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 Post subject: Congruency of binomial coefficientPosted: Fri Dec 11, 2015 8:32 pm
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Location: Larisa
Let $p$ be a prime number such that $n<p<2n$. Prove that:

$$\binom{2n}{n} \equiv 0 (\bmod p)$$

Ι don't have a solution.

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 Post subject: Re: Congruency of binomial coefficientPosted: Tue Dec 15, 2015 11:06 am

Joined: Mon Nov 09, 2015 11:52 am
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Location: Limassol/Pyla Cyprus
Well, $\binom{2n}{n} = \frac{(2n)!}{n!n!}.$ We have that $p|(2n!)$ as $p < 2n$, but $p \nmid (n!)^2$ as $p$ is prime and $p > n$. So $\binom{2n}{n} \equiv 0 \bmod p$.

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