Largest power of $p$ that divides $2n$ choose $n$?

Question

I'm working through questions that I'm not familiar with and would like some help on the following:

If $π(x)$ counts the number of primes less than or equal to $x$, how many primes are there greater than $n$ but less than or equal to $2n$ ?

And if prime $p$ satisfies $n < p ≤ 2n$, what is the largest power of $p$ that divides $2n$ choose $n$ ?

Thanks!

Answer 1

As for the first question, the number of primes less than or equal to $2n$ is $\pi(2n)$ and the number of primes less than $n$ is $\pi(n)$; thus, the number of primes less than or equal to $2n$ but not less than or equal to $n$ is $\pi(2n)-\pi(n)$.

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Secondly, recall that ${2n\choose n}=\frac{(2n)!}{n!^2}$. Since $p>n$, we know $p$ cannot divide $n!^2$. Since there's only one multiple of $p$ between $n$ and $2n$ (because $n<p$, we know $2n<2p$), we know $p$ divides $(2n)!$ only once. Thus, $p$ divides ${2n\choose n}$ exactly once.