AMC 8 2017

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.

Question 1

Which of the following values is the largest?

$\textbf{(A) }2+0+1+7\qquad\textbf{(B) }2 \times 0 +1+7\qquad\textbf{(C) }2+0 \times 1 + 7\qquad\textbf{(D) }2+0+1 \times 7\qquad\textbf{(E) }2 \times 0 \times 1 \times 7$

Solution

$\textbf{A}$

(A) $2 + 0 + 1 + 7 = 10\newline$
(B) $2 \times 0 + 1 + 7 = 8\newline$
(C) $2 + 0 \times 1 + 7 = 9\newline$
(D) $2 + 0 + 1 \times 7 = 9\newline$
(E) $2 \times 0 \times 1 \times 7 = 0$

Question 2

Alicia, Brenda, and Colby were the candidates in a recent election for student president. The pie chart below shows how the votes were distributed among the three candidates. If Brenda received $36$ votes, then how many votes were cast all together?

$\textbf{(A) }70\qquad\textbf{(B) }84\qquad\textbf{(C) }100\qquad\textbf{(D) }106\qquad\textbf{(E) }120$

Solution

$\textbf{E}$

Brenda received $36$ votes, which takes $30\%$ of the total votes. So the total number of votes is $36\div30\%=120$.

Question 3

What is the value of the expression $\sqrt{16\sqrt{8\sqrt{4}}}$?

$\textbf{(A) }4\qquad\textbf{(B) }4\sqrt{2}\qquad\textbf{(C) }8\qquad\textbf{(D) }8\sqrt{2}\qquad\textbf{(E) }16$

Solution

$\textbf{C}$

$$\sqrt{16\sqrt{8\sqrt{4}}}= \sqrt{16\sqrt{8\times 2}} = \sqrt{16\sqrt{16}}= \sqrt{16\times 4}= \sqrt{64} = 8$$

Question 4

hen $0.000315$ is multiplied by $7,928,564$ the product is closest to which of the following?

$\textbf{(A) }210\qquad\textbf{(B) }240\qquad\textbf{(C) }2100\qquad\textbf{(D) }2400\qquad\textbf{(E) }24000$

Solution

$\textbf{D}$

We can approximate $7,928,564$ to $8,000,000$, and $0.000315$ to $0.0003.$ Multiplying the two yields $2400.$ So the product is closet to $2400$.

Question 5

What is the value of the expression $$\dfrac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8}{1+2+3+4+5+6+7+8}?$$

$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$

Solution

$\textbf{B}$

The denominator $1+2+3+4+5+6+7+8=36=2\times3\times6$, which means the 2, 3, and 6 in the numerator can be canceled. The remaining numerator is $1\times4\times5\times7\times8=1120$.

Question 6

If the degree measures of the angles of a triangle are in the ratio $3:3:4$, what is the degree measure of the largest angle of the triangle?

$\textbf{(A) }18\qquad\textbf{(B) }36\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }90$

Solution

$\textbf{D}$

Since the sum of the three angles of a triangle is $180^\circ$, the largest angle, which takes $\dfrac{4}{3+3+4}=40\%$ of the $180^\circ$, is $180^\circ\times40\%=72^\circ$.

Question 7

Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$?

$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$

Solution

$\textbf{A}$

By observation, $1001$ is a factor of such kind of integer. The prime factorization of $1001=7\times11\times13$. So $11$ is a factor of $Z$.

Question 8

Malcolm wants to visit Isabella after school today and knows the street where she lives but doesn't know her house number. She tells him, "My house number has two digits, and exactly three of the following four statements about it are true."

(1) It is prime.$\newline$
(2) It is even.$\newline$
(3) It is divisible by 7.$\newline$
(4) One of its digits is 9.

This information allows Malcolm to determine Isabella's house number. What is its units digit?

$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Solution

$\textbf{D}$

Noticing that condition 1 is contradictory with condition 2 and 3. Hence, condition 1 is false. The rest three are true. Combining condition 2 and 4, 9 has to be on the tens digit. Among 90, 92, 94, 96, and 98, only 98 is divisible by 7. So the units digit is 8.

Question 9

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles?

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

Solution

$\textbf{D}$

Let $x$ be the number of all marbles. Then the number of yellow marbles is $$x-\dfrac13x-\dfrac14x-6=\dfrac{5}{12}x-6$$Since the number of marbles is a positive integer, $x$ has to be a multiple of 12. When $x=12$, the number of yellow marbles is $\dfrac{5}{12}x-6=-1$, which is impossible. When $x=24$, the number of yellow marbles is $\dfrac{5}{12}x-6=4$, which is the smallest number it could be.

Question 10

A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?

