1991 AJHSME Problems/Problem 13

Problem

How many zeros are at the end of the product $25\times 25\times 25\times 25\times 25\times 25\times 25\times 8\times 8\times 8?$

$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution

$\begin{align*} (5^2)^7\times (2^3)^3 &= 5^{14}\times 2^{9} \\ &= 5^5\times 10^9 \end{align*}$ Since $5^5$ doesn't end in a $0$ , we can conclude that the product ends in $9$ zeroes $\rightarrow \boxed{\text{C}}$ .

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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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1991 AJHSME Problems/Problem 13

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