2017 AMC 10B solution

| February 16, 2017

2017 AMC 10B solution

Problem 1

Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$ . Then she switched the digits of the result, obtaining a number between $71$ and $75$ , inclusive. What was Mary’s number?

$textbf{(A)} 11qquadtextbf{(B)} 12qquadtextbf{(C)} 13qquadtextbf{(D)} 14qquadtextbf{(E)} 15$

Solution 1

We try out each answer choice. Multiplying $12$ by $3$ , adding $11$ , and reversing the digits yields $74$ . Therefore the answer is $boxed{textbf{(B)} 12}$ .

Solution 2

Working backwards, we reverse the digits of each number from $71$ ~ $75$ and subtract $11$ from each, so we have [6, 16, 26, 36, 46] The only numbers from this list that are divisible by $3$ are $6$ and $36$ . We divide both by $3$ , yielding $2$ and $12$ . Since $2$ is not a two-digit number, the answer is $boxed{textbf{(B)} 12}$ .

Problem 2

Sofia ran $5$ laps around the $400$ -meter track at her school. For each lap, she ran the first $100$ meters at an average speed of $4$ meters per second and the remaining $300$ meters at an average speed of $5$ meters per second. How much time did Sofia take running the $5$ laps?

$textbf{(A)} text{5 minutes and 35 seconds}qquadtextbf{(B)} text{6 minutes and 40 seconds}qquadtextbf{(C)} text{7 minutes and 5 seconds}qquadtextbf{(D)} text{7 minutes and 25 seconds}$ $qquadtextbf{(E)} text{8 minutes and 10 seconds}$

Solution

If Sofia ran the first $100$ meters of each lap at $4$ meters per second and the remaining $300$ meters of each lap at $5$ meters per second, then she took $frac{100}{4}+frac{300}{5}=25+60=85$ seconds for each lap. Because she ran $5$ laps, she took a total of $5 cdot 85=425$ seconds, or $7$ minutes and $5$ seconds. The answer is $boxed{textbf{(C)} text{7 minutes and 5 seconds}}$ .

Problem 3

Real numbers $x$ , $y$ , and $z$ satisfy the inequalities $0<x<1$ , $-1<y<0$ , and $1<z<2$ . Which of the following numbers is necessarily positive?

$textbf{(A)} y+x^2qquadtextbf{(B)} y+xzqquadtextbf{(C)} y+y^2qquadtextbf{(D)} y+2y^2qquadtextbf{(E)} y+z$

Solution

Notice that $y+z$ must be positive because $|z|>|y|$ . Therefore the answer is $boxed{textbf{(E) } y+z}$ .

The other choices:

$textbf{(A)}$ As $x$ grows closer to $0$ , $x^2$ decreases and thus becomes less than $y$ .

$textbf{(B)}$ $x$ can be as small as possible ( $x>0$ ), so $xz$ grows close to $0$ as $x$ approaches $0$ .

$textbf{(C)}$ For all $-1<y<0$ , $y>y^2$ , and thus it is always negative.

$textbf{(D)}$ The same logic as above, but when $-frac{1}{2}<y<0$ this time.

Problem 4

Supposed that $x$ and $y$ are nonzero real numbers such that $frac{3x+y}{x-3y}=-2$ . What is the value of $frac{x+3y}{3x-y}$ ?

$textbf{(A)} -3qquadtextbf{(B)} -1qquadtextbf{(C)} 1qquadtextbf{(D)} 2qquadtextbf{(E)} 3$

Solution

Problem 5

Camilla had twice as many blueberry jelly beans as cherry jelly beans. After eating 10 pieces of each kind, she now has three times as many blueberry jelly beans as cherry jelly beans. How many blueberry jelly beans did she originally have?

$textbf{(A)} 10qquadtextbf{(B)} 20qquadtextbf{(C)} 30qquadtextbf{(D)} 40qquadtextbf{(E)} 50$

Solution

Denote the number of blueberry and cherry jelly beans as $b$ and $c$ respectively. Then $b = 2c$ and $b-10 = 3(c-10)$ . Substituting, we have $2c-10 = 3c-30$ , so $c=20$ , $b=boxed{textbf{(D) } 40}$ .

Problem 6

What is the largest number of solid $2text{-in} times 2text{-in} times 1text{-in}$ blocks that can fit in a $3text{-in} times 2text{-in}times3text{-in}$ box?

