AMC 8 2024
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 ones digit of\[222,222-22,222-2,222-222-22-2?\]$\textbf{(A) } 0\qquad\textbf{(B) } 2\qquad\textbf{(C) } 4\qquad\textbf{(D) } 6\qquad\textbf{(E) } 8$
$\textbf{B}$
Since we only care about the units digit, the original expression can be rewrote as $12-2-2-2-2-2=2$. Hence, the answer is B.
What is the value of this expression in decimal form?\[\frac{44}{11} + \frac{110}{44} + \frac{44}{1100}\]
$\textbf{(A) } 6.4\qquad\textbf{(B) } 6.504\qquad\textbf{(C) } 6.54\qquad\textbf{(D) } 6.9\qquad\textbf{(E) } 6.94$
$\textbf{C}$
We see that all numerators and denominators are divisible by 11. Hence, we get
\begin{align*}
\dfrac{44}{11} + \dfrac{110}{44} + \dfrac{44}{1100}&=4+\dfrac{10}{4}+\dfrac{4}{100}\\
&=4+2.5+0.04\\
&=6.54
\end{align*}
Four squares of side lengths $4$, $7$, $9$, and $10$ units are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in the color pattern white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units?
$\textbf{(A)}\ 42 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 49 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 52$
$\textbf{E}$
The area of the outer gray region is the area of the $10\times10$ square minus the area of the $9\times9$ square. The area of the inner gray region is the area of the $7\times7$ square minus the area of the $4\times4$ square. Hence, the area of all visible gray regions is\[10^2 - 9^2 + 7^2 - 4^2 = 19 + 33 = 52\]
When Yunji added all the integers from $1$ to $9$, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$
$\textbf{E}$
The sum of all integers from 1 to 9 is 45. The nearest square number less than 45 is 36. So the number she left out is $45-36=9$.
Aaliyah rolls two standard 6-sided dice. She notices that the product of the two numbers rolled is a multiple of $6$. Which of the following integers cannot be the sum of the two numbers?
$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$
$\textbf{B}$
$5$ is possible: $5=2+3$, $2\times 3$ is a multiple of 6.
$7$ is possible: $7=1+6$, $1\times 6$ is a multiple of 6.
$8$ is possible: $8=2+6$, $2\times 6$ is a multiple of 6.
$9$ is possible: $9=3+6$, $3\times 6$ is a multiple of 6.
The only integer that cannot be the sum is 6.
Sergei skated around an ice rink, gliding along different paths. The gray lines in the figures below show four of the paths labeled $P$, $Q$, $R$, and $S.$ What is the sorted order of the four paths from shortest to longest?
$\textbf{(A)}\ P,Q,R,S \qquad \textbf{(B)}\ P,R,S,Q \qquad \textbf{(C)}\ Q,S,P,R \qquad \textbf{(D)}\ R,P,S,Q \qquad \textbf{(E)}\ R,S,P,Q$
$\textbf{D}$
Notice that curved lines are always longer than the straight one that meets their endpoints. So Path P is longer than Path R. Furthermore, Path S is longer than Path P, and Path Q is longer than Path S. So the answer is D.
A $3\times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2\times 2$, $1\times 4$, and $1\times 1$, shown below. What is the minimum possible number of $1\times 1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$
$\textbf{E}$
Suppose that we need $x$ $2\times 2$ tiles, $y$ $1\times 4$ tiles and $z$ $1\times 1$ tiles to cover the $3\times 7$ rectangle without overlap. Then we get $$4x+4y+z=21$$ We see that both $4x$ and $4y$ are multiples of 4, and $21=4\times5+1$. Hence, the expression can be rewrote as $$z=4(5-x-y)+1$$ If $5-x-y=0$, we have the minimum possible number $z=1$. However, after some testing, there is no valid pair $(x,y)$ that works. So the case $5-x-y=0$ is invalid.
