2015 AMC 10B Problem 23

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.