2017 AMC 8 — Problems & Online Timed Exam

Timed

40:00

1.

Which of the following values is largest?

$2+0+1+7$

$2 \times 0 +1+7$

$2+0 \times 1 + 7$

$2+0+1 \times 7$

$2 \times 0 \times 1 \times 7$

Answer: A

Concepts:order of operations

Difficulty rating: 370

Small Hint:

Evaluate each expression using the usual order of operations.

Big Hint:

Any expression that multiplies by $0$ will lose part or all of its value.

Video solution:

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Written solution:

Option $(A)$ evaluates to $10.$

Option $(B)$ evaluates to $0 + 1 + 7 = 8.$

Option $(C)$ evaluates to $2 + 0 + 7 = 9.$

Option $(D)$ evaluates to $2 + 0 + 7 = 9.$

Option $(E)$ evaluates to $0.$

Thus, A is the correct answer.

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?

$70$

$84$

$100$

$106$

$120$

Answer: E

Concepts:percentage data and graph interpretation

Difficulty rating: 450

Small Hint:

Brenda’s slice is $30\%$ of the pie chart.

Big Hint:

If $30\%$ corresponds to $36$ votes, find $10\%$ first.

Video solution:

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Written solution:

If $36$ votes is $30\%$ of the total votes, then $10\%$ of the total votes is $12$ votes. The number of total votes would then be $10 \cdot 12 = 120.$

Thus, E is the correct answer.

3.

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

$4$

$4\sqrt{2}$

$8$

$8\sqrt{2}$

$16$

Answer: C

Concepts:radical

Difficulty rating: 560

Small Hint:

Start with the innermost square root.

Big Hint:

After simplifying $\sqrt{4}$ , the middle radical becomes easier.

Video solution:

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Written solution:

This expression can be reduced as follows: $\begin{align*} \sqrt{16\sqrt{8\sqrt{4}}} &= \sqrt{16\sqrt{16}} \\ &= \sqrt{64}\\ &= 8 \end{align*}$

Thus, C is the correct answer.

4.

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

$210$

$240$

$2{,}100$

$2{,}400$

$24{,}000$

Answer: D

Concepts:estimation

Difficulty rating: 720

Small Hint:

Round each factor to one significant digit.

Big Hint:

Think of $0.000315$ as about $3\cdot10^{-4}$ and $7{,}928{,}564$ as about $8\cdot10^6$ .

Video solution:

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Written solution:

We can approximate the product as $\begin{align*} (3 \cdot 10^{-4})(8 \cdot 10^6) &= 24 \cdot 10^2 \\ &= 2{,}400. \end{align*}$

Thus, D is the correct answer.

5.

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

$1020$

$1120$

$1220$

$2240$

$3360$

Answer: B

Concepts:factorial

Difficulty rating: 770

Small Hint:

First add the denominator.

Big Hint:

After the denominator is known, cancel factors from the numerator instead of multiplying everything out.

Video solution:

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Written solution:

The denominator is $1+2+\cdots+8=36=2\cdot3\cdot6.$ Canceling those three factors from the numerator leaves $4\cdot5\cdot7\cdot8=1120.$

Thus, B is the correct answer.

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?

$18$

$36$

$60$

$72$

$90$

Answer: D

Concepts:angle sum ratio and proportion

Difficulty rating: 770

Small Hint:

Let the three angle measures be $3x,$ $3x,$ and $4x$ .

Big Hint:

The angles of a triangle add to $180^\circ$ .

Video solution:

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Written solution:

We can let the three angles be equal to $3x,$ $3x,$ and $4x.$ Then we know that their sum equals $180.$ From this we can set $3x + 3x + 4x = 180,$ and solving this, we get $10x = 180$ and $x = 18.$

The largest angle is $4x = 72.$

Thus, D is the correct answer.

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 be a factor of $Z?$

$11$

$19$

$101$

$111$

$1111$

Answer: A

Concepts:prime factorization place value

Difficulty rating: 940

Small Hint:

Let $n$ be the three-digit block that is repeated. Express $Z$ in terms of $n$ .

Big Hint:

Factor $1001$ .

