1996 AMC 8 Solutions | LIVE by Po-Shen Loh

1.

How many positive factors of $36$ are also multiples of $4?$

$2$

$3$

$4$

$5$

$6$

Concepts:factor multiple

Difficulty rating: 560

Small Hint:

List the positive factors of $36$

Big Hint:

Among those factors, keep only the ones divisible by $4$

Solution:

The positive factors of $36$ are $1,$ $2,$ $3,$ $4,$ $6,$ $9,$ $12,$ $18,$ and $36.$ Of these, only $4,$ $12,$ and $36$ are multiples of $4.$

Thus, the correct answer is B .

2.

José, Thuy, and Kareem each start with the number $10.$ José subtracts $1$ from the number $10,$ doubles his answer, and then adds $2.$ Thuy doubles the number $10,$ subtracts $1$ from her answer, and then adds $2.$ Kareem subtracts $1$ from the number $10,$ adds $2$ to his answer, and then doubles the result. Who gets the largest final answer?

José

Thuy

Kareem

José and Thuy

Thuy and Kareem

Concepts:order of operations

Difficulty rating: 730

Small Hint:

Work out each person’s three steps in order, beginning with $10$

Big Hint:

The order of the steps matters; notice who doubles last, so the $2$ they added also gets doubled

Solution:

Starting from $10,$ José computes $9,$ $18,$ $20;$ Thuy computes $20,$ $19,$ $21;$ and Kareem computes $9,$ $11,$ $22.$

Kareem doubles last, so the $2$ he adds is doubled too, giving the largest result.

Thus, the correct answer is C .

3.

The $64$ whole numbers from $1$ through $64$ are written, one per square, on a checkerboard (an $8$ by $8$ array of $64$ squares). The first $8$ numbers are written in order across the first row, the next $8$ across the second row, and so on. After all $64$ numbers are written, the sum of the numbers in the four corners will be

$130$

$131$

$132$

$133$

$134$

Concepts:arithmetic sequence

Difficulty rating: 560

Small Hint:

The first row holds $1$ through $8;$ the last row holds $57$ through $64$

Big Hint:

The four corners are the first and last numbers of the top row and of the bottom row

Solution:

The first row is $1, 2, \ldots, 8$ and the last row is $57, 58, \ldots, 64.$ The four corners are $1,$ $8,$ $57,$ and $64.$

Their sum is $1 + 8 + 57 + 64 = 130.$

Thus, the correct answer is A .

4.

What is the value of the following expression?

$\frac{2 + 4 + 6 + \cdots + 34}{3 + 6 + 9 + \cdots + 51}$

$\dfrac{1}{3}$

$\dfrac{2}{3}$

$\dfrac{3}{2}$

$\dfrac{17}{3}$

$\dfrac{34}{3}$

Concepts:arithmetic sequence ratio and proportion

Difficulty rating: 800

Small Hint:

Factor $2$ out of the numerator and $3$ out of the denominator

Big Hint:

Both sums become the same multiple of $1 + 2 + \cdots + 17$

Solution:

The numerator is $2(1 + 2 + \cdots + 17)$ and the denominator is $3(1 + 2 + \cdots + 17).$

The common factor cancels, leaving $\dfrac{2}{3}.$

Thus, the correct answer is B .

5.

The letters $P,$ $Q,$ $R,$ $S,$ and $T$ represent numbers located on the number line as shown.

Which of the following expressions represents a negative number?

$P - Q$

$P \cdot Q$

$\dfrac{S}{Q} \cdot P$

$\dfrac{R}{P \cdot Q}$

$\dfrac{S + T}{R}$

Concepts:inequality casework

Difficulty rating: 820

Small Hint:

From the number line, $P$ and $Q$ are negative while $R,$ $S,$ and $T$ are positive

Big Hint:

A product or quotient is negative only with an odd number of negative factors; check each choice

Solution:

From the number line, $P$ and $Q$ are negative and $R,$ $S,$ and $T$ are positive.

Then $P \cdot Q$ is positive; $\dfrac{S}{Q} \cdot P$ has two negative factors, so it is positive; $\dfrac{R}{P \cdot Q}$ is positive; and $\dfrac{S + T}{R}$ is positive. Since $P$ is to the left of $Q,$ $P - Q$ is negative.

Thus, the correct answer is A .

6.

What is the smallest result that can be obtained by the following process?

