If $p$ and $q$ are primes, which binomial coefficients $\binom{pq}{n}$ are divisible by $pq$?

Question

If $p$ and $q$ are primes, which binomial coefficients $\binom{pq}{n}$, $1 \le n < pq$, are divisible by $pq$?

In particular, if $p$ and $q$ are distinct odd primes, and $n$ is even, does $pq \mid \binom{pq}{n}$?

This question is inspired by my answer here: Proving that an expression returns a real non-integer number (Number 2)

According to Kummer's theorem (https://en.wikipedia.org/wiki/Binomial_coefficient#Divisibility_properties), "the largest power of $p$ dividing $\binom{m+n}{m}$ equals $p^c$, where $c$ is the number of carries when $m$ and $n$ are added in base $p$."

If $m+n = pq$, $m = pq-n$. So, if there is at least one carry when $n$ and $pq-n$ are added in base $p$, then $p \mid \binom{pq}{n}$. So, we want to know for which $n$ there is at least one carry in base $p$ and base $q$.

An obvious generalization is to ask, if $p, q, \ldots, r$ are primes, which of the binomial coefficients $\binom{pq\cdots r}{n}$ are divisible by $pq\cdots r$. But I don't even know how to do this for two primes.

Note: The answers to this question may be useful: Divisibility of coefficients $\binom{n}{k}$ with fixed composite $n$

Answer 1

Observe, $\begin{pmatrix}15\\10\end{pmatrix}=3003$, which is not divisible by $15$. This contradicts the "in particular". Actually, the "in particular" statement is not a particular case because if $n$ is odd, then $pq-n$ is even and $\begin{pmatrix}pq\\pq-n\end{pmatrix}=\begin{pmatrix}pq\\n\end{pmatrix}$.

More generally, assume that $q>p$, then $q$ divides $(pq)!$ exactly $p$ times. On the other hand, $q$ divides $n!$ exactly $\lfloor n/q\rfloor$ times. Similarly, $q$ divides $(pq-n)!$ exactly $p-\lceil n/q\rceil$ times. Since the difference between $\lfloor n/q\rfloor$ and $\lceil n/q\rceil$ is $1$, $q$ divides the binomial coefficient exactly when $n$ is not a multiple of $q$.

A similar argument will work for studying divisibility by $p$, but one must be more careful because powers of $p$ will appear.

Answer 2

For $1 \le n \le \lfloor pq/2 \rfloor$ we have

$$ \binom{pq}{n} = \frac{pq}{1} \frac{pq-1}{2} \frac{pq-2}{3} \cdots $$

So we can divide $\binom{pq}{n}$ by $pq$ as long as $n$ does not contain neither $p$ nor $q$, thus if $\gcd(pq,n) = 1$. So at least we have

$$ \gcd(pq,n) = 1 \quad \Longrightarrow \quad pq \Bigg| \binom{pq}{n}. $$

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The general question you asked works in the same way:

$$ \gcd(p_1 p_2 p_3 \cdots p_m,n) = 1 \quad \Longrightarrow \quad p_1 p_2 p_3 \cdots p_m \Bigg| \binom{p_1 p_2 p_3 \cdots p_m}{n} $$