# 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 $\boxed{A}$

But this may be slightly complicated, so you can also do solution 2 (scroll to bottom).

## Solution 2

Make a chart with all products. 10 work out of 64. SImplify for 5/32 or $\boxed{A}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 