Now we try $5-x-y=1$, and we find it works when $x=3$ and $y=1$. So the answer is $z=4\times1+1=5$.
On Monday Taye has $\$2$. Every day, he either gains $\$3$ or doubles the amount of money he had on the previous day. How many different dollar amounts could Taye have on Thursday, 3 days later?
$\textbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7$
$\textbf{D}$
On Monday he has $\$2$. On Tuesday he might have $\$4$ or $\$5$. On Wednesday he might have $\$7$, $\$8$, or $\$10$ (the case of $\$8$ happened twice). On Thursday he might have $\$10$, $\$11$, $\$13$, $\$14$, $\$16$ or $\$20$. The answer is D.
All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28$
$\textbf{E}$
Let the number of green marbles be $x$. Then the number of red marbles should be $\dfrac12x$, and the number of blue marbles be $2x$. The total number of marbles in Maria's collection is $x+\dfrac12x+2x=\dfrac72x$. Hence, we know the answer must be a multiple of 7. The only choice that is a multiple of 7 is E.
In January 1980 the Moana Loa Observation recorded carbon dioxide $(CO_2)$ levels of 338 ppm (parts per million). Over the years the average $CO_2$ reading has increased by about 1.515 ppm each year. What is the expected $CO_2$ level in ppm in January 2030? Round your answer to the nearest integer.
$\textbf{(A)}\ 399 \qquad \textbf{(B)}\ 414 \qquad \textbf{(C)}\ 420 \qquad \textbf{(D)}\ 444 \qquad \textbf{(E)}\ 459$
$\textbf{B}$
After 50 years, the $CO_2$ level will increase by $1.515\times 50=75.75\approx 76$ ppm. So the $CO_2$ level in January 2030 will be $76 + 338 = 414$ ppm.
The coordinates of $\triangle ABC$ are $A(5,7)$, $B(11,7)$, and $C(3,y)$, with $y>7$. The area of $\triangle ABC$ is 12. What is the value of $y$?
$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$
$\textbf{D}$
The triangle has base $11-5=6$, and height $y-7$. So the area of the triangle is $$\dfrac12\times6(y-7)=12\rightarrow y=11$$
Rohan keeps a total of $90$ guppies in $4$ fish tanks.
$\bullet$ There is $1$ more guppy in the $2$nd tank than the $1$st tank.$\newline$
$\bullet$ There are $2$ more guppies in the 3rd tank than the $2$nd tank.$\newline$
$\bullet$ There are $3$ more guppies in the 4th tank than the $3$rd tank.
How many guppies are in the $4$th tank?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$
$\textbf{E}$
Let the number of guppies in the $1$st tank be $x$. Then the numbers of guppies in the $2$nd, $3$rd and $4$th tanks are $x+1$, $x+3$ and $x+6$, respectively. The total number of guppies is $x+(x+1)+(x+3)+(x+6)=4x+10=90$. Hence, we get $x=20$. So the number of guppies in the $4$th tank is $x+6=26$.
Buzz Bunny is hopping up and down a set of stairs, one step at a time. In how many ways can Buzz start on the ground, make a sequence of $6$ hops, and end up back on the ground? (For example, one sequence of hops is up-up-down-down-up-down.)
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 12$
$\textbf{B}$
Any sequence can be written as some re-arrangement of $UUUDDD$. Clearly we must start going up, and end going down. Noticing that Buzz can not go down when he is on the ground, we can list all valid sequences: $UUUDDD$, $UUDUDD$, $UUDDUD$, $UDUUDD$, $UDUDUD$. Thus, the answer is B.
The one-way routes connecting towns $A,M,C,X,Y,$ and $Z$ are shown in the figure below (not drawn to scale). The distances in kilometers along each route are marked. Traveling along these routes, what is the shortest distance from A to Z in kilometers?
