## AMC 8 2018

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**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.

An amusement park has a collection of scale models, with ratio $1 : 20$, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica to the nearest whole number?

$\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

$\textbf{A}$

The height of the model is $289\times\dfrac{1}{20}=14.45\ \text{feet}$. The nearest whole number is $14$.

What is the value of the product\[\left(1+\frac{1}{1}\right)\times\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times\left(1+\frac{1}{5}\right)\times\left(1+\frac{1}{6}\right)?\]

$\textbf{(A) }\dfrac{7}{6}\qquad\textbf{(B) }\dfrac{4}{3}\qquad\textbf{(C) }\dfrac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8$

$\textbf{D}$

By adding up the numbers in each of the $6$ parentheses, we get $$\dfrac{2}{1} \times \dfrac{3}{2} \times \dfrac{4}{3} \times \dfrac{5}{4} \times \dfrac{6}{5} \times \dfrac{7}{6}$$ Most of the terms cancel out, with only$\dfrac{7}{1}$ remains. So the answer is $7$.

Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?

$\textbf{(A) } \textrm{Arn}\qquad\textbf{(B) }\textrm{Bob}\qquad\textbf{(C) }\textrm{Cyd}\qquad\textbf{(D) }\textrm{Dan}\qquad \textbf{(E) }\textrm{Eve}$

$\textbf{D}$

The five numbers which cause people to leave the circle are $7, 14, 17, 21,$ and $27.$ The most straightforward way to do this would be to draw out the circle with the people, and cross off people as you count. Arn counts $7$ so he leaves first. Then, Cyd, Fon, Bob, and Eve count $14$, $17$, $21$, and $27$, respectively. So the last one standing is Dan.

The twelve-sided figure shown has been drawn on $1 \textrm{ cm}\times 1 \textrm{ cm}$ graph paper. What is the area of the figure in $\textrm{cm}^2$?

$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$

$\textbf{C}$

There are 9 whole squares and 8 triangles in the graph. The area of each square is 1, and the triangle is half the square. So the total area is $9+8/2=13$.

What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$?

$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$

$\textbf{E}$

Rearranging the terms in pair, we get $(1-2)+(3-4)+(5-6)+\dots+(2017-2018)+2019$. The number of pairs is $2018/2=1009$. So the answer is $-1009+2019=1010$.

On a trip to the beach, Anh traveled 50 miles on the highway and 10 miles on a coastal access road. He drove three times as fast on the highway as on the coastal road. If Anh spent 30 minutes driving on the coastal road, how many minutes did his entire trip take?

$\textbf{(A) }50\qquad\textbf{(B) }70\qquad\textbf{(C) }80\qquad\textbf{(D) }90\qquad \textbf{(E) }100$

$\textbf{C}$

The speed on coastal road is $10/0.5=20$ miles per hour. Hence, the speed on the highway is $3\times20=60$ miles per hour. To travel 50 miles on the highway, the time is $\dfrac{50}{60}$ hours, or $50$ minutes. The time for the entire trip is $50+30=80$ minutes.

The $5$-digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$. What is the remainder when this number is divided by $8$?

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

$\textbf{B}$

We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$. This means $2+0+1+8+U$ must be divisible by $9$. The only possible value for $U$ then must be $7$. Since we are looking for the remainder when divided by $8$, we can ignore the thousands and above. The remainder when $187$ is divided by $8$ is $3$.

Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.

What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?

$\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00$

$\textbf{C}$

The mean number of days is the total number of days divided by the total number of students. The total number of days is $1\times 1+2\times 3+3\times 2+4\times 6+5\times 8+6\times 3+7\times 2=109$. The total number of students is $1+3+2+6+8+3+2=25$. Hence, the mean number is $109/25=4.36$.

Jenica is tiling the floor of her 12 foot by 16 foot living room. She plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will she use?

