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1998 AHSME Problems/Problem 13

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Problem

Walter rolls four standard six-sided dice and finds that the product of the numbers of the upper faces is 144. Which of he following could not be the sum of the upper four faces?

\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18

Solution

We have 144 = 2^4 3^2.

As the numbers on the dice are less than 9, the two 3s must come from different dice. This leaves us with three cases: (6,6,a,b), (6,3,a,b), and (3,3,a,b).

In the first case we have ab=2^2, leading to the solutions (6,6,4,1) and (6,6,2,2).

In the second case we have ab=2^3, leading to the only solution (6,3,4,2).

In the third case we have ab=2^4, leading to the only solution (3,3,4,4).

We found all four possibilities for the numbers on the upper faces of the dice. The sums of these numbers are 17, 16, 15, and 14. Therefore the answer is \mathrm{(E)}.

See also

1998 AHSME (Problems)
Preceded by
Problem 12
Followed by
Problem 14
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