$\textbf{(A) }\dfrac{1}{10}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{3}{10}\qquad\textbf{(D) }\dfrac{2}{5}\qquad\textbf{(E) }\dfrac{1}{2}$

Solution

$\textbf{C}$

There are $\dbinom{5}{3}=10$ ways to choose three cards from five. When 4 is the largest value selected, we can only choose the three cards from 1, 2, 3, and 4, which has $\dbinom{4}{3}=4$ ways. Noticing the situation that 1, 2, and 3 are selected fails to meet the condition, we have only $4-1=3$ ways to make 4 the largest. So the probability is $\dfrac{3}{10}$.

Question 11

A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?

$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$

Solution

$\textbf{C}$

Noticing that the two diagonals share the center tile, each diagonal contains $(37+1)/2=19$ tiles. The number of tiles to cover the floor is $19^2=361$.

Question 12

The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?

$\textbf{(A) }2\textrm{ and }19\qquad\textbf{(B) }20\textrm{ and }39\qquad\textbf{(C) }40\textrm{ and }59\qquad\textbf{(D) }60\textrm{ and }79\qquad\textbf{(E) }80\textrm{ and }124$

Solution

$\textbf{D}$

The lowest common multiple of 4, 5, and 6 is 60. Since the reminder is the same for all three divisor, the integer is $60+1=61$, which lies between 60 and 79.

Question 13

Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?

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

Solution

$\textbf{B}$

When someone wins a game, his component must lose a game. So the total number of games won must be equal to the total number of games lost. Since the total number of lost is $2+3+3=8$, Kyler won $8-4-3=1$.

Question 14

Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?

$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$

Solution

$\textbf{C}$

Let the correct rate of the half of the problems they solved together be $x$. By the overall correct of Chloe, we get $$\dfrac{80\%+x}{2}=88\%\Rightarrow x=96\%$$ Since Zoe had correct answers to $90\%$ of the problems she solved alone, the overall correct rate of Zoe is $\dfrac{90\%+96\%}2=93\%$.

Question 15

In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from one letter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.

$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }12\qquad\textbf{(D) }24\qquad\textbf{(E) }36$

Solution

$\textbf{D}$

From $A$ to $M$ we have 4 choices. Then from $M$ to $C$ we have 3 choice. Similarly, from $C$ to $8$ we have 2 choices. So the number of total path is $4\times3\times2=24$.

Question 16

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$?

$\textbf{(A) }\dfrac{3}{4}\qquad\textbf{(B) }\dfrac{3}{2}\qquad\textbf{(C) }2\qquad\textbf{(D) }\dfrac{12}{5}\qquad\textbf{(E) }\dfrac{5}{2}$

Solution

$\textbf{D}$

The equal perimeter can be wrote as $$\overline{AC}+\overline{CD}+\overline{AD}=\overline{AB}+\overline{BD}+\overline{AD}\Rightarrow\overline{CD}-\overline{BD}=1$$ Since $\overline{BD}+\overline{CD}=\overline{BC}=5$, we get $\overline{BD}=2$. For $\triangle ABD$, the base $\overline{AB}=4$, the distance from $D$ to $AB$ is $\dfrac25$ of the length of $AC$. So the height of $\triangle ABD$ is $3\cdot\dfrac25=\dfrac65$. The area of $\triangle ABD$ is $\dfrac12\cdot4\cdot\dfrac65=\dfrac{12}{5}$.

Question 17

Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?

$\textbf{(A) }9\qquad\textbf{(B) }27\qquad\textbf{(C) }45\qquad\textbf{(D) }63\qquad\textbf{(E) }81$

Solution

$\textbf{C}$

Let $x$ be the number of chests. In the first attempt, we know the number of gold coins is $9(x-2)$. In the second attempt, the number of gold coins is $6x+3$. So $$9(x-2)=6x+3\Rightarrow x=7$$ Hence, the number of gold coins is $6x+3=45$.

Question 18

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=12$, $BC=4$, $CD=3$, and $AD=13$.

What is the area of quadrilateral $ABCD$?

$\textbf{(A) }12\qquad\textbf{(B) }24\qquad\textbf{(C) }26\qquad\textbf{(D) }30\qquad\textbf{(E) }36$

Solution

$\textbf{B}$

Since $\angle BCD$ is a right angle, $\overline{BD}=\sqrt{4^2+3^2}=5$. Noticing that the length of sides of $\triangle ABD$, 5, 12, 13, satisfies $5^2+12^2=13^2$. We realize that $\triangle ABD$ is also a right triangle. The area of the quadrilateral $ABCD$ is the area of $\triangle ABD$ subtracts the area of $\triangle BCD$, which is $$\dfrac12AB\cdot BD-\dfrac12BC\cdot CD=\dfrac12\times12\times5-\dfrac12\times4\times3=24$$

Question 19

For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?

$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Solution

$\textbf{D}$

We notice that $$98!+99!+100!=98!(1+99+99\times100)=98!\times10000$$ Thinking about $98!$, the multiples of 5 (like $5, 10, \ldots, 95$) contribute $95/5=19$ factors of 5, while the multiples of 25 (25, 50, and 75) contribute 1 more factor 5 than others. So $98!$ contains $19+3=22$ factors of 5. Now think about $10000=10^4=2^4\times5^4$. So 10,000 contains 4 factors of 5. In total, $n=22+4=26$.

