2017 AMC 8 Exam Problems

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All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

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1.

Which of the following values is largest?

2+0+1+7 2+0+1+7

2×0+1+7 2 \times 0 +1+7

2+0×1+7 2+0 \times 1 + 7

2+0+1×7 2+0+1 \times 7

2×0×1×7 2 \times 0 \times 1 \times 7

Answer: A
Video solution:
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Written solution:

Option (A) evaluates to 10.10.

Option (B) evaluates to 0+1+7=8.0 + 1 + 7 = 8.

Option (C) evalutes to 2+0+7=9.2 + 0 + 7 = 9.

Option (D) evaluates to 2+0+7=9.2 + 0 + 7 = 9.

Option (E) evalutes to 0.0.

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 3636 votes, then how many votes were cast all together?

70 70

84 84

100 100

106 106

120 120

Answer: E
Video solution:
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Written solution:

If 3636 votes is 30%30\% of the total votes, then 10%10\% of the total votes is 1212 votes. The number of total votes would then be 1012=120.10 \cdot 12 = 120.

Thus, E is the correct answer.

3.

What is the value of the expression 1684?\sqrt{16\sqrt{8\sqrt{4}}}?

4 4

42 4\sqrt{2}

8 8

82 8\sqrt{2}

16 16

Answer: C
Video solution:
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Written solution:

This expression can be reduced as follows: 1684=1616=64=8 \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 0.0003150.000315 is multiplied by 7,928,5647,928,564 the product is closest to which of the following?

210 210

240 240

2100 2100

2400 2400

24000 24000

Answer: D
Video solution:
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Written solution:

We can approximate the product as (3104)(8106)=24102=2400. \begin{align*} (3 \cdot 10^{-4})(8 \cdot 10^6) &= 24 \cdot 10^2 \\ &= 2400. \end{align*}

Thus, D is the correct answer.

5.

What is the value of the expression 123456781+2+3+4+5+6+7+8?\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 1020

1120 1120

1220 1220

2240 2240

3360 3360

Answer: B
Video solution:
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Written solution:

Looking at the denominator independently, 1+2+3++8=36,1 + 2 + 3 + \cdots + 8 = 36, so the desired answer is 1234567836=4578=1120. \begin{gather*} \dfrac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{36} = \\ 4 \cdot 5 \cdot 7 \cdot 8 = 1120. \end{gather*}

Thus, B is the correct answer.

6.

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

18 18

36 36

60 60

72 72

90 90

Answer: D
Video solution:
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Written solution:

We can let the three angles be equal to 3x,3x,3x, 3x, and 4x.4x. Then we know that their sum equals 180.180. From this we can set 3x+3x+4x=180,3x + 3x + 4x = 180, and solving this, we get 10x=18010x = 180 and x=18.x = 18.

The largest angle is 4x=72.4x = 72.

Thus, D is the correct answer.

7.

Let ZZ 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?Z?

11 11

19 19

101 101

111 111

1111 1111

Answer: A
Video solution:
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Written solution:

We can let ZZ have the form abcabc.abcabc. Then, we get that Z=1001abc=71113abc.\begin{align*} Z = 1001 \cdot abc = 7 \cdot 11 \cdot 13 \cdot abc. \end{align*} This means that 1111 is a factor of Z.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 4

6 6

7 7

8 8

9 9

Answer: D
Video solution:
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Written solution:

If a number is even, it is divisible by 2.2. This means that the house number is either divisible by 22 or by 7.7. This rules out the possibility that the house number is prime. This means that the house number is divisible by 22 and by 7,7, and one of its digits is 9.9.

The divisibility conditions show that the house number is divisble by 14,14, so if we look at all the two-digit multiples of 14,14, the only one with a digit of 99 is 98.98. Therefore, units digit is 8.8.

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 Macy could have?

1 1

2 2

3 3

4 4

5 5

Answer: D
Video solution:
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Written solution:

If the number of marbles is divisble by both 33 and 4,4, then the number must be divisible by 12.12. If we test 12,12, we get that there are 44 blue marbles and 33 red marbles. This leaves a maximum of 55 green marbles, which is not possible.

If there are 2424 marbles, then there are 88 blue marbles and 66 red marbles. To find the number of yellow marbles, we get 24866=4.24 - 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?

110 \dfrac{1}{10}

15 \dfrac{1}{5}

310 \dfrac{3}{10}

25 \dfrac{2}{5}

12 \dfrac{1}{2}

Answer: C
Video solution:
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Written solution:

The number of ways to choose 33 cards from 55 is (53)=10.{5 \choose 3} = 10. If 44 is the largest value selected, then the other two cards have to be chosen from 1,2,3.{1, 2, 3}. There are (32)=3{3 \choose 2} = 3 ways to do this. The probability is then 310.\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?