• Choose three different numbers from the set $\{3, 5, 7, 11, 13, 17\}.$

• Add two of these numbers.

• Multiply their sum by the third number.

$15$

$30$

$36$

$50$

$56$

Concepts:optimization casework

Difficulty rating: 820

Small Hint:

To keep the result small, use the three smallest numbers $3,$ $5,$ and $7$

Big Hint:

The smallest number should be the multiplier, since it scales the sum the least

Solution:

Use the three smallest numbers $3,$ $5,$ and $7.$ The choices are $3(5 + 7) = 36,$ $5(3 + 7) = 50,$ and $7(3 + 5) = 56.$

Making the smallest number the multiplier gives the least result, $36.$

Thus, the correct answer is C .

7.

Brent has goldfish that quadruple (become four times as many) every month, and Gretel has goldfish that double every month. If Brent has $4$ goldfish at the same time that Gretel has $128$ goldfish, then in how many months from that time will they have the same number of goldfish?

$4$

$5$

$6$

$7$

$8$

Concepts:geometric sequence exponent

Difficulty rating: 860

Small Hint:

Each month Brent’s count multiplies by $4$ and Gretel’s by $2$

Big Hint:

The ratio of Gretel’s to Brent’s count starts at $128 : 4 = 32$ and halves each month

Solution:

Brent’s counts are $4,$ $16,$ $64,$ $256,$ $1024,$ $4096,$ and Gretel’s are $128,$ $256,$ $512,$ $1024,$ $2048,$ $4096.$

They are equal after $5$ months, when both have $4096.$

Thus, the correct answer is B .

8.

Points $A$ and $B$ are $10$ units apart. Points $B$ and $C$ are $4$ units apart. Points $C$ and $D$ are $3$ units apart. If $A$ and $D$ are as close as possible, then the number of units between them is

$0$

$3$

$9$

$11$

$17$

Concepts:optimization

Difficulty rating: 930

Small Hint:

Place all the points on one line so that $C$ and $D$ are pulled back toward $A$

Big Hint:

With everything collinear, $AD = AB - BC - CD$

Solution:

The distance is smallest when the points are collinear with $C$ and $D$ toward $A:$ take $A = 0,$ $B = 10,$ $C = 6,$ $D = 3.$

Then $AD = 10 - 4 - 3 = 3.$

Thus, the correct answer is B .

9.

If $5$ times a number is $2,$ then $100$ times the reciprocal of the number is

$2.5$

$40$

$50$

$250$

$500$

Concepts:linear equation fraction

Difficulty rating: 730

Small Hint:

The number satisfies $5n = 2,$ so $n = \dfrac{2}{5}$

Big Hint:

The reciprocal of $\dfrac{2}{5}$ is $\dfrac{5}{2};$ multiply it by $100$

Solution:

The number is $\dfrac{2}{5},$ whose reciprocal is $\dfrac{5}{2}.$

Then $100 \cdot \dfrac{5}{2} = 250.$

Thus, the correct answer is D .

10.

When Walter drove up to the gasoline pump, he noticed that his gasoline tank was $\dfrac18$ full. He purchased $7.5$ gallons of gasoline for $\$10.$ With this additional gasoline, his gasoline tank was then $\dfrac58$ full. The number of gallons of gasoline his tank holds when it is full is

$8.75$

$10$

$11.5$

$15$

$22.5$

Concepts:fraction ratio and proportion

Difficulty rating: 860

Small Hint:

Going from $\dfrac18$ to $\dfrac58$ full is an increase of $\dfrac12$ of a tank

Big Hint:

If half a tank is $7.5$ gallons, double it for a full tank

Solution:

The increase is $\dfrac58 - \dfrac18 = \dfrac12$ of a tank, which equals $7.5$ gallons.

So a full tank holds $2 \cdot 7.5 = 15$ gallons.

Thus, the correct answer is D .

11.

Let $x$ be the number

$0.\underbrace{0000\ldots00001}_{1996 \text{ zeros}},$

where there are $1996$ zeros after the decimal point. Which of the following expressions represents the largest number?

$3 + x$

$3 - x$

$3 \cdot x$

$\frac{3}{x}$

$\frac{x}{3}$

Concepts:estimation

Difficulty rating: 860

Small Hint:

$x$ is a tiny positive number, just above $0$

Big Hint:

Dividing $3$ by a tiny number gives a huge result, while the other choices stay near $3$ or near $0$

Solution:

Since $x$ is a very small positive number, $3 + x$ and $3 - x$ are near $3,$ while $3 \cdot x$ and $\frac{x}{3}$ are near $0.$

But $\frac{3}{x}$ is $3$ followed by $1997$ zeros, far larger than any other choice.

