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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.

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    $\begingroup$ $v_p({2n \choose n}) = \sum_{k \ge 1} \left[\lfloor \frac{2n}{p^k}\rfloor - 2\lfloor \frac{n}{p^k}\rfloor\right]$ $\endgroup$ – mathworker21 Aug 17 at 10:25
  • $\begingroup$ @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 $\endgroup$ – Cloud JR Aug 17 at 10:31
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    $\begingroup$ use what you're given in problem statement. try to figure it out for a bit $\endgroup$ – mathworker21 Aug 17 at 10:36
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Since $n < p^{j} \leq 2n$, it is clear that $\bigl[\frac{n}{p^j}\bigr]=0$. Also since $n<p^{j}$, $2n < 2p^{j} \le p^{j+1}$ 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 $$v_{p}\left(\binom{2n}{n}\right)=\sum_{\substack{k\geq 1 \\ k\neq i}} \left\{\left[\frac{2n}{2^k}\right]-2\left[\frac{n}{p^k}\right]\right\} + \biggl[\frac{2n}{p^j}\biggr]-2\biggl[\frac{n}{p^j}\biggr] \geq 1$$

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  • $\begingroup$ Your answer was written very poorly (with all due respect). I hope you don't mind me editing it to make the writing better. $\endgroup$ – mathworker21 Aug 17 at 10:46
  • $\begingroup$ Emm. I don't understand what was poor with my answer. Also i don't recognize your edited part. $\endgroup$ – crskhr Aug 17 at 10:47

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