SOLUTION: What is the smallest natural k for which k! is divisible by 2016? The topic is on modulo arithmetic and I really hope you can solve it:)

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Question 923015: What is the smallest natural k for which k! is divisible by 2016?

The topic is on modulo arithmetic and I really hope you can solve it:)
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
The prime factorization of 2016 is 2^5 * 3^2 * 7.

Therefore k must be at least 7, but 7! only has four 2's in its prime factorization (2, 2*2, 2*3) but enough 3's. Setting k = 8 gives the desired result, as 8! = 2016*20.