## AMC 8 2000

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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.

Aunt Anna is $42$ years old. Caitlin is $5$ years younger than Brianna, and Brianna is half as old as Aunt Anna. How old is Caitlin?

$\text{(A)}\ 15 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 37$

$\textbf{B}$

Brianna is $42/2=21$ years old. Caitlin is $21-5=16$ years old.

Which of these numbers is less than its reciprocal?

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

$\textbf{A}$

The reciprocal of $-2$ is $-\dfrac12$. Apparently $-2<-\dfrac12$.

How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi?$

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$

How many whole numbers lie in the interval between $\dfrac{5}{3}$ and $2\pi?$

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$

In $1960$ only $5\%$ of the working adults in Carlin City worked at home. By $1970$ the "at-home" work force increased to $8\%$. In $1980$ there were approximately $15\%$ working at home, and in $1990$ there were $30\%$. The graph that best illustrates this is

$\textbf{E}$

The coordinates of each point in graph E coordinate with the description of the question.

Each principal of Lincoln High School serves exactly one $3$-year term. What is the maximum number of principals this school could have during an 8-year period?

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 8$

$\textbf{C}$

If the first year of the $8$-year period was the final year of a principal's term, then in the next six years two more principals would serve, and the last year of the period would be the first year of the fourth principal's term. Therefore, the maximum number of principals who can serve during an $8$-year period is $4$ if the terms are divided $1\ |\ 2\ 3\ 4\ |\ 5\ 6\ 7\ |\ 8$.

Figure $ABCD$ is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded $L$-shaped region is

$\text{(A)}\ 7 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

$\textbf{A}$

The area of the shaded region is the area of a $4\times4$ square subtracting the area of a $3\times3$ square. So the answer is $4\times4-3\times3=7$.

What is the minimum possible product of three different numbers of the set $\{-8,-6,-4,0,3,5,7\}$?

$\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0$

$\textbf{B}$

The minimum product should be a negative integer. We can choose 3 negative numbers to get a negative product, or 1 negative number with 2 positive numbers. By comparing $(-8)\times(-6)\times(-4) =-192$ and $(-8)\times5\times7= -280$, the answer is $-280$.

Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is

$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 41 \qquad \text{(E)}\ 53$

$\textbf{D}$

The total number of all dots of the 3 dice is $3(1+2+3+4+5+6)=63$, and the total number of all visible dots in this view is $1+2+3+4+6+5+1=22$. So the total number of dots which are not visible in this view is $63-22=41$.

Three-digit powers of $2$ and $5$ are used in this “cross-number” puzzle. What is the only possible digit for the outlined square?

$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

$\textbf{D}$

The $3$-digit powers of $5$ are $125$ and $625$ in the column, so space $2$ is filled with a $2$. The only $3$-digit power of $2$ beginning with $2$ is $256$ in the row, so the outlined block is filled with a $6$.

Ara and Shea were once the same height. Since then Shea has grown $20\%$ while Ara has grown half as many inches as Shea. Shea is now 60 inches tall. How tall, in inches, is Ara now?

$\text{(A)}\ 48 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 52 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$

$\textbf{E}$

Let the original height of Ara and Shea be $h$. Now Shea is $1.2h$ and Ara is $1.1h$, so Ara is $\dfrac{11}{12}$ the height of Shea. Given that Shea is 60 inches now, the height of Ara should be $60\times\dfrac{11}{12}=55$ inches.

The number $64$ has the property that it is divisible by its units digit. How many whole numbers between 10 and 50 have this property?

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

$\textbf{C}$

Casework by the units digit $u$ will help organize the answer.

$u=0$ gives no solutions, since no real numbers are divisible by $0$.

$u=1$ has $4$ solutions, since all positive integers are divisible by $1$.

$u=2$ has $4$ solutions, since every number ending in $2$ is divisible by $2$.

$u=3$ has $1$ solution: $33$. $33\pm 10$ or $33\pm 20$ will retain the units digit, but will stop the number from being divisible by $3$. $33\pm 30$ is the smallest multiple of $10$ that will keep the number divisible by $3$, but those numbers are $3$ and $63$, which are out of the range of the problem.

