. When unfolded, the paper will match which of the following figures?

$\textbf{E}$

Noticing that the 4 triangles we cut is at the center of the large square. When unfolded, 4 blank triangles will form a inclined little square at the center of the paper.

$\textit{Wind chill}$ is a measure of how cold people feel when exposed to wind outside. A good estimate for wind chill can be found using this calculation\[(\text{wind chill}) = (\text{air temperature}) - 0.7 \times (\text{wind speed}),\]where temperature is measured in degrees Fahrenheit $(^{\circ}\text{F})$ and the wind speed is measured in miles per hour (mph). Suppose the air temperature is $36^{\circ}\text{F}$ and the wind speed is $18$ mph. Which of the following is closest to the approximate wind chill?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 35$

$\textbf{B}$

According to the definition, we have $\text{wind chill}=36-0.7\times18=23.4\approx23$.

The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

$\textbf{D}$

By filling out the grid, we see that 3 of these four numbers are prime.

A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?

$\textbf{(A)}\ 1250 \qquad \textbf{(B)}\ 1500 \qquad \textbf{(C)}\ 1750 \qquad \textbf{(D)}\ 1800 \qquad \textbf{(E)}\ 2000$

$\textbf{B}$

According to the sample, $\dfrac{30}{180}=\dfrac16$ of fish are trout. Since the ratio of trout to the total number of fish is the same in both the sample and the lake, we get the number of fish in the lake $250\div\dfrac16=1500$.

The digits $2, 0, 2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18$

$\textbf{C}$

If we take 0 as the base, then the value of the expression is 0. Hence, we must take 0 as the exponent in order to get the maximum of the expression. When the exponent is 0, no matter what value the base is (except 0), we have $\text{base}^0=1$. So we can just take the smallest value, 2, as the base of 0 to avoid wasting the largest value, 3. Now we have only $\{2,3\}$ left. Since $3^2>2^3$, the maximum value of the expression is $2^0\times3^2=9$.

A rectangle, with sides parallel to the $x$-axis and $y$-axis, has opposite vertices located at $(15, 3)$ and $(16, 5)$. A line drawn through points $A(0, 0)$ and $B(3, 1)$. Another line is drawn through points $C(0, 10)$ and $D(2, 9)$. How many points on the rectangle lie on at least one of the two lines?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

$\textbf{B}$

The line passing through point $A$ and $B$ is $y=\dfrac13x$. Line $\overline{AB}$ intersects line $x=15$ at point $(15,5)$, which is the top left vertex of the rectangle. Furthermore, line $\overline{AB}$ will not intersect any other point on the rectangle.

The line passing through point $C$ and $D$ is $y=-\dfrac12x+10$. Line $\overline{CD}$ intersects line $x=15$ at point $(15,2.5)$, which is below the bottom left vertex of the rectangle. So line $\overline{CD}$ will not intersects any point on the rectangle.

Finally, we can conclude that only 1 point on the rectangles lies on at least one of the two lines.

Lola, Lolo, Tiya, and Tiyo participated in a ping pong tournament. Each player competed against each of the other three players exactly twice. Shown below are the win-loss records for the players. The numbers $1$ and $0$ represent a win or loss, respectively. For example, Lola won five matches and lost the fourth match. What was Tiyo’s win-loss record?

$\textbf{(A)}\ \texttt{000101} \qquad \textbf{(B)}\ \texttt{001001} \qquad \textbf{(C)}\ \texttt{010000} \qquad \textbf{(D)}\ \texttt{010101} \qquad \textbf{(E)}\ \texttt{011000}$

$\textbf{A}$

For each game, where there is a win, there is a loss. Hence, the number of 1 and 0 must be the same in each column. For example, there are two 1 and one 0 in the first column of the records, so the first number of Tiyo's record is 0. Similarly, we get the whole win-loss record of Tiya 000101.

Malaika is skiing on a mountain. The graph below shows her elevation, in meters, above the base of the mountain as she skis along a trail. In total, how many seconds does she spend at an elevation between $4$ and $7$ meters?

$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 14$

$\textbf{B}$

The time intervals between $4$ and $7$ meters are marked in red. We get three intervals from $t=2\ \text{s}$ to $t=4\ \text{s}$, from $t=6\ \text{s}$ to $t=10\ \text{s}$, and from $t=12\ \text{s}$ to $t=14\ \text{s}$. So the total time is $2+4+2=8\ \text{s}$.

