Prime Numbers and Combinatoric Division True or false, for all primes p > 2, there are an infinite number of positive integers n for 
which p does not divide ${2n \choose n}$. 
I think the answer would be no, but I am not sure. Since n! has all numbers less than n as factors I would assume that p would be among those. Is my reasoning correct?
 A: The answer is actually yes.  For a hint, see Legendre's Formula.
Solution:
For $p$ to not divide $\binom{2n}{n}$, $v_p((2n)!)$ must be exactly $v_p(n!) + v_p(n!) = 2v_p(n!)$ (where $v_a(b)$ is defined as the largest nonnegative integer $k$ such that $a^k$ divides b).
So the question is equivalent to, given a prime $p > 2$, do there exist an infinite number of $n$ such that $v_p((2n)!) = 2v_p(n!)$?  I claim the answer is yes.  Consider $n = p^i$ for some positive integer $i$.  Then by Legendre's Formula, $$v_p(n!) = \Bigg\lfloor\dfrac{n}{p}\Bigg\rfloor + \Bigg\lfloor\dfrac{n}{p^2}\Bigg\rfloor + \cdots + \Bigg\lfloor\dfrac{n}{p^i}\Bigg\rfloor = p^{i-1}+p^{i-2}+\cdots+1,$$$$v_p(2n) = \Bigg\lfloor\dfrac{2n}{p}\Bigg\rfloor + \Bigg\lfloor\dfrac{2n}{p^2}\Bigg\rfloor + \cdots + \Bigg\lfloor\dfrac{2n}{p^i}\Bigg\rfloor \\ = 2p^{i-1}+2p^{i-2}+\cdots+2 = 2v_p(n!),$$ as desired.  Since there are infinitely many powers of $p$, the conclusion follows.
The key idea here is that there are no powers of $p$ between $p^i$ and $2p^i$.  This only works when $p > 2$, and fails for $p = 2$.  In fact, $2$ divides $\binom{2n}{n}$ for every positive integer $n$, which you can use $v_2$ and Legendre's formula to prove.