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 30 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 32$
$\textbf{A}$
Starting from A, the first step should be $A\rightarrow M$ or $A \rightarrow X$. $A\rightarrow X$ is always a better choice because if we really want to get to town M, we can travel $5+2=7$ kilometers along $A\rightarrow X\rightarrow M$ rather than come to M directly in 8 kilometers. So the first step should be $A\rightarrow X$.
Now starting from X, the next step should be $X\rightarrow M$ or $X\rightarrow Y$. Again, $X\rightarrow M$ is always a better choice because if we really want to get to town Y, we can travel $2+6=8$ kilometers along $X\rightarrow M\rightarrow Y$ rather than come to Y directly in 10 kilometers. So the second step should be $X\rightarrow M$.
There are several ways from M to Z, including $M\rightarrow Z$, $M\rightarrow C\rightarrow Z$, $M\rightarrow Y\rightarrow Z$ and $M\rightarrow Y\rightarrow C\rightarrow Z$. Comparing all the routes, $M\rightarrow Y\rightarrow C\rightarrow Z$ is the shortest.
So the shortest route from A to Z is $A\rightarrow X\rightarrow M\rightarrow Y\rightarrow C\rightarrow Z$. The answer is $5+2+6+5+10=28$ kilometers.
Let the letters $F$,$L$,$Y$,$B$,$U$,$G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation \[8\times\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\] What is the value of $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$?
$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$
$\textbf{C}$
We see that $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=1001\times \underline{F}~\underline{L}~\underline{Y}$ and $\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}=1001\times \underline{B}~\underline{U}~\underline{G}$. So the original expression equals to $$8\times 1001\times \underline{F}~\underline{L}~\underline{Y}= 1001\times \underline{B}~\underline{U}~\underline{G}$$ which means $$8\times\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}$$ We know that $8\times 125=1000$. So we try $\underline{F}~\underline{L}~\underline{Y} =124$ first. Then we have $8\times124=992$, where the digits 2 and 9 repeat.
Trying $\underline{F}~\underline{L}~\underline{Y} =123$, we get $8\times123=984$. Now letters $F$,$L$,$Y$,$B$,$U$,$G$ represent distinct digits. So the answer is $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}=123+984=1107$.
Minh enters the numbers $1$ through $81$ into the cells of a $9 \times 9$ grid in some order. She calculates the product of the numbers in each row and column. What is the least number of rows and columns that could have a product divisible by $3$?
$\textbf{(A) } 8\qquad\textbf{(B) } 9\qquad\textbf{(C) } 10\qquad\textbf{(D) } 11\qquad\textbf{(E) } 12$
$\textbf{D}$
For a row or column to have a product divisible by $3$, there must be at least one multiple of $3$ in the row or column. From $1$ to $81$, there are $81/3=27$ multiples of $3$. To create the least amount of rows and columns with multiples of $3$, we must find a way to keep them all together. If we put $25$ of these numbers in a $5\times5$ grid, there would be $5$ rows and $5$ columns with products divisible by $3$. However, we have $27$ numbers, so $2$ numbers remain to put in the $9\times9$ grid. If we put both numbers in the $6$th column, but one in the first row, and one in the second row (next to the $5\times5$ already filled), we would have a total of $6$ columns now, and still $5$ rows with products that are multiples of $3$. So the answer is $5+6=11$.
A chess king is said to attack all squares one step away from it, horizontally, vertically, or diagonally. For instance, a king on the center square of a $3\times3$ grid attacks all 8 other squares, as shown below. Suppose a white king and a black king are placed on different squares of $3\times3$ grid so that they do not attack each other. In how many ways can this be done?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 27 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 32$
$\textbf{E}$
There are 4 corner squares in the grid, where a king on each corner square has 5 unreachable squares. There are 4 edge squares in the grid, where a king on each edge square has 3 unreachable squares. So we have $4\times5+4\times3=32$ ways to arrange the two kings.
Three concentric circles centered at $O$ have radii of $1$, $2$, and $3$. Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$, as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?