148 148

324 324

361 361

1296 1296

1369 1369

Answer: C
Video solution:
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Written solution:

3737 tiles on both diagonals imply that there are 1919 tiles on each diagonal, since one tile overlaps in the middle. The total number of tiles would then be 192=36119^2 = 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 2\text{ and }19

20 and 39 20\text{ and }39

40 and 59 40\text{ and }59

60 and 79 60\text{ and }79

80 and 124 80\text{ and }124

Answer: D
Video solution:
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Written solution:

If a number leaves a remainder of 11 when divided by 4,5, and 6,4, 5, \text{ and } 6, then it is one more than the least common multiple of these numbers. The least common multiple is 60,60, so the smallest such positive integer is 60+1=61.60 + 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 0

1 1

2 2

3 3

4 4

Answer: B
Video solution:
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Written solution:

Since there are no ties, the number of wins has to equal the number of losses. The number of losses is 2+3+3=8,2 + 3 + 3 = 8, so the number of games Kyler won is 843=1.8 - 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%80\% of the problems she solved alone, but overall 88%88\% of her answers were correct. Zoe had correct answers to 90%90\% of the problems she solved alone. What was Zoe's overall percentage of correct answers?

89 89

92 92

93 93

96 96

98 98

Answer: C
Video solution:
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Written solution:

Since the answer should be same regardless of the number of problems, we can assume that there were 100100 problems on the assignment. 80%80\% of 5050 is 40,40, so Chloe answered 4040 questions correctly alone. 88%88\% of 100100 is 88,88, so Chloe answered 8888 questions correctly in total. This means that Chloe answered 8840=4888 - 40 = 48 together with Zoe.

90%90\% of 5050 is 45,45, so Zoe answered 4545 questions correct by herself. We know that she answered 4848 questions correct with Chloe, so she answered 45+48=9345 + 48 = 93 correctly in total. This means that her overall percentage is 93%.93\%.

Thus, C is the correct answer.

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 only allows 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 8

9 9

12 12

24 24

36 36

Answer: D
Video solution:
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Written solution:

Starting from A,A, there are 44 ways to reach an M.M. From each M,M, there are 33 ways to reach a C.C. From each C,C, there are 22 ways to reach an 8.8. Multiplying all these possibilities, we get 432=24.4 \cdot 3 \cdot 2 = 24.

Thus, D is the correct answer.

16.

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

34 \dfrac{3}{4}

32 \dfrac{3}{2}

2 2

125 \dfrac{12}{5}

52 \dfrac{5}{2}

Answer: D
Video solution:
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Written solution:

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

ACD\triangle ACD and ABD\triangle ABD have the same altitudes, so their areas are proportional to their bases. This means that the area of ABD\triangle ABD is 25\dfrac{2}{5} the area of ABC,\triangle ABC, which is 25342=125.\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 9

27 27

45 45

63 63

81 81

Answer: C
Video solution:
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Written solution:

Let nn be the number of treasure chests and gg be the number of gold coins. Then 9(n2)=g9(n - 2) = g and 6n+3=g.6n + 3 = g. Solving this system yields n=7,n = 7, so the number of gold coins is 67+3=45.6 \cdot 7 + 3 = 45.

Thus, C is the correct answer.

18.

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

12 12

24 24

26 26

30 30

36 36

Answer: B
Video solution:
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Written solution:

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

Then the area of ABCDABCD is equal to area(BDA)area(BCD)=121251243=306=24.\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,M, the notation M!M! denotes the product of the integers 11 through M.M. What is the largest integer nn for which 5n5^n is a factor of the sum: 98!+99!+100!98!+99!+100!

23 23

24 24

25 25

26 26

27 27

Answer: D
Video solution:
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Written solution:

The expression can be factored as follows:

98!+99!+100!=98!+9998!+1009998!=98!(1+99+10099)=98!(100+10099)=98!100(1+99)=98!1002\begin{align*} &98! + 99! + 100! \\ &= 98! + 99 \cdot 98! + 100 \cdot 99 \cdot 98! \\ &= 98!(1 + 99 + 100 \cdot 99) \\ &= 98!(100 + 100 \cdot 99) \\ &= 98! \cdot 100 \cdot (1 + 99) \\ &= 98! \cdot 100^2 \end{align*}

Each 100100 possesses two factors of 5.5. The number of factors of 55 in 98!98! is 98/5+98/25=19+3=22.\lfloor 98 / 5 \rfloor + \lfloor 98 / 25 \rfloor= 19 + 3 = 22. This counts the number of numbers divisible by 55 and the number of numbers divisible by 2525 to get both factors of 5.5. The total number of factors of 55 is 2+2+22=26.2 + 2 + 22 = 26.

Thus, D is the correct answer.

20.

An integer between 10001000 and 9999,9999, inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?

1475 \dfrac{14}{75}

56225 \dfrac{56}{225}

107400 \dfrac{107}{400}

725 \dfrac{7}{25}

925 \dfrac{9}{25}

Answer: B
Video solution:
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Written solution:

Since the number is odd, the last digit is odd, giving 55 possibilities. The thousands digit cannot be zero or the the number we already got, so that gives 88 possibilities. Similarly, the hundreds digit has 88 possibilities, and the tens digit has 77 possibilities. This gives a total of 5887=2240,5 \cdot 8 \cdot 8 \cdot 7 = 2240, making the probability 22409000=56225.\dfrac{2240}{9000} = \dfrac{56}{225}.

