# $p$ | $2n\choose n$ if there exists a positive integer $j$ with $n < p^j\leq 2n$

If there exists a positive integer $$j$$ with $$n < p^j\leq 2n$$ then $$p\mid$$ $${2n}\choose {n}$$

I tried something but I can't able to prove, the numerator has more prime power than the denominator. please give me a hint to start with.

• $v_p({2n \choose n}) = \sum_{k \ge 1} \left[\lfloor \frac{2n}{p^k}\rfloor - 2\lfloor \frac{n}{p^k}\rfloor\right]$ – mathworker21 Aug 17 at 10:25
• @mathworker21 $v_p({2n \choose n}) = \sum_{k \ge 1} \left[\lfloor \frac{2n}{p^k}\rfloor - 2\lfloor \frac{n}{p^k}\rfloor\right] \geq 0$ how to prove it is atleast 1 – Cloud JR Aug 17 at 10:31
• use what you're given in problem statement. try to figure it out for a bit – mathworker21 Aug 17 at 10:36

Since $$n < p^{j} \leq 2n$$, it is clear that $$\bigl[\frac{n}{p^j}\bigr]=0$$. Also since $$n, $$2n < 2p^{j} \le p^{j+1}$$ which gives $$\bigl[\frac{2n}{p^j}\bigr]=1$$, which gives us $$\bigl[\frac{2n}{p^j}\bigr]-2\bigl[\frac{n}{p^j}\bigr]=1$$. So $$v_{p}\left(\binom{2n}{n}\right)=\sum_{\substack{k\geq 1 \\ k\neq i}} \left\{\left[\frac{2n}{2^k}\right]-2\left[\frac{n}{p^k}\right]\right\} + \biggl[\frac{2n}{p^j}\biggr]-2\biggl[\frac{n}{p^j}\biggr] \geq 1$$