$textbf{(A)} 3qquadtextbf{(B)} 4qquadtextbf{(C)} 5qquadtextbf{(D)} 6qquadtextbf{(E)} 7$

Solution

We find that the volume of the larger block is $18$ , and the volume of the smaller block is $4$ . Dividing the two, we see that only a maximum of $4$ $2$ by $2$ by $1$ blocks can fit inside the $3$ by $3$ by $2$ block. Therefore, the answer is $boxed{textbf{(B) }4}$ .

Problem 7

Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend’s house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend’s house. In kilometers rounded to the nearest tenth, how far did Samia walk?

$textbf{(A)} 2.0qquadtextbf{(B)} 2.2qquadtextbf{(C)} 2.8qquadtextbf{(D)} 3.4qquadtextbf{(E)} 4.4$

Solution

Problem 8

Points $A(11, 9)$ and $B(2, -3)$ are vertices of $triangle ABC$ with $AB=AC$ . The altitude from $A$ meets the opposite side at $D(-1, 3)$ . What are the coordinates of point $C$ ?

$textbf{(A)} (-8, 9)qquadtextbf{(B)} (-4, 8)qquadtextbf{(C)} (-4, 9)qquadtextbf{(D)} (-2, 3)qquadtextbf{(E)} (-1, 0)$

Solution

Since $AB = AC$ , then $triangle ABC$ is isosceles, so $BD = CD$ . Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = boxed{textbf{(C) } (-4,9)}$ .

Problem 9

A radio program has a quiz consisting of $3$ multiple-choice questions, each with $3$ choices. A contestant wins if he or she gets $2$ or more of the questions right. The contestant answers randomly to each question. What is the probability of winning?

$textbf{(A)} frac{1}{27}qquadtextbf{(B)} frac{1}{9}qquadtextbf{(C)} frac{2}{9}qquadtextbf{(D)} frac{7}{27}qquadtextbf{(E)} frac{1}{2}$

Problem 10

The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, 5)$ . What is $c$ ?

$textbf{(A)} -13qquadtextbf{(B)} -8qquadtextbf{(C)} 2qquadtextbf{(D)} 8qquadtextbf{(E)} 13$

Solution

Placeholder

Problem 11

At Typico High School, $60%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

$textbf{(A)} 10%qquadtextbf{(B)} 12%qquadtextbf{(C)} 20%qquadtextbf{(D)} 25%qquadtextbf{(E)} 33frac{1}{3}%$

Solution

$60% cdot 20% = 12%$ of the people that claim that they dislike dancing actually like it, and $40% cdot 90% = 36%$ . Therefore, the answer is $frac{12%}{12%+36%} = boxed{textbf{(D) } 25%}$ .

Problem 12

Elmer’s new car gives $50%$ percent better fuel efficiency. However, the new car uses diesel fuel, which is $20%$ more expensive per liter than the gasoline the old car used. By what percent will Elmer save money if he uses his new car instead of his old car for a long trip?

$textbf{(A) } 20% qquad textbf{(B) } 26tfrac23% qquad textbf{(C) } 27tfrac79% qquad textbf{(D) } 33tfrac13% qquad textbf{(E) } 66tfrac23%$

Solution

Suppose that his old car runs at $x$ km per liter. Then his new car runs at $frac{3}{2}x$ km per liter, or $x$ km per $frac{2}{3}$ of a liter. Let the cost of the old car’s fuel be $c$ , so the trip in the old car takes $xc$ dollars, while the trip in the new car takes $frac{2}{3}cdotfrac{6}{5}xc = frac{4}{5}xc$ He saves $frac{frac{1}{5}xc}{xc} = boxed{textbf{(A)} 20%}$ .

Problem 13

There are students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are students taking yoga, taking bridge, and taking painting. There are students taking at least two classes. How many students are taking all three classes?

Solution

By PIE, the answer is .

Problem 14

An integer is selected at random in the range . What is the probablilty that the remainder when $N^{16}$ is divided by is ?

Solution

By Fermat’s Little Theorem, $N^{16} = (N^4)^4 equiv 1 text{ (mod 5)}$ when N is relatively prime to 5. However, this happens with probability $boxed{textbf{(D) } frac 45}$ .

Solution 2

Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits . The pattern for is , no matter what power, so doesn’t work. Likewise, the pattern for is always . Doing the same for the rest of the digits, we find that the units digits of $1^{16}$ , $2^{16}$ , $3^{16}$ , $4^{16}$ , $6^{16}$ , $7^{16}$ , $8^{16}$ and $9^{16}$ all have the remainder of when divided by $boxed{textbf{(D) } frac 45}$ .

Problem 15

Rectangle $ABCD$ has $AB=3$ and $BC=4$ . Point $E$ is the foot of the perpendicular from $B$ to diagonal $overline{AC}$ . What is the area of $triangle ABC$ ?