$u=4$ has $2$ solutions: $24$ and $44$. Adding or subtracting $10$ will kill divisibility by $4$, since $10$ is not divisible by $4$.

$u=5$ has $4$ solutions. Every number ending in $5$ is divisible by $5$.

$u=6$ has $1$ solution: $36$. $36\pm 10$ or $36\pm 20$ will kill divisibility by $3$, and thus kill divisibility by $6$.

$u=7$ has no solutions. The first multiples of $7$ that end in $7$ are $7$ and $77$, but both are outside of the range of this problem.

$u=8$ has $1$ solution: $48$. $\pm 10, \pm 20, \pm 30$ will all kill divisibility by $8$ since $10, 20,$ and $30$ are not divisible by $8$.

$u=9$ has no solutions. $9$ and $99$ are the smallest multiples of $9$ that end in $9$.

In conclusion, we have $0 + 4 + 4 + 1 + 2 + 4 + 1 + 0 + 1 + 0 = 17$ solutions.

A block wall 100 feet long and 7 feet high will be constructed using blocks that are 1 foot high and either 2 feet long or 1 foot long (no blocks may be cut). The vertical joins in the blocks must be staggered as shown, and the wall must be even on the ends. What is the smallest number of blocks needed to build this wall?

$\text{(A)}\ 344 \qquad \text{(B)}\ 347 \qquad \text{(C)}\ 350 \qquad \text{(D)}\ 353 \qquad \text{(E)}\ 356$

$\textbf{D}$

Since the bricks are $1$ foot high, there will be $7$ rows. To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $100/2 = 50$ bricks to construct. Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks in the middle, and $2$ 1-foot bricks on each end for a total of $51$ bricks.

Four rows of $50$ bricks and three rows of $51$ bricks totals $4\times 50 + 3\times 51 = 200 + 153 = 353$ bricks.

In triangle $CAT$, we have $\angle ACT = \angle ATC$ and $\angle CAT = 36^\circ$. If $\overline{TR}$ bisects $\angle ATC$, then $\angle CRT =$

$\text{(A)}\ 36^\circ \qquad \text{(B)}\ 54^\circ \qquad \text{(C)}\ 72^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 108^\circ$

$\textbf{C}$

In $\triangle ACT$, $\angle ACT=\angle ATC=\dfrac{180^\circ-36^\circ}{2}=72^\circ$. Since $\overline{TR}$ bisects $\angle ATC$, we have $\angle CTR=\dfrac12\angle ATC=36^\circ$. Hence, in $\triangle CRT$, $\angle CRT=180^\circ-72^\circ-36^\circ=72^\circ$.

What is the units digit of $19^{19} + 99^{99}$?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

$\textbf{D}$

Since we only care about the units digit, we focus on $9^{19} + 9^{99}$. We found the pattern of the units digit that odd powers of 9 have a units digit of 9, and even powers of 9 have a units digit of 1. So the given number can be simplified as $9+9=18$. Hence, the units digit is 8.

Triangles $ABC$, $ADE$, and $EFG$ are all equilateral. Points $D$ and $G$ are midpoints of $\overline{AC}$ and $\overline{AE}$, respectively. If $AB = 4$, what is the perimeter of figure $ABCDEFG$?

$\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 21$

$\textbf{C}$

The large triangle $ABC$ has sides of length $4$. The medium triangle $ADE$ has sides of length $2$. The small triangle $EFG$ has sides of length $1$. There are $3$ segment sizes, and all segments depicted are one of these lengths.

Starting at $A$ and going clockwise, the perimeter is:$$AB + BC + CD + DE + EF + FG + GA=4 + 4 + 2 + 2 + 1 + 1 + 1=15$$

In order for Mateen to walk a kilometer (1000m) in his rectangular backyard, he must walk the length 25 times or walk its perimeter 10 times. What is the area of Mateen's backyard in square meters?