Harold made a plum pie to take on a picnic. He was able to eat only $\dfrac{1}{4}$ of the pie, and he left the rest for his friends. A moose came by and ate $\dfrac{1}{3}$ of what Harold left behind. After that, a porcupine ate $\dfrac{1}{3}$ of what the moose left behind. How much of the original pie still remained after the porcupine left?

$\textbf{(A)}\ \dfrac{1}{12} \qquad \textbf{(B)}\ \dfrac{1}{6} \qquad \textbf{(C)}\ \dfrac{1}{4} \qquad \textbf{(D)}\ \dfrac{1}{3} \qquad \textbf{(E)}\ \dfrac{5}{12}$

$\textbf{D}$

Harold left $\dfrac34$ of the pie. Then the moose left $\left(1-\dfrac13\right)\times\dfrac34=\dfrac12$ of the pie. After that, the porcupine left $\left(1-\dfrac13\right)\times\dfrac12=\dfrac13$ of the pie.

NASA’s Perseverance Rover was launched on July $30,$ $2020.$ After traveling $292{,}526{,}838$ miles, it landed on Mars in Jezero Crater about $6.5$ months later. Which of the following is closest to the Rover’s average interplanetary speed in miles per hour?

$\textbf{(A)}\ 6{,}000 \qquad \textbf{(B)}\ 12{,}000 \qquad \textbf{(C)}\ 60{,}000 \qquad \textbf{(D)}\ 120{,}000 \qquad \textbf{(E)}\ 600{,}000$

$\textbf{C}$

It takes the Rover $6.5\times30\times24=4680\ \text{hours}$ to travel 292,526,838 miles. So its speed is $$\dfrac{292,526,828}{4680}\approx\dfrac{300,000,000}{5000}=60,000\ \text{mph}$$

The figure below shows a large white circle with a number of smaller white and shaded circles in its interior. What fraction of the interior of the large white circle is shaded?

$\textbf{(A)}\ \dfrac{1}{4} \qquad \textbf{(B)}\ \dfrac{11}{36} \qquad \textbf{(C)}\ \dfrac{1}{3} \qquad \textbf{(D)}\ \dfrac{19}{36} \qquad \textbf{(E)}\ \dfrac{5}{9}$

$\textbf{B}$

The radius of the large white circle is 3. So the area of the large white circle is $\pi\times3^2=9\pi$. The shaded area is the area of three little circles of radius 0.5 pluses the area of a circle of radius 2, then subtracts two circles of radius 1. So the shaded area is $\pi\times0.5^2\times3+\pi\times2^2-\pi\times1^2\times2=2.75\pi$. Hence, the ratio of the shaded area to the large white circle is $\dfrac{2.75\pi}{9\pi}=\dfrac{11}{36}$.

Along the route of a bicycle race, $7$ water stations are evenly spaced between the start and finish lines, as shown in the figure below. There are also $2$ repair stations evenly spaced between the start and finish lines. The $3$rd water station is located $2$ miles after the $1$st repair station. How long is the race in miles?

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 96$

$\textbf{D}$

Let the distance of the race be $d$. The location of the 3rd water station is $\dfrac3{7+1}d=\dfrac38d$. And the location of the 1st repair station is $\dfrac1{2+1}d=\dfrac13d$. Hence, we get $$\dfrac38d-\dfrac13d=2\rightarrow d=48\ \text{miles}$$

Nicolas is planning to send a package to his friend Anton, who is a stamp collector. To pay for the postage, Nicolas would like to cover the package with a large number of stamps. Suppose he has a collection of $5$-cent, $10$-cent, and $25$-cent stamps, with exactly $20$ of each type. What is the greatest number of stamps Nicolas can use to make exactly $\$7.10$ in postage? (Note: The amount $\$7.10$ corresponds to $7$ dollars and $10$ cents. One dollar is worth $100$ cents.)

$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 46 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 54\qquad \textbf{(E)}\ 55$

$\textbf{E}$

To achieve the greatest number of stamps, we start with the 5-cent stamps. 20 5-cent stamps are worth $\$1$, which is far away from $\$7.10$. 20 10-cent stamps are worth $\$2$. The value of all 5-cent and 10-cent stamps is still $7.10-2-1=\$4.10$ away from the postage. If we add 16 25-cent stamps, we are still $4.10-16\cdot0.25=\$0.1$ away from the postage. Hence, we need to add one 25-cent stamp, and then subtract one 5-cent stamp and one 10-cent stamp to achieve the rest $\$0.1$. Finally, we need 19 5-cent, 19 10-cent, and 17 25-cent stamps to pay for the postage. The total number of stamps is $19+19+17=55$.

