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 August 25th, 2013, 03:36 PM #1 Newbie   Joined: Aug 2013 Posts: 1 Thanks: 0 Factorial and primes Hi, Does anybody where can I find the proof of the following: Let (n,k) be the standard combination (choose k among n). Then if (n,1)=(n,2)=...=(n,n-1)=0 (mod p), then n=p^t for a non-zero integer t, where p is a prime. Thanks
 August 29th, 2013, 09:10 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Factorial and primes It's fairly easy to prove the contrapositive (hence proving the original statement). Suppose that n has a factor other than p. Then $\frac{n!}{p!(n-p)!}$ would not be divisible by p, since the denominator would have the same number of factors of p as the numerator. And if n is less than p, then it is obvious that no combination of n objects would be divisible by p since p would not appear in the numerator or denominator.

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