$\text{(A)}\ 40 \qquad \text{(B)}\ 200 \qquad \text{(C)}\ 400 \qquad \text{(D)}\ 500 \qquad \text{(E)}\ 1000$

$\textbf{C}$

The length of the rectangular backyard is $1000/25=40\ \text{m}$, and the perimeter of the yard is $1000/10=100\ \text{m}$. So the width of the yard is $\dfrac{100}{2}-40=10\ \text{m}$. Therefore, the area of the yard is $40\times10=400\ \text{m}^2$.

The operation $\otimes$ is defined for all nonzero numbers by $a\otimes b = \dfrac{a^2}{b}$. Determine $[(1\otimes 2)\otimes 3] - [1\otimes (2\otimes 3)]$.

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

$\textbf{A}$

\begin{align*}

[(1\otimes 2)\otimes 3]-[1\otimes (2\otimes 3)]&=[\dfrac{1^2}{2}\otimes 3]-[1\otimes \frac{2^2}{3}]\\

&=[\frac{1}{2}\otimes 3]-[1\otimes \frac{4}{3}]\\

&=[\frac{(\frac{1}{2})^{2}}{3}]-[\frac{1^2}{(\frac{4}{3})}]\\

&=\frac{1}{12} - \frac{3}{4}\\

&=-\frac{2}{3}

\end{align*}

Consider these two geoboard quadrilaterals. Which of the following statements is true?

$\text{(A)}\ \text{The area of quadrilateral I is more than the area of quadrilateral II.}$

$\text{(B)}\ \text{The area of quadrilateral I is less than the area of quadrilateral II.}$

$\text{(C)}\ \text{The quadrilaterals have the same area and the same perimeter.}$

$\text{(D)}\ \text{The quadrilaterals have the same area, but the perimeter of I is more than the perimeter of II.}$

$\text{(E)}\ \text{The quadrilaterals have the same area, but the perimeter of I is less than the perimeter of II.}$

$\textbf{E}$

The area of figure I is $1\times1=1$. The area of figure II is the sum of two triangles $\dfrac12(1)(1)+\dfrac12(1)(1)=1$. The perimeter of figure 1 is $1+\sqrt2+1+\sqrt2=2+2\sqrt2$. The perimeter of figure II is $\sqrt5+\sqrt2+1+\sqrt2=1+2\sqrt2+\sqrt5>2+2\sqrt2$. So the answer is $E$.

Three circular arcs of radius 5 units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region?

$\text{(A)}\ 25 \qquad \text{(B)}\ 10 + 5\pi \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 50 + 5\pi \qquad \text{(E)}\ 25\pi$

$\textbf{C}$

Draw line $\overline{BD}$. Then draw $\overline {CO}$, where $O$ is the center of the semicircle. You have two quarter circles on top, and two quarter circle-sized "bites" on the bottom. Move the pieces from the top to fit in the bottom like a jigsaw puzzle. You now have a rectangle with length $\overline {BD}$ and height $\overline {AO}$, which are equal to $10$ and $5$, respectively. Thus, the total area is $50$.

You have nine coins: a collection of pennies, nickels, dimes, and quarters having a total value of $\$1.02$, with at least one coin of each type. How many dimes must you have?

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

$\textbf{A}$

Since you have one coin of each type, $1 + 5 + 10 + 25 = 41$ cents are already determined, leaving you with a total of $102 - 41 = 61$ cents remaining for $5$ coins.

In order to get 61 cents, you must have $1$ more penny. If you had more than $1$ penny, you must have at least $6$ pennies to leave a multiple of $5$ for the nickels, dimes, and quarters. But you only have $5$ more coins to assign.

Now you have $61 - 1 = 60$ cents remaining for $4$ coins, which may be nickels, dimes, or quarters. If you have 2 more quarters, the rest 2 coins should be 2 nickels. If you have only 1 more quarter, the value of the rest 3 coins is 35 cents, which is impossible for nickels and dimes. Similarly, it is impossible that you have no more quarter.

In conclusion, the 9 coins are 2 pennies, 3 nickels, 1 dime and 3 quarters. The answer is $\text{A}$.

