## Online Course Discussion Forum

### AMC 10A Question

Hello!

I am not sure as to how to solve for problem 25 on the most recent 2019 AMC 10A problem set.The problem is as follows:

For how many integers nn between 11 and 5050, inclusive, is

(n2−1)!(n!)n(n2−1)!(n!)n

an integer? (Recall that 0!=10!=1.)(A) 31(B) 32(C) 33(D) 34(E) 35

Can someone please provide a solution and explanation as to why the answer is (D)?Thanks!

Hey Oliver!

You might want to check out the review we posted online. Dr. Wang goes over the both the 2019 AMC 10A and 12A. You can find it here. Use the enrollment key AMC1012AF to get in.

Best,

Areteem Staff

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