$\textbf{(A)}\ 108 \qquad \textbf{(B)}\ 120 \qquad \textbf{(C)}\ 135 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 150$
$\textbf{A}$
Let $\angle{BOC}=x^\circ$. The area of the shaded region between the two smaller circles is $\pi\times2^2-\pi\times1^2=3\pi$. The area of the shaded region between the two larger circles is $\dfrac{x}{360}$ of the area between the two larger circle, which is $\pi\times3^2-\pi\times2^2=5\pi$. Hence, the total area of shade regions is $3\pi+\dfrac{x}{360}\cdot5\pi$.
Since the shaded and unshaded regions are equal in area, and the sum of them is the area of the largest circle, so the area of the shaded region is half the area of the largest circle. Thus, we have $$3\pi+\dfrac{x}{360}\cdot5\pi=\dfrac12\times\pi\times3^2$$ Finally, we get $x=108$.
Jordan owns 15 pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction?
$\textbf{(A) } 0\qquad\textbf{(B) } \dfrac{1}{5} \qquad\textbf{(C) } \dfrac{4}{15} \qquad\textbf{(D) } \dfrac{1}{3} \qquad\textbf{(E) } \dfrac{2}{5}$
$\textbf{C}$
Jordan has $15\times\dfrac35=9$ pairs of red sneakers, and $15-9=6$ pairs of white sneakers. For the 15 pairs of sneakers, $15\times\dfrac23=10$ of them are high-top. We would want as many white high-top sneakers as possible, so we set $6$ high-top sneakers to be white. Then, we have $10-6=4$ red high-top sneakers. So the answer is $\dfrac{4}{15}$.
Any three vertices of the cube $PQRSTUVW,$ shown in the figure below, can be connected to form a triangle. (For example, vertices $P, Q,$ and $R$ can be connected to form $\triangle{PQR}$.) How many of these triangles are equilateral and contain $P$ as a vertex?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }6$
$\textbf{D}$
The points closet to $P$ are $Q$, $S$ and $W$. If we choose any two of them as the other two vertices to form a triangle, for example, $\triangle PQS$. The triangle is isosceles, but not equilateral.
The second closet points to $P$ are $R$, $T$ and $V$. If we choose any two of them as the other two vertices to form a triangle, for example, $\triangle PRT$. The triangle is equilateral. We have $\dbinom32=3$ ways to choose two of them. So the answer is D.
A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was $3 : 1$. Then $3$ green frogs moved to the sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4 : 1$. What is the difference between the number of green frogs and the number of yellow frogs now?
$\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24$
$\textbf{E}$
Let the number of yellow frogs in the beginning be $x$, and green frogs be $3x$. After moving position, there are $3x+2$ green frogs and $x-2$ yellow frogs now. The ratio is $$\dfrac{3x+2}{x-2}=4$$ Hence, we get $x=10$. The difference between the number of green frogs and the number of yellow frogs now is $(3x+2)-(x-2)=2x+4=24$.
A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.
$\textbf{(A) } 300\qquad\textbf{(B)} 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800$
$\textbf{B}$
The area of the cross section of the tape is $\pi\times2^2-\pi\times1^2=3\pi$ square inches. When unfolded, the cross section of the tape is a long and thin rectangle with width 0.015 inches. So the length of the rectangle is $\dfrac{3\pi}{0.015}=200\pi\approx600$ inches.
$\textbf{C}$
The segment connecting point $(2000,3000)$ to point $(5000,8000)$ will pass the same number of cells as the segment connecting point $(0,0)$ to point $(3000,5000)$. Since the greatest common divisor of 3000 and 5000 is 1000, the segment connecting point $(0,0)$ to point $(3000,5000)$ can be divided into 1000 segments connecting point $(0,0)$ to point $(3,5)$, where each segment looks like this:
Each segment will pass 7 cells. So the total number of cells the whole segment will pass is $7000$.
$\textbf{B}$
The area of the artwork is the area of the left triangle pluses the area of the right triangle, then subtracts the area of their intersection, which is also a triangle.