Video solution:

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Written solution:

Let $n$ be the three-digit number formed by either repeated block. Then $\begin{aligned} Z&=1000n+n\\ &=1001n\\ &=7\cdot11\cdot13\cdot n. \end{aligned}$ Therefore, $11$ must be a factor of $Z.$

Thus, A is the correct answer.

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.

$(2)$ It is even.

$(3)$ It is divisible by $7.$

$(4)$ One of its digits is $9.$

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

$4$

$6$

$7$

$8$

$9$

Answer: D

Concepts:logical deduction divisibility

Difficulty rating: 1240

Small Hint:

Since exactly three statements are true, only one statement can fail.

Big Hint:

A two-digit number cannot be both prime and an even multiple greater than $2$ , and it cannot be both prime and divisible by $7$ .

Video solution:

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Written solution:

Statements $(1)$ and $(2)$ cannot both be true, because the only even prime is not two-digit. Statements $(1)$ and $(3)$ also cannot both be true, because a two-digit number divisible by $7$ would not be prime. Since only one statement is false, statement $(1)$ must be false, while statements $(2),$ $(3),$ and $(4)$ are true.

The house number is divisible by $2$ and $7,$ so it is divisible by $14.$ Among the two-digit multiples of $14,$ the only one with a digit of $9$ is $98.$ Therefore, the units digit is $8.$

Thus, D is the correct answer.

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 that Marcy could have?

$1$

$2$

$3$

$4$

$5$

Answer: D

Concepts:fraction least common multiple

Difficulty rating: 1070

Small Hint:

The total number of marbles must be divisible by both $3$ and $4$ .

Big Hint:

Test the smallest multiples of $12$ until there is room for the $6$ green marbles.

Video solution:

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Written solution:

If the number of marbles is divisible by both $3$ and $4,$ then the number must be divisible by $12.$ If we test $12,$ we get that there are $4$ blue marbles and $3$ red marbles. This leaves a maximum of $5$ green marbles, which is not possible.

If there are $24$ marbles, then there are $8$ blue marbles and $6$ red marbles. To find the number of yellow marbles, we get $24 - 8 - 6 - 6 = 4.$

Thus, D is the correct answer.

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?

$\dfrac{1}{10}$

$\dfrac{1}{5}$

$\dfrac{3}{10}$

$\dfrac{2}{5}$

$\dfrac{1}{2}$

Answer: C

Concepts:basic probability combinations

Difficulty rating: 1020

Small Hint:

Count all ways to choose $3$ cards from the $5$ cards.

Big Hint:

If $4$ is the largest selected card, the other two selected cards must come from $1,$ $2,$ and $3$ .

Video solution:

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Written solution:

The number of ways to choose $3$ cards from $5$ is ${5 \choose 3} = 10.$ If $4$ is the largest value selected, then the other two cards have to be chosen from $\{1, 2, 3\}.$ There are ${3 \choose 2} = 3$ ways to do this. The probability is then $\dfrac{3}{10}.$

Thus, C is the correct answer.

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?

$148$

$324$

$361$

$1296$

$1369$

Answer: C

Concepts:diagonal perfect square

Difficulty rating: 1140

Small Hint:

In an odd-by-odd square, the two diagonals share exactly the center tile.

Big Hint:

If each diagonal has $n$ tiles, the two diagonals together contain $2n-1$ tiles.

Video solution:

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Written solution:

$37$ tiles on both diagonals imply that there are $19$ tiles on each diagonal, since one tile overlaps in the middle. The total number of tiles would then be $19^2 = 361$ since the number of tiles in each row is equal to the number of tiles in one diagonal.

Thus, C is the correct answer.

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?

$2$ and $19$

$20$ and $39$

$40$ and $59$

$60$ and $79$

$80$ and $124$

Answer: D

Concepts:least common multiple modular arithmetic

Difficulty rating: 1020

Small Hint:

Numbers with remainder $1$ after division by $4,$ $5,$ and $6$ are one more than multiples of their least common multiple.

Big Hint:

Find the least common multiple of $4,$ $5,$ and $6$ , then add $1$ .

