1970 AHSME Problems/Problem 23

Problem

The number $10!$ ( $10$ is written in base $10$ ), when written in the base $12$ system, ends with exactly $k$ zeros. The value of $k$ is

$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

Solution

A number in base $b$ that ends in exactly $k$ zeros will be divisible by $b^k$ , but not by $b^{k+1}$ . Thus, we want to find the highest $k$ for which $12^k | 10!$ .

There are $4$ factors of $3$ : $3, 6, 9$ , and an extra factor from $9$ .

There are $8$ factors of $2$ : $2, 4, 6, 8, 10$ , an extra factor from $4, 8$ , and a third extra factor from $8$ .

So, $2^8 \cdot 3^4 = 4^4 \cdot 3^4 = 12^4$ will divide $10!$ . Thus, the answer is $\fbox{D}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AHSME Problems and Solutions

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1970 AHSME Problems/Problem 23

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