Oh god help plz...?

Question

Find all positive intergers n (with n>1) for which n! is divisible by

n
sigma i
i=1

PS: I can't figure out how to express the sum symbol

what is the answer and how do u do it??

Answer 1

Any n such that n+1 is not prime

∑ i = 1+2+3+...+n
= n*(n+1)/2

n! = n*(n-1)*(n-2)...(1)

n! / ∑ = 2*(n-1)! / (n+1)
This is an integer if there exists an (n-k) such that (n+1) is divisible by (n-k). So as long as n+1 is not prime, the condition will be met.

Answer 2

It is easy to prove that if n is odd, then the sum of the integers from 1 to n divides n!, because the sum from 1 to n is n(n + 1)/2. When n is odd, n + 1 is even and (n + 1)/2 is <= n - 1. Therefore n cancels the largest factor in n!, and (n + 1)/ 2 cancels some other factor of n!.

If n is even, the above argument is not available. It is obvious that if n is even and n + 1 is prime, then n(n + 1)/2 cannot divide n!, because n + 1 is larger than all the factors of n!. However, if n is even and n + 1 is composite, it appears that n(n + 1)/2 divides n!, but I don't have a proof.