Video solution:

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Written solution:

If a number leaves a remainder of $1$ when divided by $4,$ $5,$ and $6,$ then it is one more than the least common multiple of these numbers. The least common multiple is $60,$ so the smallest such positive integer is $60 + 1 = 61.$

Thus, D is the correct answer.

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?

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:logical deduction

Difficulty rating: 1020

Small Hint:

Every recorded win contributes one recorded loss to another player.

Big Hint:

Compare total wins and total losses across all three players.

Video solution:

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Written solution:

Across all players, every win is matched by one loss; drawn games, if any, contribute neither. The number of losses is $2+3+3=8,$ so the number of games Kyler won is $8-4-3=1.$

Thus, B is the correct answer.

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?

$89$

$92$

$93$

$96$

$98$

Answer: C

Concepts:percentage

Difficulty rating: 1370

Small Hint:

Assume there are $100$ homework problems to make the percentages concrete.

Big Hint:

Use Chloe’s overall score to determine how many jointly solved answers were correct.

Video solution:

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Written solution:

Since the answer is the same regardless of the number of problems, we can assume that there were $100$ problems on the assignment. $80\%$ of $50$ is $40,$ so Chloe answered $40$ questions correctly alone. $88\%$ of $100$ is $88,$ so Chloe answered $88$ questions correctly in total. This means that Chloe answered $88 - 40 = 48$ together with Zoe.

$90\%$ of $50$ is $45,$ so Zoe answered $45$ questions correctly by herself. We know that she answered $48$ questions correctly with Chloe, so she answered $45 + 48 = 93$ correctly in total. This means that her overall percentage is $93\%.$

Thus, C is the correct answer.

15.

In the arrangement of letters and numerals below, by how many different paths can one spell $\mathrm{AMC8}$ ? 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.

$8$

$9$

$12$

$24$

$36$

Answer: D

Concepts:multiplication principle

Difficulty rating: 1220

Small Hint:

Count choices one step at a time from the central $A$ .

Big Hint:

From each $M$ , count adjacent $C$ ’s, then from each $C$ , count adjacent $8$ ’s.

Video solution:

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Written solution:

Starting from $A,$ there are $4$ ways to reach an $M.$ From each $M,$ there are $3$ ways to reach a $C.$ From each $C,$ there are $2$ ways to reach an $8.$ Multiplying all these possibilities, we get $4 \cdot 3 \cdot 2 = 24.$

Thus, D is the correct answer.

16.

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

$\dfrac{3}{4}$

$\dfrac{3}{2}$

$2$

$\dfrac{12}{5}$

$\dfrac{5}{2}$

Answer: D

Concepts:area ratio perimeter

Difficulty rating: 1420

Small Hint:

Use the equal-perimeter condition to find $BD$ and $CD$ .

Big Hint:

Once $BC$ is split, the two triangles have the same altitude from $A$ , so compare areas by base lengths.

Video solution:

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Written solution:

The only way to split $\overline{BC}$ into two parts such that the two triangles have the same perimeter is if $\overline{CD} = 3$ and $\overline{BD} = 2.$

$\triangle ACD$ and $\triangle ABD$ have the same altitudes, so their areas are proportional to their bases. This means that the area of $\triangle ABD$ is $\dfrac{2}{5}$ the area of $\triangle ABC,$ which is $\dfrac{2}{5} \cdot 3 \cdot \dfrac{4}{2} = \dfrac{12}{5}.$

Thus, D is the correct answer.

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?

$9$

$27$

$45$

$63$

$81$

Answer: C

Concepts:system of equations

Difficulty rating: 1240

Small Hint:

Let $n$ be the number of treasure chests.

Big Hint:

Write one equation for the $9$ -coins attempt and another for the $6$ -coins attempt.

Video solution:

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Written solution:

Let $n$ be the number of treasure chests and $g$ be the number of gold coins. Then $9(n - 2) = g$ and $6n + 3 = g.$ Solving this system yields $n = 7,$ so the number of gold coins is $6 \cdot 7 + 3 = 45.$

Thus, C is the correct answer.