The length of the right-angle side of the left triangle is $8\sqrt2$. So the area of the left triangle is $\dfrac12\times8\sqrt2\times8\sqrt2=64$. Similarly, the length of the right-angle side of the right triangle is $12\sqrt2$. So the area of the right triangle is $\dfrac12\time12\sqrt2\times12\sqrt2=144$. The length of right-angle side of the intersection triangle is $\sqrt2h$. So the area of the intersection triangle is $\dfrac12\times\sqrt2h\times\sqrt2h=h^2$. Thus, the total area of the artwork is $$64+144-h^2=183$$ Hence, we get $h=5$.
A small airplane has $4$ rows of seats with $3$ seats in each row. Eight passengers have boarded the plane and are distributed randomly among the seats. A married couple is next to board. What is the probability there will be $2$ adjacent seats in the same row for the couple?
$\textbf{(A)}\ \dfrac{8}{15} \qquad \textbf{(B)}\ \dfrac{32}{55} \qquad \textbf{(C)}\ \dfrac{20}{33} \qquad \textbf{(D)}\ \dfrac{34}{55} \qquad \textbf{(E)}\ \dfrac{8}{11}$
$\textbf{C}$
Suppose the eight passengers who have boarded the plane are indistinguishable. We have 4 seat left for the couple now. Now the question is, if the 4 empty seats are placed, what is the probability that there are 2 adjacent seats open.
We have $\dbinom{12}{4}=495$ ways to choose the 4 empty seats from $4\times3=12$ seats in total. The 4 empty seats can be arranged in 4 rows like $(3,1,0,0)$, $(2,2,0,0)$, $(2,1,1,0)$ or $(1,1,1,1)$. We will discuss the number of ways there are 2 adjacent seats open in each case.
Case 1: $(3,1,0,0)$. There are 4 ways to choose which row has 3 empty seats, then 3 ways to choose which row has 1 empty seat, and 3 ways to choose the position of the one seat (left, middle or the right). Hence, we have $4\times3\times3=36$ arrangements in this case. Since there are always 2 adjacent empty seats in the row of 3 empty seats, all 36 arrangements are valid.
Case 2: $(2,2,0,0)$. There might be 2 pairs of adjacent empty seats in this case, or just 1 pair of adjacent empty seats. We will discuss them separately.
Case 2.1: 2 pairs of adjacent empty seats in $(2,2,0,0)$. We have $\dbinom42=6$ ways to choose which 2 rows have empty seats. To make the 2 empty seats in each row to be adjacent, we have 2 arrangements in each row (the middle and the left, or the middle and the right). Hence, we have $6\times2\times2=24$ arrangements in total.
Case 2.2: only 1 pair of adjacent empty seats in $(2,2,0,0)$. We have 4 ways to choose which row has 2 adjacent empty seats, then 3 ways to choose which row has 2 empty seats but not adjacent. For the row of 2 adjacent empty seats, we have 2 arrangements (the middle and the left, or the middle and the right). For the row of 2 empty seats but not adjacent, we have only 1 arrangement (the left and the right). So in total we have $4\times3\times2\times1=24$ arrangements.
Case 3: $(2,1,1,0)$. There are 4 ways to choose which row has 2 empty seats, and $\dbinom32=3$ ways to choose which 2 rows have 1 empty seat each. For the row of 2 empty seats, the make the seats adjacent, we have 2 arrangements (the middle and the left, or the middle and the right). For the row of 1 empty seat, we have 3 arrangements (the left, the middle or the right). So we have $4\times3\times2\times3\times3=216$ arrangements in this case.
Case 4: $(1,1,1,1)$. In the case, we can not get 2 adjacent empty seats in any row.
In total, we have $36+24+24+216+0=300$ arrangements to have 2 adjacent empty seats. So the probability is $\dfrac{300}{495}=\dfrac{20}{33}$.