$\textbf{(A) }48\qquad\textbf{(B) }87\qquad\textbf{(C) }91\qquad\textbf{(D) }96\qquad \textbf{(E) }120$

$\textbf{B}$

Along the 12-foot border she need 12 one-foot by one-foot tiles, and 16 tiles for the 16-foot border. So in total $(12+16)\times2=56$. Noticing that the 4 corners is counted twice, actually she need $56-4=52$ one-foot by one-foot tiles. The area of the center is 10 feet by 14 feet, so she need $5\times7=35$ two-foot by two-foot tiles. The total number of tiles is $52+35=87$.

The $\textit{harmonic mean}$ of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4?

$\textbf{(A) }\dfrac{3}{7}\qquad\textbf{(B) }\dfrac{7}{12}\qquad\textbf{(C) }\dfrac{12}{7}\qquad\textbf{(D) }\dfrac{7}{4}\qquad \textbf{(E) }\dfrac{7}{3}$

$\textbf{C}$

The sum of the reciprocals is $\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{4}= \dfrac{7}{4}$. Their average is $\dfrac{7}{12}$. Taking the reciprocal of this gives the harmonic mean $\dfrac{12}{7}$.

Abby, Bridget, and four of their classmates will be seated in two rows of three for a group picture, as shown.

\begin{eqnarray*}

\textrm{X}&\quad\textrm{X}\quad&\textrm{X} \\ \textrm{X}&\quad\textrm{X}\quad&\textrm{X}

\end{eqnarray*}

If the seating positions are assigned randomly, what is the probability that Abby and Bridget are adjacent to each other in the same row or the same column?

$\textbf{(A) } \dfrac{1}{3} \qquad \textbf{(B) } \dfrac{2}{5} \qquad \textbf{(C) } \dfrac{7}{15} \qquad \textbf{(D) } \dfrac{1}{2} \qquad \textbf{(E) } \dfrac{2}{3}$

$\textbf{C}$

There are a total of $6!$ ways to arrange the kids. If Abby and Bridge are in the first row, we have 4 ways to make them adjacent, and $4!$ ways to arrange the other kids. Similarly, there are $4\times4!$ ways in the second row. If Abby and Bridge are in the first column, we have 2 ways to make them adjacent, and $4!$ ways to arrange the other kids. Similarly, there are $2\times4!$ ways in the second column, and the same for the third column. So the probability is $$\dfrac{2\times(4\times4!)+3\times(2\times4!)}{6!}=\dfrac{7}{15}$$

The clock in Sri's car, which is not accurate, gains time at a constant rate. One day as he begins shopping he notes that his car clock and his watch (which is accurate) both say 12:00 noon. When he is done shopping, his watch says 12:30 and his car clock says 12:35. Later that day, Sri loses his watch. He looks at his car clock and it says 7:00. What is the actual time?

$\textbf{(A) }5:50\qquad\textbf{(B) }6:00\qquad\textbf{(C) }6:30\qquad\textbf{(D) }6:55\qquad \textbf{(E) }8:10$

$\textbf{B}$

For every $35$ minutes the clock passes, the watch passes $30$ minutes. That means that the clock is $\dfrac{7}{6}$ as fast as the watch. When the clock passes 7 hours to 7:00, the watch passes $7\times\dfrac67=6$ hours, which means the actual time is 6:00.

Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?

$\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$

$\textbf{A}$

Apparently the score of the first 4 tests is less than 82, and the score of the last is greater than 82. Every time the score of the first 4 tests drops by 1 point from 82, the score of the last test must increase 4 points to balance the lost. For example, if the score of the first 4 tests is $82-1=81$, than the last test must be $82+4=86$. Hence, the score of the last test could be 86, 90, 94, or 98 (it can not be greater than 100). Only these 4 values are possible.

Let $N$ be the greatest five-digit number whose digits have a product of $120$. What is the sum of the digits of $N$?

$\textbf{(A) }15\qquad\textbf{(B) }16\qquad\textbf{(C)17} \qquad\textbf{(D)}18\qquad\textbf{(E)}20$

$\textbf{D}$

First, we consider whether 9 could be the first digit of the five-digit number. However, it doesn't work because 9 is not a factor of 120. Then we confirm 8 is the first digit because 8 is a factor of 120. The rest 4 digits have a product of 120/8=15. Similarly, the second digit is 5. The rest 3 digits have a product of 15/5=3, which is apparently 3, 1, and 1. So the five-digit number is 85311. The sum of the digits is $8+5+3+1+1=18$.

