# Show that if $p$ is a prime satisfying $n<p<2n$ then $\binom{2n}{n}\equiv 0 (mod\space p)$.

Show that if $p$ is a prime satisfying $n<p<2n$ then $\binom{2n}{n}\equiv 0 (mod\space p)$.

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 $(n+1)(n+2)...(n+n)\equiv0 (mod\space p)$. That is $\frac{(2n)!}{n!}\equiv0 (mod\space p)$.

Therefore $\frac{n!}{n!}\frac{(2n)!}{n!}\equiv0 (mod\space p)$. But note that $g.c.d.(n!,p)=1$. So $\frac{(2n)!}{n!n!}\equiv0 (mod\space \frac{p}{g.c.d.(n!,p)})$. Hence the result.

Is the above solution correct? If not, can someone please give me a hint?

• You have a correct idea that is central to a proof. Your finish is a little uneven, in that the $\gcd(n!,p) = 1$ could be developed earlier, and it would be useful in showing $\binom{2n}{n} \equiv 0 \bmod p$. – hardmath Jan 19 '17 at 3:15
• Yes I see it now. Thank you. :) – Janitha357 Jan 19 '17 at 3:32
• How do you go from "Therefore, ...." to "So ....". Is there a result that allows us to go from *** mod something to **** mod something / something else. ? (as you did) – Clclstdnt Jan 8 at 16:51

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 $4.$
• "Divisibility by $p$ is then automatic for all entries in a triangle descending and traveling inwards." I don't understand this part. Please explain. – Janitha357 Jan 19 '17 at 3:30