1997 AJHSME Problems/Problem 20

Problem

A pair of 8-sided dice have sides numbered 1 through 8. Each side has the same probability (chance) of landing face up. The probability that the product of the two numbers that land face-up exceeds 36 is

$\text{(A)}\ \dfrac{5}{32} \qquad \text{(B)}\ \dfrac{11}{64} \qquad \text{(C)}\ \dfrac{3}{16} \qquad \text{(D)}\ \dfrac{1}{4} \qquad \text{(E)}\ \dfrac{1}{2}$

Solution

There are $8 \times 8 = 64$ combinations to examine.

If one die is $1$ , then even with an $8$ on the other die, no combinations will work.

If one die is $2$ , then even with an $8$ on the other die, no combinations will work.

If one die is $3$ , then even with an $8$ on the other die, no combinations will work.

If one die is $4$ , then even with an $8$ on the other die, no combinations will work.

If one die is $5$ , then the other die must be an $8$ to have a product over $36$ . Thus, $(5,8)$ works.

If one die is $6$ , then the other die must be either $7$ or $8$ to have a product over $36$ . Thus, $(6, 7)$ and $(6, 8)$ both work.

If one die is $7$ , then the other die can be $6, 7,$ or $8$ to have a product over $36$ . Thus, $(7, 6)$ , $(7, 7)$ , and $(7, 8)$ all work.

If one die is $8$ , then the other die can be $5, 6, 7,$ or $8$ to have a product over $36$ . Thus, $(8, 5), (8,6), (8,7),$ and $(8,8)$ work.

There are a total of $1 + 2 + 3 + 4 = 10$ combinations that work out of a total of $64$ possibilities.

Thus, the answer is $\frac{10}{64} = \frac{5}{32}$ , and the answer is $\boxed{A}$

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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1997 AJHSME Problems/Problem 20

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