1998 AHSME Problems/Problem 13

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 $3$s 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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