Keiko tosses one penny and Ephraim tosses two pennies. The probability that Ephraim gets the same number of heads that Keiko gets is:

$\text{(A)}\ \dfrac{1}{4} \qquad \text{(B)}\ \dfrac{3}{8} \qquad \text{(C)}\ \dfrac{1}{2} \qquad \text{(D)}\ \dfrac{2}{3} \qquad \text{(E)}\ \dfrac{3}{4}$

$\textbf{B}$

Let $K(n)$ be the probability that Keiko gets $n$ heads, and $E(n)$ be the probability that Ephraim gets $n$ heads.

$$K(0) = \dfrac{1}{2}$$

$$K(1) = \dfrac{1}{2}$$

$$K(2) = 0\ \text{(Keiko has only one penny!)}$$

$$E(0) = \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{1}{4}$$

$$E(1) = \dfrac{1}{2}\cdot\dfrac{1}{2} + \dfrac{1}{2}\cdot\dfrac{1}{2} = 2\cdot\dfrac{1}{4} = \dfrac{1}{2}\ \text{(because Ephraim can get HT or TH)}$$

$$E(2) = \dfrac{1}{2}\cdot\dfrac{1}{2} = \dfrac{1}{4}$$

The probability that Keiko gets $0$ heads and Ephraim gets $0$ heads is $K(0)\cdot E(0)$. Similarly for $1$ head and $2$ heads. Thus, we have:$$P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2)=\dfrac{1}{2}\cdot\dfrac{1}{4} + \dfrac{1}{2}\cdot\dfrac{1}{2} + 0 = \dfrac{3}{8}$$

A cube has edge length 2. Suppose that we glue a cube of edge length 1 on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to

$\text{(A)}\ 10 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 17 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 25$

$\textbf{C}$

The surface area of the large cube is $6\times2\times2=24$, and the surface area of the little cube is $6\times1\times1=6$. When they are glued together, we need to subtract 2 contact surfaces of area $1\times1=1$ each. So the total surface area of the new figure is $24+6-2=28$, which is $\dfrac{28-24}{24}=\dfrac16\approx 17\%$ larger than the large cube.

There is a list of seven numbers. The average of the first four numbers is 5, and the average of the last four numbers is 8. If the average of all seven numbers is $6\dfrac{4}{7}$, then the number common to both sets of four numbers is

$\text{(A)}\ 5\dfrac{3}{7} \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 6\dfrac{4}{7} \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 7\dfrac{3}{7}$

$\textbf{B}$

The sum of the first 4 numbers is $4\times5=20$. The sum of the last 4 numbers is $4\times8=32$. The sum of all 7 numbers is $7\times6\dfrac47=46$. Hence, the number common to both sets, which is the 4th number, is $20+32-46=6$.

If $\angle A = 20^\circ$ and $\angle AFG = \angle AGF$, then $\angle B + \angle D =$

$\text{(A)}\ 48^\circ \qquad \text{(B)}\ 60^\circ \qquad \text{(C)}\ 72^\circ \qquad \text{(D)}\ 80^\circ \qquad \text{(E)}\ 90^\circ$

$\textbf{D}$

In $\triangle BFD$, $\angle B+\angle D=180^\circ-\angle BFD=\angle AFG$. In $\triangle AFG$, $\angle AFG=\dfrac{180^\circ-\angle A}{2}=\dfrac{180^\circ-20^\circ}{2}=80^\circ$. So the answer is D.

The area of rectangle $ABCD$ is 72. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is

$\text{(A)}\ 21 \qquad \text{(B)}\ 27 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 40$

$\textbf{B}$

Let the midpoints of $\overline{BC}$ and $\overline{CD}$ be $E$ and $F$. The area of $\triangle ABE$ is 1/4 of the area of rectangle $ABCD$. The area of $\triangle ADF$ is also 1/4 of the area of rectangle $ABCD$. The area of $\triangle CEF$ is 1/8 of the area of rectangle $ABCD$. So the area of $\triangle AEF$ is $1-\dfrac14-\dfrac14-\dfrac18=\dfrac38$ of the area of rectangle $ABCD$. The answer is $72\cdot\dfrac38=27$.