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AMC 10A Question

 
 
NiOliver的头像
AMC 10A Question
NiOliver - 2019年02月8日 Friday 19:24
 

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

(n21)!(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!

 
ReynosoDavid的头像
Re: AMC 10A Question
ReynosoDavid - 2019年02月8日 Friday 20:33
 

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

NiOliver的头像
Re: AMC 10A Question
NiOliver - 2019年02月16日 Saturday 12:59
 
Alright, thank you so much Mr. David!