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 $3$ 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 $\mathrm{(E)}$ .

1998 AHSME (Problems)
Preceded by Problem 12	Followed by Problem 14
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1998 AHSME Problems/Problem 13

From AoPSWiki

Problem

Solution

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