Question 20

An integer between $1000$ and $9999$, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

$\textbf{(A) }\dfrac{14}{75}\qquad\textbf{(B) }\dfrac{56}{225}\qquad\textbf{(C) }\dfrac{107}{400}\qquad\textbf{(D) }\dfrac{7}{25}\qquad\textbf{(E) }\dfrac{9}{25}$

Solution

$\textbf{B}$

There are $5$ options for the last digit, as the integer must be odd. The first digit now has $8$ options left (it can't be $0$ or the same as the last digit). The second digit also has $8$ options left (it can't be the same as the first or last digit). Finally, the third digit has $7$ options (it can't be the same as the three digits that are already chosen). Since there are $9000$ total integers, the probability is\[\dfrac{8 \times 8 \times 7 \times 5}{9000} =\dfrac{56}{225}\]

Question 21

Suppose $a$, $b$, and $c$ are nonzero real numbers, and $a+b+c=0$. What are the possible value(s) for $\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|}$?

$\textbf{(A) }0\qquad\textbf{(B) }1\text{ and }-1\qquad\textbf{(C) }2\text{ and }-2\qquad\textbf{(D) }0,2,\text{ and }-2\qquad\textbf{(E) }0,1,\text{ and }-1$

Solution

$\textbf{A}$

Since $a+b+c=0$, and none of $a, b, c$ is zero, we have two cases. In case 1, two of them are positive and the other one is negative, we have $$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|}=1+1-1-1=0$$ In case 2, two of the them are negative and the other one is positive, we have $$\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|}=-1-1+1+1=0$$ So the only possible answer is 0.

Question 22

In the right triangle $ABC$, $AC=12$, $BC=5$, and angle $C$ is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

$\textbf{(A) }\dfrac{7}{6}\qquad\textbf{(B) }\dfrac{13}{5}\qquad\textbf{(C) }\dfrac{59}{18}\qquad\textbf{(D) }\dfrac{10}{3}\qquad\textbf{(E) }\dfrac{60}{13}$

Solution

$\textbf{D}$

Let $O$ be the center of the semicircle, and $r$ be the radius. Since $OD\perp BD$, $OC\perp BC$, and $OD=OC$, we know $OB$ bisects $\angle ABC$. Hence, we get $\triangle OBD\cong\triangle OBC$, $BD=BC=5$. In the right triangle $AOD$, $OA=12-r$, $OD=r$, $AD=AB-BD=13-5=8$. By the Pythagorean Theorem, $(12-r)^2=r^2+8^2\Rightarrow r=\dfrac{10}{3}$.

Question 23

Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 5 minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?

$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$

Solution

$\textbf{C}$

Let the time in minutes to travel one mile in the first day is $x$. Then the times to travel one mile in the next three days are $x+5$, $x+10$, and $x+15$. Given that Linda traveled for one hour (60 minutes) per day, and for each day her distance traveled was an integer number of miles, 60 had to be divisible by $x$, $x+5$, $x+10$, and $x+15$. After trying all the factors of 60, we get $x=5$.

Question 24

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$

Solution

$\textbf{D}$

We use Principle of Inclusion-Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is\[\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor\] To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days, which is\[\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor\] We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \dfrac{365}{60} \right \rfloor.$ Therefore, our answer is \begin{align*}365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor &= 365 - 285+72 - 6\\ &=146\end{align*}

Question 25

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $\angle TUS = 60^\circ$. Arcs $\stackrel{\huge\frown}{TR}$ and $\stackrel{\huge\frown}{SR}$ are each one-sixth of a circle with radius 2. What is the area of the region shown?

$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\dfrac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\dfrac{2\pi}{3}\qquad\textbf{(E) }4+\dfrac{4\pi}{3}$

Solution

$\textbf{B}$

Let the centers of the circles containing arcs $\stackrel{\huge\frown}{SR}$ and $\stackrel{\huge\frown}{TR}$ be $X$ and $Y$, respectively. Since arcs $\stackrel{\huge\frown}{SR}$ and $\stackrel{\huge\frown}{TR}$ are both 1/6 of a circle, we get $\angle SXR=\angle TYR=60^\circ$. Considering that $\angle TUS=60^\circ$, we get $\angle SXR+\angle TYR+\angle TUS=180^\circ$. Hence, $UXY$ is a triangle. The required area is the area of $\triangle UXY$ subtracts the area of two sectors of the circles, which is $$\dfrac12 UX\cdot UY\sin\angle XUT-2\times\dfrac16\pi r^2=\dfrac12\times4\times4\times\sin60^\circ-2\times\dfrac16\times\pi\times(2)^2=4\sqrt3-\dfrac43\pi$$