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 $ABCD?$

$12$

$24$

$26$

$30$

$36$

Answer: B

Concepts:Pythagorean Theorem Pythagorean Triple area

Difficulty rating: 1430

Small Hint:

First find $BD$ using the small right triangle.

Big Hint:

Then recognize the larger triangle as a $5$ - $12$ - $13$ right triangle and subtract the small triangle’s area.

Video solution:

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Written solution:

Since $\angle BCD$ is a right angle, we can apply the Pythagorean theorem to $\triangle BCD$ to get that $\overline{BD} = 5.$ We also get that $\angle DBA$ is right since the sides of $\triangle BDA$ form a Pythagorean triple.

Then the area of $ABCD$ is equal to $\begin{align*} \text{area}(\triangle &BDA) - \text{area}(\triangle BCD) \\ &= \dfrac{1}{2} \cdot 12 \cdot 5 - \dfrac{1}{2} \cdot 4 \cdot 3 \\ &= 30 - 6 \\ &= 24. \end{align*}$

Thus, B is the correct answer.

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: $98!+99!+100!$

$23$

$24$

$25$

$26$

$27$

Answer: D

Concepts:factorial Legendre’s Formula

Difficulty rating: 1640

Small Hint:

Factor $98!$ out of all three terms.

Big Hint:

After factoring, count the powers of $5$ in $98!$ and in $100^2$ .

Video solution:

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Written solution:

Factor out $98!$ : $\begin{aligned} 98!+99!+100! &=98!\bigl(1+99\\ &\qquad+100\cdot99\bigr)\\ &=98!\cdot100^2. \end{aligned}$

The exponent of $5$ in $98!$ is $\left\lfloor\frac{98}{5}\right\rfloor+\left\lfloor\frac{98}{25}\right\rfloor$ , which equals $19+3=22.$ The factor $100^2$ contributes four more factors of $5$ , so the total exponent is $22+4=26.$

Thus, D is the correct answer.

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?

$\dfrac{14}{75}$

$\dfrac{56}{225}$

$\dfrac{107}{400}$

$\dfrac{7}{25}$

$\dfrac{9}{25}$

Answer: B

Concepts:basic probability multiplication principle

Difficulty rating: 1550

Small Hint:

Count favorable four-digit numbers directly, then divide by the number of four-digit integers.

Big Hint:

Choose the odd units digit first, then choose the nonzero thousands digit, then the remaining two digits.

Video solution:

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Written solution:

Since the number is odd, the last digit is odd, giving $5$ possibilities. The thousands digit cannot be zero or the number we already got, so that gives $8$ possibilities. Similarly, the hundreds digit has $8$ possibilities, and the tens digit has $7$ possibilities. This gives a total of $5 \cdot 8 \cdot 8 \cdot 7 = 2{,}240,$ making the probability $\dfrac{2{,}240}{9{,}000} = \dfrac{56}{225}.$

Thus, B is the correct answer.

21.

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

$0$

$1$ and $-1$

$2$ and $-2$

$0$ , $2$ , and $-2$

$0$ , $1$ , and $-1$

Answer: A

Concepts:absolute value casework

Difficulty rating: 1510

Small Hint:

Because $a+b+c=0$ , the three numbers cannot all have the same sign.

Big Hint:

Each fraction $\dfrac{x}{|x|}$ is just the sign of $x$ .

Video solution:

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Written solution:

Because $a+b+c=0$ and none of the numbers is zero, their signs are either two positive and one negative or two negative and one positive. In the first case, the first three sign fractions sum to $1$ , while $\dfrac{abc}{|abc|}=-1$ . In the second case, the first three sum to $-1$ , while $\dfrac{abc}{|abc|}=1$ . Either way, the entire expression equals $0$ .

Thus, A is the correct answer.

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?

$\dfrac{7}{6}$

$\dfrac{13}{5}$

$\dfrac{59}{18}$

$\dfrac{10}{3}$

$\dfrac{60}{13}$

Answer: D

Concepts:tangent line similarity

Difficulty rating: 1640

Small Hint:

Draw the radius to the point where the semicircle touches the hypotenuse.

Big Hint:

Use tangent lengths from the same external point, then set up similar triangles.