Thus, B is the correct answer.

21.

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

0 0

1 and 1 1\text{ and }-1

2 and 2 2\text{ and }-2

0,2, and 2 0,2,\text{ and }-2

0,1, and 1 0,1,\text{ and }-1

Answer: A
Video solution:
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Written solution:

Since the sum is 0,0, all the numbers cannot be positive or negative. This gives two cases: two are positive and one is negative or vice versa. Also note that if x<0,x < 0, xx=1,\dfrac{x}{|x|} = -1, and if x>0,x > 0, xx=1.\dfrac{x}{|x|} = 1.

Case I:\textbf{Case I:} two are positive and one is negative

WLOG, let a,b>0a, b > 0 and c<0.c < 0. Then abc<0.abc < 0. This means that aa=bb=1\dfrac{a}{|a|} = \dfrac{b}{|b|} = 1 and cc=abcabc=1.\dfrac{c}{|c|} = \dfrac{abc}{|abc|} = -1. Then aa+bb+cc+abcabc=0. \dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|} = 0.

Case II:\textbf{Case II:} two are negative and one is positive

WLOG, let a,b<0a, b < 0 and c>0.c > 0. Then abc>0.abc > 0. This means that aa=bb=1\dfrac{a}{|a|} = \dfrac{b}{|b|} = -1 and cc=abcabc=1.\dfrac{c}{|c|} = \dfrac{abc}{|abc|} = 1. Then aa+bb+cc+abcabc=0. \dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|} = 0.

Either way, aa+bb+cc+abcabc\dfrac{a}{|a|}+\dfrac{b}{|b|}+\dfrac{c}{|c|}+\dfrac{abc}{|abc|} equals 0.0.

Thus, A is the correct answer.

22.

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

76 \dfrac{7}{6}

135 \dfrac{13}{5}

5918 \dfrac{59}{18}

103 \dfrac{10}{3}

6013 \dfrac{60}{13}

Answer: D
Video solution:
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Written solution:

Let OO be the center of the inscribed semicircle and DD be the tangent point of the semicircle on AB.\overline{AB}. Then BD=5BD = 5 since BD\overline{BD} and BC\overline{BC} are tangents to the semicircle. Then AD=8AD = 8 and OD=r.OD = r. OD\overline{OD} is perpendicular toAB\overline{AB} so ADOBCA,\triangle ADO \sim \triangle BCA, so r8=512.\dfrac{r}{8} = \dfrac{5}{12}. Solving this, we get r=103.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 55 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 10

15 15

25 25

50 50

82 82

Answer: C
Video solution:
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Written solution:

Linda traveled for 6060 minutes every day. Since one mile was traveled in an integer amount of minutes each day, her mph every day must be a factor of 60.60. The factors of 6060 are 1,2,3,4,5,6,10,12,15,20,30,1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60.60. The only sequence of four of these numbers that differ by 55 are 5,10,15,5, 10, 15, and 20.20. For the four days, she traveled 605+6010+6015+6020=25\dfrac{60}{5} + \dfrac{60}{10} + \dfrac{60}{15} + \dfrac{60}{20} = 25 miles.

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 78

80 80

144 144

146 146

152 152

Answer: D
Video solution:
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Written solution:

In a 6060 day period, the first child calls 2020 times, the second child calls 1515 times, and the third child calls 1212 days. 20+15+12=4720 + 15 + 12 = 47 overcounts, however. The first and second children call on the same day 60/12=560 / 12 = 5 times. The first and third children call on the same day 60/15=460 / 15 = 4 times. The second and third children call on the same day 60/20=360 / 20 = 3 times. Subtracting these from 4747 yields 47543=35.47 - 5 - 4 - 3 = 35.

The 60th60^{\text{th}} is added in thrice and subtracted out thrice, so we need to add it back in. This means that for every 6060 days, Mrs. Sanders receives a call 3636 days, which means that she does not receive a call on 2424 days. There are 66 6060 day periods, and there are no calls the 361st361^{\text{st}} or 362nd362^{\text{nd}} day, which results in 246+2=14624 \cdot 6 + 2 = 146 total days with no phone calls.

Thus, D is the correct answer.

25.

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

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

33π 3\sqrt{3}-\pi

434π3 4\sqrt{3}-\dfrac{4\pi}{3}

23 2\sqrt{3}

432π3 4\sqrt{3}-\dfrac{2\pi}{3}

4+4π3 4+\dfrac{4\pi}{3}

Answer: B
Video solution:
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Written solution:

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

The area of this region is the area of an equilaterial triangle with side length of 44 minus the area of two-sixths of a circle with radius 2.2. The area for an equilateral triangle with side length ss is s234.\dfrac{s^2\sqrt{3}}{4}. This means that the total area is 423413π22=4343π.\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.