Thus, the correct answer is D .

12.

What number should be removed from the list

$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11$

so that the average of the remaining numbers is $6.1?$

$4$

$5$

$6$

$7$

$8$

Concepts:mean work backwards

Difficulty rating: 820

Small Hint:

The eleven numbers add up to $66$

Big Hint:

Ten numbers averaging $6.1$ must total $61;$ find what was removed

Solution:

The sum of $1$ through $11$ is $66.$ For ten numbers to average $6.1,$ their sum must be $10 \cdot 6.1 = 61.$

So the removed number is $66 - 61 = 5.$

Thus, the correct answer is B .

13.

In the fall of $1996,$ a total of $800$ students participated in an annual school clean-up day. The organizers of the event expect that in each of the years $1997,$ $1998,$ and $1999,$ participation will increase by $50\%$ over the previous year. The number of participants the organizers expect in the fall of $1999$ is

$1200$

$1500$

$2000$

$2400$

$2700$

Concepts:percentage geometric sequence

Difficulty rating: 930

Small Hint:

A $50\%$ increase multiplies the number by $1.5$

Big Hint:

Apply the factor $1.5$ three times, for $1997,$ $1998,$ and $1999$

Solution:

Each year multiplies the count by $1.5:$ $800 \to 1200 \to 1800 \to 2700.$

So $800 \cdot 1.5^3 = 2700$ participants are expected in $1999.$

Thus, the correct answer is E .

14.

Six different digits from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ are placed in the squares in the figure shown so that the sum of the entries in the vertical column is $23$ and the sum of the entries in the horizontal row is $12.$ The sum of the six digits used is

$27$

$29$

$31$

$33$

$35$

Concepts:digits logical deduction

Difficulty rating: 1090

Small Hint:

Three distinct digits summing to $23$ must be $6,$ $8,$ and $9$

Big Hint:

The shared square is counted in both sums, so the total of the six digits is $23 + 12$ minus the shared digit

Solution:

Three distinct digits from $1$ through $9$ summing to $23$ must be $6,$ $8,$ and $9.$ The row’s other three digits are at least $1 + 2 + 3 = 6,$ so the shared square (belonging to both the column and the row) is at most $12 - 6 = 6.$ Hence the shared digit is $6.$

The six digits are then $6,$ $8,$ $9,$ $1,$ $2,$ and $3,$ whose sum is $29.$ Equivalently, $23 + 12 - 6 = 29.$

Thus, the correct answer is B .

15.

The remainder when the product $1492 \cdot 1776 \cdot 1812 \cdot 1996$ is divided by $5$ is

$0$

$1$

$2$

$3$

$4$

Concepts:units digit modular arithmetic

Difficulty rating: 930

Small Hint:

The remainder upon division by $5$ depends only on the units digit of the product

Big Hint:

Multiply the units digits $2,$ $6,$ $2,$ and $6$ and look at that product’s units digit

Solution:

The units digit of the product equals the units digit of $2 \cdot 6 \cdot 2 \cdot 6 = 144,$ which is $4.$

A number ending in $4,$ when divided by $5,$ leaves remainder $4.$

Thus, the correct answer is E .

16.

What is the value of the following expression?

$\begin{aligned} &1 - 2 - 3 + 4 + 5 - 6 - 7 \\ &\quad {}+ 8 + 9 - 10 - 11 + 12 \\ &\quad {}+ 13 - \cdots + 1992 \\ &\quad {}+ 1993 - 1994 \\ &\quad {}- 1995 + 1996 \end{aligned}$

$-998$

$-1$

$0$

$1$

$998$

Concepts:summation pairing and grouping

Difficulty rating: 1060

Small Hint:

Group the terms in blocks of four starting from the first: $(1 - 2 - 3 + 4),$ $(5 - 6 - 7 + 8),$ and so on

Big Hint:

Each block of four sums to $0;$ check that the $1996$ terms divide evenly into such blocks

Solution:

Grouping in blocks of four gives $1 - 2 - 3 + 4 = 0,$ $5 - 6 - 7 + 8 = 0,$ and so on.

