Let $p$ be a prime number such that $n<p<2n$. Prove that:
$$\binom{2n}{n} \equiv 0 (\bmod p)$$
Solution
\[\binom{2n}{n} = \frac{(2n)!}{n!n!}.\] We have that $p|(2n!)$ as $p < 2n$, but $p \nmid (n!)^2$ as $p$ is prime and $p > n$. So $\binom{2n}{n} \equiv 0 \bmod p$.
The exercise can also be found in mathimatikoi.org
$$\binom{2n}{n} \equiv 0 (\bmod p)$$
Solution
\[\binom{2n}{n} = \frac{(2n)!}{n!n!}.\] We have that $p|(2n!)$ as $p < 2n$, but $p \nmid (n!)^2$ as $p$ is prime and $p > n$. So $\binom{2n}{n} \equiv 0 \bmod p$.
The exercise can also be found in mathimatikoi.org
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