Prove that $\binom{2n}{n}$ is not divisible by $p$

Question

Let $n$ be an integer greater than $5$. I would like to prove that if $p$ is a prime such that $ \frac{2}{3}n < p \leq n$ then $\binom{2n}{n}$ is not a multiple of $p$.

I think I could assume that $\binom{2n}{n}$ is a multiple of $p$ but I do not really know what to make of it.

Answer 1

Recall that $${2n\choose n}=\frac{(2n)!}{n!n!} $$ and that the exponent $k=v_p(m!)$ of $p$ such that $p^k\mid m!$ and $p^{k+1}\mid m!$ is given by $$ v_p(m!)=\left\lfloor \frac mp\right\rfloor+\left\lfloor \frac m{p^2}\right\rfloor+\left\lfloor \frac m{p^3}\right\rfloor+\ldots$$ From the given conditions, $v_p(n!)=1+0+\ldots =1$ and $v_p((2n)!)=2+\lfloor \frac {2}{p}\rfloor+\ldots$, which can only be $>2$ if $p=2$, but $p>\frac23n\ge \frac{10}3>2$. Hence $v_p$ of the binomial is $2-1-1=0$.