$textbf{(A)} 1qquadtextbf{(B)} frac{42}{25}qquadtextbf{(C)} frac{28}{15}qquadtextbf{(D)} 2qquadtextbf{(E)} frac{54}{25}$

Problem 16

How many of the base-ten numerals for the positive integers less than or equal to $2017$ contain the digit $0$ ?

$textbf{(A)} 469qquadtextbf{(B)} 471qquadtextbf{(C)} 475qquadtextbf{(D)} 478qquadtextbf{(E)} 481$

Solution

We can use complementary counting. There are $2017$ positive integers in total to consider, and there are $9$ one-digit integers, $9 cdot 9 = 81$ two digit integers without a zero, $9 cdot 9 cdot 9$ three digit integers without a zero, and $9 cdot 9 cdot 9 = 1458$ three-digit integers without a zero. Therefore, the answer is $2017 - 9 - 81 - 729 - 729 = boxed{textbf{(A) }469}$ .

Problem 17

Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3$ , $23578$ , and $987620$ are monotonous, but $88$ , $7434$ , and $23557$ are not. How many monotonous positive integers are there?

$textbf{(A)} 1024qquadtextbf{(B)} 1524qquadtextbf{(C)} 1533qquadtextbf{(D)} 1536qquadtextbf{(E)} 2048$

Solution

The number of one-digit numbers that work is $binom{10}{1}$ , and the number of two-digit integers that work is $binom{10}{2} + binom{9}2$ . We use similar logic for three-digit integers, four digit integers, etc. Summing, we have $2^{10}+2^9 - 9 - 1 - 1$ , and we need to subtract another 1 for the 0 case, so the answer is $2^{10}+2^9 - 9 - 1 - 1 - 1 = boxed{textbf{(B) }1524}$ .

Problem 18

In the figure below, 3 of the 6 disks are to be painted blue, 2 are to be painted red, and 1 is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

$textbf{(A)} 6qquadtextbf{(B)} 8qquadtextbf{(C)} 9qquadtextbf{(D)} 12qquadtextbf{(E)} 25$

Solution

First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$ . For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = boxed{textbf{(D) } 12}$ .

Problem 19

Let $ABC$ be an equilateral triangle. Extend side $overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3AB$ . Similarly, extend side $overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3BC$ , and extend side $overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3CA$ . What is the ratio of the area of $triangle A'B'C'$ to the area of $triangle ABC$ ?

$textbf{(A)} 9:1qquadtextbf{(B)} 16:1qquadtextbf{(C)} 25:1qquadtextbf{(D)} 36:1qquadtextbf{(E)} 37:1$

Solution

Note that by symmetry, $triangle A'B'C'$ is also equilateral. Therefore, we only need to find one of the sides of $A'B'C'$ to determine the area ratio. WLOG, let $AB = BC = CA = 1$ . Therefore, $BB' = 3$ and $BC' = 4$ . Also, $angle B'BC' = 120^{circ}$ , so by the Law of Cosines, $B'C' = sqrt{37}$ . Therefore, the answer is $(sqrt{37})^2 : 1^2 = boxed{textbf{(E) } 37 : 1}$

Problem 20

The number $21!=51,090,942,171,709,440,000$ has over $60,000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$textbf{(A)} frac{1}{21}qquadtextbf{(B)} frac{1}{19}qquadtextbf{(C)} frac{1}{18}qquadtextbf{(D)} frac{1}{2}qquadtextbf{(E)} frac{11}{21}$

Solution

We note that the only thing that affects the parity of the factor are the powers of 2. Since there are $10+5+2+1 = 18$ factors of 2, and since only haveing a 0 power of 2 makes the factor odd, then the answer is $boxed{textbf{(B) } frac 1{19}}$ .

Problem 21

In $triangle ABC$ , $AB=6$ , $AC=8$ , $BC=10$ , and $D$ is the midpoint of $overline{BC}$ . What is the sum of the radii of the circles inscibed in $triangle ADB$ and $triangle ADC$ ?

$textbf{(A)} sqrt{5}qquadtextbf{(B)} frac{11}{4}qquadtextbf{(C)} 2sqrt{2}qquadtextbf{(D)} frac{17}{6}qquadtextbf{(E)} 3$

Solution

We note that by the converse of the Pythagorean Theorem, $triangle ABC$ is a right triangle with a right angle at $A$ . Therefore, $AD = BD = CD = 5$ , and $[ADB] = [ADC] = 12$ . Since $A = rs$ , the inradius of $triangle ADB$ is $frac{12}{(5+5+6)/2} = frac 32$ , and the inradius of $triangle ADC$ is $frac{12}{(5+5+8)/2} = frac 43$ . Adding the two together, we have $boxed{textbf{(D) } frac{17}6}$ .

