The smallest number $m$ such that $10$ divides $m!$ is $m=5$.
The smallest number $m$ such that $25$ divides $m!$ is $m=10$.
Let $s(n)$ be the smallest number $m$ such that $n$ divides $m!$.
So $s(10)=5$ and $s(25)=10$.
Let $S(n)$ be $\sum s(i)$ for $2 \le i \le n$.
$S(100)=2012$.
Find $S(10^8)$.
Solution
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