Viswam walks half a mile to get to school each day. His route consists of $10$ city blocks of equal length and he takes $1$ minute to walk each block. Today, after walking $5$ blocks, Viswam discovers he has to make a detour, walking $3$ blocks of equal length instead of $1$ block to reach the next corner. From the time he starts his detour, at what speed, in mph, must he walk, in order to get to school at his usual time?

$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4.2 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 4.8 \qquad \textbf{(E)}\ 5$

$\textbf{B}$

The distance of each block is 0.05 mile. From the time he starts his detour, he need to walk 7 blocks in 5 minutes. So his speed is $\dfrac{7\times0.05\ \text{mile}}{1/12\ \text{h}}=4.2\ \text{mph}$.

The letters $\text{P}, \text{Q},$ and $\text{R}$ are entered into a $20\times20$ table according to the pattern shown below. How many $\text{P}$s, $\text{Q}$s, and $\text{R}$s will appear in the completed table?

$\textbf{(A)}~132\text{ Ps, }134\text{ Qs, }134\text{ Rs}\newline$

$\textbf{(B)}~133\text{ Ps, }133\text{ Qs, }134\text{ Rs}\newline$

$\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}\newline$

$\textbf{(D)}~134\text{ Ps, }132\text{ Qs, }134\text{ Rs}\newline$

$\textbf{(E)}~134\text{ Ps, }133\text{ Qs, }133\text{ Rs}$

$\textbf{C}$

The letter of the diagonal from top left to bottom right is Q. So the number of Q is $20+2\times(17+14+11+8+5+2)=134$. By symmetry, the numbers of P and R are the same. Since there are $20\times20=400$ grids in the table, the number of P (or R) is $(400-134)/2=133$.

A regular octahedron has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the right of $Q$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

$\textbf{A}$

Face 6 is on the left side of $Q$. Face 5 is directly below face 6. Face 2,3,4,5 share the same vertex, and thus make the bottom half of the octahedron. Face 7 is on the opposite position of $Q$ on the top half of the octahedron. Hence, face 1 must be on the right side of $Q$.

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting point?

$\textbf{(A)}\ 405 \qquad \textbf{(B)}\ 407 \qquad \textbf{(C)}\ 409 \qquad \textbf{(D)}\ 411 \qquad \textbf{(E)}\ 413$

$\textbf{D}$

When Greta jumps 405 time to the right, she is on the 2025th pad, which is 2 pads away from the 2023th pad. Then she jumps one more times to the right. $2+5=7$ is still not a multiple of 3. Hence, she jumps one more time to the right. Now $2+5+5=12$ is a multiple of 3. So she can jumps 4 time to the left and finally reaches the 2023th pad. The total number of jumps is $405+2+4=411$.

An equilateral triangle is placed inside a larger equilateral triangle so that the region between them can be divided into three congruent trapezoids, as shown below. The side length of the inner triangle is $\dfrac23$ the side length of the larger triangle. What is the ratio of the area of one trapezoid to the area of the inner triangle?

$\textbf{(A) } 1 : 3 \qquad \textbf{(B) } 3 : 8 \qquad \textbf{(C) } 5 : 12 \qquad \textbf{(D) } 7 : 16 \qquad \textbf{(E) } 4 : 9$

$\textbf{C}$

Let the side length of the inner triangle be $a$. The area of the inner triangle is $$\dfrac12a^2\sin60^\circ=\dfrac{\sqrt3}{4}a^2$$ The side length of the larger triangle $\dfrac32a$. So the area of the larger triangle is $$\dfrac{\sqrt3}{4}\left(\dfrac32a\right)^2=\dfrac{9\sqrt3}{16}a^2$$ Since the three trapezoids are congruent, the area of one trapezoid is $\dfrac13$ of the difference of the areas of two triangle $$\dfrac13\left(\dfrac{9\sqrt3}{16}a^2-\dfrac{\sqrt3}{4}a^2\right)=\dfrac{5\sqrt3}{48}a^2$$Hence, the ratio of the area of one trapezoid to the area of the inner triangle is $$\dfrac{5\sqrt3}{48}a^2:\dfrac{\sqrt3}{4}a^2=5:12$$

Two integers are inserted into the list $3,3,8,11,28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of two additional numbers?