In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of $1$ square unit, then what is the area of the shaded region, in square units?

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

$\textbf{D}$

Let the radii of the small circles be $r$. Then the radius of the large circle is $2r$. The sum of the areas of the two small circles is $\pi r^2+\pi r^2=2\pi r^2=1$. The area of the shaded region is the area of the large circle subtract by the areas of the two small circles, which is $\pi(2r)^2-2\pi r^2=2\pi r^2=1$.

Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?

$\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880$

$\textbf{C}$

Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just $5!$. However, when we treat the two Arabic books as a whole, the order of the two books can be arranged in $2!$ ways. Similarly, the four Spanish books can be arranged in $4!$ ways. So finally we have $5!\times2!\times4!=5760$ ways to arrange the nine books.

Bella begins to walk from her house toward her friend Ella's house. At the same time, Ella begins to ride her bicycle toward Bella's house. They each maintain a constant speed, and Ella rides 5 times as fast as Bella walks. The distance between their houses is $2$ miles, which is $10,560$ feet, and Bella covers $2 \dfrac{1}{2}$ feet with each step. How many steps will Bella take by the time she meets Ella?

$\textbf{(A) }704\qquad\textbf{(B) }845\qquad\textbf{(C) }1056\qquad\textbf{(D) }1760\qquad \textbf{(E) }3520$

$\textbf{A}$

Since the speed of Ella is 5 times faster than Bella, when Bella moves 2.5 feet in the time of one step, Ella moves $5\times2.5=12.5$ feet. Together they moves $2.5+12.5=15$ feet in the time of one step. Given that the total distance is $10,560$ feet, Bella moves $10560/15=704$ steps when she meets Ella.

How many positive factors does $23,232$ have?

$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

$\textbf{E}$

First we need to find the prime factorization of $23,232$. For such a large number $23,232$, noticing that it is a even number, and $2+3+2+3+2=12$ is a multiple of 3, we can make steps one by one. Finally we get $23232=2^6\times3^1\times11^2$. Now we just add one to the powers and multiply. Therefore, the answer is $(6+1)\times(1+1)\times(2+1)=42$.

In a sign pyramid a cell gets a “$+$” if the two cells below it have the same sign, and it gets a “$-$” if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a “$+$” at the top of the pyramid?

$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

$\textbf{C}$

Let the plus sign represent $1$ and the negative sign represent $-1$. The four numbers on the bottom are $a$, $b$, $c$, and $d$, which are either $1$ or $-1$. The top cell is $ab^3c^3d=1$. Since the square of either $1$ or $-1$ is $1$, the equation can be simplified as $abcd=1$. Among the 4 numbers, we can choose none of them to be $-1$, or 2 of them to be $-1$, or 4 of them to be $-1$. So there are $\dbinom{4}{0}+\dbinom{4}{2}+\dbinom{4}{4}=8$ ways to fill out the bottom cells.

In $\triangle ABC,$ a point $E$ is on $\overline{AB}$ with $AE=1$ and $EB=2.$ Point $D$ is on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$ and point $F$ is on $\overline{BC}$ so that $\overline{EF} \parallel \overline{AC}.$ What is the ratio of the area of $CDEF$ to the area of $\triangle ABC?$

$\textbf{(A) } \dfrac{4}{9} \qquad \textbf{(B) } \dfrac{1}{2} \qquad \textbf{(C) } \dfrac{5}{9} \qquad \textbf{(D) } \dfrac{3}{5} \qquad \textbf{(E) } \dfrac{2}{3}$

$\textbf{A}$

Since $\overline{AE}=\dfrac13\overline{AB}$, and $\overline{BE}=\dfrac23\overline{AB}$, the area of $\triangle ADE$ is $\left(\dfrac13\right)^2=\dfrac19$ of the area of $\triangle ABC$, and the area of $\triangle BEF$ is $\left(\dfrac23\right)^2=\dfrac49$ of the are of $\triangle ABC$. Hence, the ratio of the area of $CDEF$ to the area of $\triangle ABC$ is $1-\dfrac19-\dfrac49=\dfrac49$.