Video solution:

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Written solution:

Let $O$ be the center of the inscribed semicircle and $D$ be the tangent point of the semicircle on $\overline{AB}.$ Then $BD = 5$ since $\overline{BD}$ and $\overline{BC}$ are tangents to the semicircle. Then $AD = 8$ and $OD = r.$ $\overline{OD}$ is perpendicular to $\overline{AB}$ so $\triangle ADO \sim \triangle BCA,$ so $\dfrac{r}{8} = \dfrac{5}{12}.$ Solving this, we get $r = \dfrac{10}{3}.$

Thus, D is the correct answer.

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?

$10$

$15$

$25$

$50$

$82$

Answer: C

Concepts:factor arithmetic sequence

Difficulty rating: 1610

Small Hint:

Each day’s minutes per mile must divide $60$ .

Big Hint:

List the divisors of $60$ and find four of them spaced $5$ apart.

Video solution:

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Written solution:

Linda traveled for $60$ minutes every day. Since one mile was traveled in an integer amount of minutes each day, her minutes per mile every day must be a factor of $60.$ The factors of $60$ are $1,$ $2,$ $3,$ $4,$ $5,$ $6,$ $10,$ $12,$ $15,$ $20,$ $30,$ and $60.$ The only sequence of four of these numbers that differ by $5$ are $5,$ $10,$ $15,$ and $20.$ For the four days, she traveled $\dfrac{60}{5} + \dfrac{60}{10} + \dfrac{60}{15} + \dfrac{60}{20} = 25$ miles in total.

Thus, C is the correct answer.

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?

$78$

$80$

$144$

$146$

$152$

Answer: D

Concepts:inclusion-exclusion least common multiple

Difficulty rating: 1800

Small Hint:

Calls repeat every $\operatorname{lcm}(3,4,5)$ days.

Big Hint:

Count days with at least one call in each $60$ -day cycle, then handle the last few days of $2017.$

Video solution:

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Written solution:

In a $60$ -day period, the first child calls $20$ times, the second child calls $15$ times, and the third child calls $12$ times. $20 + 15 + 12 = 47$ overcounts, however. The first and second children call on the same day $\frac{60}{12} = 5$ times. The first and third children call on the same day $\frac{60}{15} = 4$ times. The second and third children call on the same day $\frac{60}{20} = 3$ times. Subtracting these from $47$ yields $47 - 5 - 4 - 3 = 35.$

The $60$ th day is added in thrice and subtracted out thrice, so we need to add it back in. This means that for every $60$ days, Mrs. Sanders receives a call $36$ days, which means that she does not receive a call on $24$ days. There are $6$ $60$ -day periods, and there are no calls on the $361$ st or $362$ nd day, which results in $24 \cdot 6 + 2 = 146$ total days with no phone calls.

Thus, D is the correct answer.

25.

In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length $2,$ and $m\angle TUS = 60^\circ.$

Arcs $\overset{\large\frown}{TR}$ and $\overset{\large\frown}{SR}$ are each one-sixth of a circle with radius $2.$ What is the area of the region shown?

$3\sqrt{3}-\pi$

$4\sqrt{3}-\dfrac{4\pi}{3}$

$2\sqrt{3}$

$4\sqrt{3}-\dfrac{2\pi}{3}$

$4+\dfrac{4\pi}{3}$

Answer: B

Concepts:equilateral triangle sector area decomposition

Difficulty rating: 1750

Small Hint:

Extend the two straight sides to form an equilateral triangle.

Big Hint:

Subtract two $60^\circ$ sectors of radius $2$ from that equilateral triangle.

Video solution:

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Written solution:

We can extend $\overline{SU}$ and $\overline{TU}$ to form the following picture.

The area of this region is the area of an equilateral triangle with side length of $4$ minus the area of two-sixths of a circle with radius $2.$ The area for an equilateral triangle with side length $s$ is $\dfrac{s^2\sqrt{3}}{4}.$ This means that the total area is $\dfrac{4^2 \sqrt{3}}{4} - \dfrac{1}{3} \pi \cdot 2^2 = 4 \sqrt{3} - \dfrac{4}{3} \pi.$

Thus, B is the correct answer.