$\textbf{(A)}\ 56 \qquad \textbf{(B)}\ 57 \qquad \textbf{(C)}\ 58 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 61$

$\textbf{D}$

The range of the original list is $28-3=25$. To double the range, we can set one of the two inserted integers as 53. The median of the original list is 8. To keep the median unchanged, the second inserted integer must be no more than 8. The mode of the original list is 3. To keep the mode unchanged, the maximum value of the second integer is 7. Hence, the maximum sum of the two additional integers is $53+7=60$.

Alina writes the numbers $1, 2, \dots , 9$ on separate cards, one number per card. She wishes to divide the cards into $3$ groups of $3$ cards so that the sum of the numbers in each group will be the same. In how many ways can this be done?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

$\textbf{C}$

The sum of all cards is $1+2+\dots+9=45$. So the sum of each group is 15. Now think about the group with number 9. The rest two cards have a sum of 6. Hence, this group can be $(1,5,9)$ or $(2,4,9)$.

In the $(1,5,9)$ case, think about another group with number 8. The sum of the rest two cards is 7. Therefore, we have only $(3,4,8)$.

In the case of $(2,4,9)$, again think about the group with number 8. The only possible choice is $(1,6,8)$.

Finally, we can conclude that there are 2 ways for the answer.

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$. What is the first term?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$

$\textbf{D}$

Let the first two terms be $a$ and $b$. The sequence is $a,b,ab,ab^2,a^2b^3,a^3b^5,\dots$. The prime factorization of the sixth term is $a^3b^5=4000=5^3\times2^5$. Hence, the first term is $a=5$.

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ gray and white tiles shown below on the right.

What is the probability that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids? Below is an example of such tiling.

$\textbf{(A) } \dfrac{1}{1024} \qquad \textbf{(B) } \dfrac{1}{256} \qquad \textbf{(C) } \dfrac{1}{64} \qquad \textbf{(D) } \dfrac{1}{16} \qquad \textbf{(E) } \dfrac{1}{4}$

$\textbf{C}$

Since all 9 grids can be chose randomly from 4 patterns, we have $4^9$ choices in total. However, there are $4$ cases that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids. In each case, there are $4^5$ ways to arrange the rest 5 grids. So the probability is $\dfrac{4\cdot4^5}{4^9}=\dfrac1{4^3}=\dfrac1{64}$.

Isosceles triangle $ABC$ has equal side lengths $AB$ and $BC$. In the figures below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height $h$ of $\triangle ABC$?

$\textbf{(A)}\ 14.6 \qquad \textbf{(B)}\ 14.8 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 15.2 \qquad \textbf{(E)}\ 15.4$

$\textbf{A}$

Let the area of $\triangle ABC$ be $a$. The area of the white triangle in the left figure is $\dfrac{11^2}{h^2}a$. So the area of the shaded portion in the left figure is $\left(1-\dfrac{11^2}{h^2}\right)a$. Similarly, the area of the shaded triangle in the right figure is $\dfrac{(h-5)^2}{h^2}a$. Hence, we get $$\left(1-\dfrac{11^2}{h^2}\right)a=\dfrac{(h-5)^2}{h^2}a\rightarrow h=14.6$$

Fifteen integers $a_1, a_2, a_3, \dots, a_{15}$ are arranged in order on a number line. The integers are equally spaced and have the property that\[1 \le a_1 \le 10, 13 \le a_2 \le 20, \text{ and } 241 \le a_{15}\le 250.\]What is the sum of digits of $a_{14}?$

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$

$\textbf{A}$

Let the common difference of the arithmetic sequence be $d$. Hence, we have $$241\le a_{15}=a_1+14d\le250\rightarrow241-a_1\le14d\le250-a_1$$ Since $1\le a_1\le10$, we have $241-a_1\ge241-10=231$ and $250-a_1\le 250-1=249$. So $$231\le241-a_1\le14d\le250-a_1\le249\rightarrow16.5\le d\le17.8$$ The common difference must be an integer, so $d=17$. Using $13\le a_2\le 20$, we get $13\le a_1+17\le20$. So $$a_1\le3$$ However, by $241\le a_{15}\le250$, we get $241\le a_1+17\cdot14\le 250$. So $$a_1\ge 3$$ Now we know that $a_1=3$. Therefore, $a_{14}=a_1+13d=3+13\times17=224$. The sum of digits of $a_{14}$ is $2+2+4=8$.