There are $\frac{1996}{4} = 499$ such blocks, each equal to $0,$ so the total is $0.$

Thus, the correct answer is C .

17.

Figure $OPQR$ is a square. Point $O$ is the origin, and point $Q$ has coordinates $(2, 2).$ What are the coordinates for $T$ so that the area of triangle $PQT$ equals the area of square $OPQR?$

$(-6, 0)$

$(-4, 0)$

$(-2, 0)$

$(2, 0)$

$(4, 0)$

Concepts:coordinate geometry triangle area

Difficulty rating: 1090

Small Hint:

The square has side $2,$ so its area is $4;$ also $P = (2, 0)$ and the base $PQ$ is vertical with length $2$

Big Hint:

If $T = (t, 0),$ its horizontal distance from line $PQ$ is $\lvert 2 - t\rvert$

Solution:

Since $OPQR$ is a square with $O = (0, 0)$ and $Q = (2, 2),$ we have $P = (2, 0)$ and $R = (0, 2),$ so the area is $2^2 = 4.$

Triangle $PQT$ has vertical base $PQ$ of length $2,$ and $T = (t, 0)$ lies on the $x$ -axis. Its area is $\tfrac12 \cdot 2 \cdot \lvert 2 - t\rvert = \lvert 2 - t\rvert.$ Setting $\lvert 2 - t\rvert = 4$ gives $t = -2$ or $t = 6.$ Only $t = -2$ appears among the choices, so $T = (-2, 0).$

Thus, the correct answer is C .

18.

Ana’s monthly salary was $\$2000$ in May. In June she received a $20\%$ raise. In July she received a $20\%$ pay cut. After the two changes in June and July, Ana’s monthly salary was

$\$1920$

$\$1980$

$\$2000$

$\$2020$

$\$2040$

Concepts:percentage

Difficulty rating: 960

Small Hint:

A $20\%$ raise multiplies by $1.2;$ a $20\%$ cut multiplies by $0.8$

Big Hint:

Apply both factors in turn: $2000 \cdot 1.2 \cdot 0.8$

Solution:

After the raise, the salary is $2000 \cdot 1.2 = 2400.$

After the cut, it is $2400 \cdot 0.8 = 1920.$

Thus, the correct answer is A .

19.

The pie charts below indicate the percent of students who prefer golf, bowling, or tennis at East Junior High School and West Middle School. The total number of students at East is $2000$ and at West, $2500.$ In the two schools combined, the percent of students who prefer tennis is

$30\%$

$31\%$

$32\%$

$33\%$

$34\%$

Concepts:percentage mean

Difficulty rating: 1090

Small Hint:

Find the actual number of tennis fans at each school, not just the percents

Big Hint:

Add the two counts and divide by the total of $4500$ students

Solution:

East has $0.22 \cdot 2000 = 440$ tennis fans, and West has $0.40 \cdot 2500 = 1000.$

Together $1440$ of the $4500$ students prefer tennis, which is $\dfrac{1440}{4500} = 32\%.$

Thus, the correct answer is C .

20.

Suppose there is a special key on a calculator that replaces the number $x$ currently displayed with the number given by the formula $\frac{1}{1 - x}.$ For example, if the calculator is displaying $2$ and the special key is pressed, then the calculator will display $-1$ since $\frac{1}{1 - 2} = -1.$ Now suppose that the calculator is displaying $5.$ After the special key is pressed $100$ times in a row, the calculator will display

$-0.25$

$0$

$0.8$

$1.25$

$5$

Concepts:function modular arithmetic

Difficulty rating: 1280

Small Hint:

Apply the key a few times starting from $5$ and watch for a repeating cycle

Big Hint:

The values cycle with period $3;$ find where $100$ lands in the cycle

Solution:

Starting from $5:$ $\frac{1}{1 - 5} = -0.25,$ then $\frac{1}{1 + 0.25} = 0.8,$ then $\frac{1}{1 - 0.8} = 5.$ The values repeat with period $3.$

Since $100 = 3 \cdot 33 + 1,$ the $100$ th press gives the same result as the first press, $-0.25.$

Thus, the correct answer is A .

21.

How many subsets containing three different numbers can be selected from the set $\{89, 95, 99, 132, 166, 173\}$ so that the sum of the three numbers is even?

