Show that if $p$ is a prime satisfying $n<p<2n$ then $\binom{2n}{n}\equiv 0 \pmod p$.
The following is the attempted solution.
Let $n\in \mathbb{Z}$. Suppose $p$ is a prime such that $n<p<2n$. Then $p\in\{n+1,n+2,...,2n-1\}$. But $\binom{2n}{n}=\frac{(2n)!}{n!n!}=\frac{(n+1)(n+2)...(n+n)}{n!}$.
And since $p\in\{n+1,n+2,...,2n-1\}$ $\space$ we have $(n+1)(n+2)...(n+n)\equiv0 \pmod p$. That is $\frac{(2n)!}{n!}\equiv0 \pmod p$.
Therefore $\frac{n!}{n!}\frac{(2n)!}{n!}\equiv0 \pmod p$. But note that $g.c.d.(n!,p)=1$. So $\frac{(2n)!}{n!n!}\equiv0 \pmod{\frac{p}{g.c.d.(n!,p)}}$. Hence the result.
Is the above solution correct? If not, can someone please give me a hint?
I like the part about how all the entries in line $p,$ with $p $ PRIME, are divisible by $p$ except the endpoints. Divisiblity by $p$ is then automatic for all entries in a triangle descending and travelling inwards. Let me paste a picture and then identify an example...
Well, here is the easy one to see: look at all the entries in rows 5,6,7,8 that are divisible by $5,$ a kind of upside down triangle with apex at $70,$ which is $8$ choose $4.$