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Divisibility of factorials

Problem 549 Published on Sunday, 28th February 2016, 10:00 am; Solved by 1981;
Difficulty rating: 15%

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 ≤ in.
S(100)=2012.

Find S(108).