Question

Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n.$ What is the sum of the digits of $s?$

$7$

$8$

$9$

$10$

$11$

Solution:

The number of trailing zeros is the number of factors of $5$ . For $5\le n\le9$ , $n!$ has $k=1$ zero. We need $(2n)!$ to have $3$ zeros, which happens when $15\le2n\le19$ . Thus $n=8,9$ .

For $10\le n\le14$ , $n!$ has $k=2$ zeros. We need $(2n)!$ to have $6$ zeros, which happens when $25\le2n\le29$ . Thus $n=13,14$ .

These are the four least possible values, so $s=8+9+13+14=44$ . The sum of the digits of $s$ is $8$ .

Thus, the correct answer is B.

2015 AMC 10B Problem 23

Solution:

Problem 23 in Other Years