Online Course Discussion Forum

AMC 10A Question

 
 
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AMC 10A Question
by Oliver Ni - Friday, 8 February 2019, 7:24 PM
 

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!

 
Picture of David Reynoso
Re: AMC 10A Question
by David Reynoso - Friday, 8 February 2019, 8:33 PM
 

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

Picture of Oliver Ni
Re: AMC 10A Question
by Oliver Ni - Saturday, 16 February 2019, 12:59 PM
 
Alright, thank you so much Mr. David!