How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?

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

$\textbf{E}$

Looking at the reminders, we notice that they are all 4 less than the divisors as $6-2=4$, $9-5=4$ and $11-7=4$. This means we are looking for a value that is four less than a multiple of $6$, $9$, and $11$. The least common multiple of these numbers is $11\times3^{2}\times2=198$, so the numbers that fulfill this can be written as $198k-4$, where $k$ is a positive integer. This value is only a three-digit integer when $k$ is $1, 2, 3, 4$ or $5$, which gives $194, 392, 590, 788,$ and $986$ respectively. Thus, we have $5$ values.

Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD,$ and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F.$ The area of quadrilateral $AFED$ is $45.$ What is the area of $ABCD?$

$\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144$

$\textbf{B}$

Let the area of $\triangle CEF$ be $x$. Thus, the area of triangle $\triangle ACD$ is $45+x$ and the area of the square is $2(45+x) = 90+2x$. By AA similarity, $\triangle CEF \sim \triangle ABF$ with a 1:2 ratio, so the area of triangle $\triangle ABF$ is $4x$. Now, consider trapezoid $ABED$. Its area is $45+4x$, which is three-fourths the area of the square. We set up an equation in $x$: \[45+4x = \frac{3}{4}\left(90+2x\right)\] By solving the equation, we get $x = 9$. The area of square $ABCD$ is $90+2x=90 + 2 \times 9 =108$.

From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?

$\textbf{(A) } \dfrac{2}{7} \qquad \textbf{(B) } \dfrac{5}{42} \qquad \textbf{(C) } \dfrac{11}{14} \qquad \textbf{(D) } \dfrac{5}{7} \qquad \textbf{(E) } \dfrac{6}{7}$

$\textbf{D}$

We have $\dbinom{8}{3}=56$ ways to choose 3 vertices from 8. When at least one side of the triangle is the also a side of the octagon, at least two vertices are adjacent. So here we have 2 cases: only 2 vertices are adjacent, or all 3 vertices are adjacent.

In case 1, we have 8 choices for the 2 adjacent vertices. When the position of the 2 adjacent vertices is fixed, the rest one has only 4 choices since it can not be placed next to the 2 adjacent vertices. Thus we have $8\times4=32$ choices in case 1.

In case 2, we have simply 8 choices to make 3 vertices adjacent.

Finally, the probability is $\dfrac{32+8}{56}=\dfrac57$.

In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of edges $\overline{FB}$ and $\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$

$\textbf{(A) } \dfrac{5}{4} \qquad \textbf{(B) } \dfrac{4}{3} \qquad \textbf{(C) } \dfrac{3}{2} \qquad \textbf{(D) } \dfrac{25}{16} \qquad \textbf{(E) } \dfrac{9}{4}$

$\textbf{C}$

Let the side length of the cube be 1, the area of a face is 1. For the rhombus $EJCI$, the length of the diagonal $\overline{CE}=\sqrt3$, and the length of another diagonal $\overline{JI}=\sqrt2$. So the area of the rhombus $EJCI$ is $$\dfrac12\overline{CE}\cdot\overline{JI}=\dfrac{\sqrt6}{2}$$ The ratio $R=\dfrac{\sqrt6}{2}$, thus $R^2=\dfrac32$.

How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive?

$\textbf{(A) }4\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }57\qquad \textbf{(E) }58$

$\textbf{E}$

Noticing that $2^8+1=257$, which is greater than $6^3=216$, but less than $7^3=343$. So $7^3$ is the minimum perfect cube. We also notice that $2^{18}+1=\left(2^6\right)^3+1=64^3+1$. Thus $64^3$ is the maximum perfect cube. The number of perfect cube is $64-7+1=58$.