$6$

$8$

$10$

$12$

$24$

Concepts:parity combinations

Difficulty rating: 1200

Small Hint:

With only two even numbers available, an even sum must use two odd numbers and one even

Big Hint:

There are $4$ odd and $2$ even numbers; choose $2$ of the $4$ odds and $1$ of the $2$ evens

Solution:

The set has $4$ odd numbers— $89,$ $95,$ $99,$ and $173$ —and $2$ even numbers— $132$ and $166.$ A sum of three is even only with two odds and one even, since three evens is impossible with just two available.

The count is $\binom{4}{2} \cdot \binom{2}{1} = 6 \cdot 2 = 12.$

Thus, the correct answer is D .

22.

The horizontal and vertical distances between adjacent points equal $1$ unit. The area of triangle $ABC$ is

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

$1$

$\dfrac{5}{4}$

Concepts:Pick’s Theorem triangle area

Difficulty rating: 1140

Small Hint:

Assign coordinates to $A,$ $B,$ and $C$ and use the determinant formula for area

Big Hint:

Alternatively use Pick’s theorem, area $= I + \tfrac{B}{2} - 1,$ with $I = 0$ interior points and $B = 3$ boundary points

Solution:

Taking $A = (0, 0),$ $B = (3, 2),$ and $C = (4, 3),$ the determinant formula gives $\frac12\left\lvert 3\cdot3-2\cdot4\right\rvert=\frac12.$

Equivalently, by Pick’s theorem with no interior lattice points and $3$ boundary points, the area is $0 + \tfrac32 - 1 = \tfrac12.$

Thus, the correct answer is B .

23.

The manager of a company planned to distribute a $\$50$ bonus to each employee from the company fund, but the fund contained $\$5$ less than what was needed. Instead the manager gave each employee a $\$45$ bonus and kept the remaining $\$95$ in the company fund. The amount of money in the company fund before any bonuses were paid was

$\$945$

$\$950$

$\$955$

$\$990$

$\$995$

Concepts:linear equation

Difficulty rating: 1090

Small Hint:

Let $n$ be the number of employees; the fund equals $50n - 5$ and also $45n + 95$

Big Hint:

Set the two expressions for the fund equal to solve for $n,$ then compute the fund

Solution:

Let $n$ be the number of employees. The fund is $50n - 5$ (five dollars short of $50$ each) and also $45n + 95.$

Setting $50n - 5 = 45n + 95$ gives $5n = 100,$ so $n = 20.$ The fund is $45 \cdot 20 + 95 = 995.$

Thus, the correct answer is E .

24.

The measure of angle $ABC$ is $50^\circ.$ $\overline{AD}$ bisects angle $BAC,$ and $\overline{DC}$ bisects angle $BCA.$ The measure of angle $ADC$ is

$90^\circ$

$100^\circ$

$115^\circ$

$122.5^\circ$

$125^\circ$

Concepts:angle chasing angle bisector

Difficulty rating: 1150

Small Hint:

In triangle $ABC,$ angles $BAC$ and $BCA$ add to $180^\circ - 50^\circ = 130^\circ$

Big Hint:

The bisected halves at $A$ and $C$ add to $65^\circ;$ use the angle sum in triangle $ADC$

Solution:

In triangle $ABC,$ $\begin{aligned} \angle BAC + \angle BCA &= 180^\circ - 50^\circ \\ &= 130^\circ. \end{aligned}$

The bisectors give $\angle DAC + \angle DCA = \tfrac{130^\circ}{2} = 65^\circ.$ In triangle $ADC,$ $\angle ADC = 180^\circ - 65^\circ = 115^\circ.$

Thus, the correct answer is C .

25.

A point is chosen at random from within a circular region. What is the probability that the point is closer to the center of the region than it is to the boundary of the region?

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Concepts:geometric probability area ratio

Difficulty rating: 1260

Small Hint:

A point is closer to the center than the boundary exactly when its distance from the center is less than half the radius

Big Hint:

Compare the area of the inner circle of radius $\tfrac12$ to the whole circle of radius $1$

Solution:

Take the radius to be $1.$ A point at distance $r$ from the center is closer to the center than to the boundary when $r \lt 1 - r,$ i.e. $r \lt \tfrac12.$

The favorable region is a circle of radius $\tfrac12,$ with area $\pi(\tfrac12)^2 = \tfrac{\pi}{4},$ out of the total area $\pi.$ The probability is $\dfrac{\frac{\pi}{4}}{\pi} = \dfrac14.$

Thus, the correct answer is A .