Problem 22

The diameter $overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$ . Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$ . Segment $overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$ . What is the area of $triangle ABC$ ?

$textbf{(A)} frac{120}{37}qquadtextbf{(B)} frac{140}{39}qquadtextbf{(C)} frac{145}{39}qquadtextbf{(D)} frac{140}{37}qquadtextbf{(E)} frac{120}{31}$

Solution 1

Notice that $ADE$ and $ABC$ are right triangles. Then $AE = sqrt{7^2+5^2} = sqrt{74}$ . $sin{DAE} = frac{5}{sqrt{74}} = sin{BAE} = sin{BAC} = frac{BC}{4}$ , so $BC = frac{20}{sqrt{74}}$ . We also find that $AC = frac{28}{sqrt{74}}$ , and thus the area of $ABC$ is $frac{frac{20}{sqrt{74}}cdotfrac{28}{sqrt{74}}}{2} = frac{frac{560}{74}}{2} = boxed{textbf{(D) } frac{140}{37}}$ .

Solution 2

We note that $triangle ACB ~ triangle ADE$ by $AA$ similarity. Also, since the area of $triangle ADE = frac{7 cdot 5}2 = frac{35}2$ and $AE = sqrt{74}$ , $frac{[ABC]}{[ADE]} = frac{[ABC]}{frac{35}2} = left(frac{4}{sqrt{74}}right)^2$ , so the area of $triangle ABC = boxed{textbf{(D) } frac{140}{37}}$ .

Problem 23

Let $N=123456789101112dots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$ ?

$textbf{(A)} 1qquadtextbf{(B)} 4qquadtextbf{(C)} 9qquadtextbf{(D)} 18qquadtextbf{(E)} 44$

Solution

We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, $N equiv 4 text{ (mod 5)}$ . The remainder when $N$ is divided by $9$ is $1+2+3+4 cdot 1+0+1+1 cdot 4+3+4+4$ , but since $10 equiv 1 text{ (mod 9)}$ , we can also write this as $1+2+3 cdot 10+11+12 cdot 43 + 44 = frac{44 cdot 45}2 = 22 cdot 45$ , which has a remainder of 0 mod 9. Therefore, by CRT, the answer is $boxed{textbf{(C) } 9}$ .

Note: the sum of the digits of $N$ is $270$

Solution 2

Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by $9$ . From $1$ thru $9$ , the sum is $45$ . $10$ thru $19$ , the sum is $55$ , $20$ thru $30$ is $65$ , and $30$ thru $40$ is $75$ . Thus the sum of the digits is $45+55+65+75+4+5+6+7+8 = 240+30 = 270$ , and thus $N$ is divisible by $9$ . The solution proceeds as above.

Problem 24

The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?

$textbf{(A)} 48qquadtextbf{(B)} 60qquadtextbf{(C)} 108qquadtextbf{(D)} 120qquadtextbf{(E)} 169$

Solution

WLOG, let the centroid of $triangle ABC$ be $I = (-1,-1)$ . By symmetry, $A = (1,1)$ , so $AI = BI = CI = 2sqrt{2}$ , so since $triangle AIB$ is isosceles and $angle AIB = 120^{circ}$ , then by Law of Cosines, $AB = 2sqrt{6}$ . Therefore, the area of the triangle is $frac{(2sqrt{6})^2sqrt{3}}4 = 6sqrt{3}$ , so the square of the area of the triangle is $boxed{textbf{(C) } 108}$ .

Problem 25

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?

$textbf{(A)} 92qquadtextbf{(B)} 94qquadtextbf{(C)} 96qquadtextbf{(D)} 98qquadtextbf{(E)} 100$

Solution 1

Let the sum of the scores of Isabella’s first $6$ tests be $S$ . Since the mean of her first $7$ scores is an integer, then $S + 95 equiv 0 text{ (mod 7)}$ , or $S equiv 3 text{ (mod 7)}$ . Also, $S equiv 0 text{ (mod 6)}$ , so by CRT, $S equiv 24 text{ (mod 42)}$ . We also know that $91 cdot 6 leq S leq 100 cdot 6$ , so by inspection, $S = 570$ . However, we also have that the mean of the first $5$ integers must be an integer, so the sum of the first $5$ test scores must be an multiple of $5$ , which implies that the $6$ th test score is $boxed{textbf{(E) } 100}$ .