# Difference between revisions of "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 $\fbox{D}$.