# Problem 549: Divisibility of factorials
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
∑s(i) for 2 ≤ i ≤ n. S(100)=2012. Find S(108).