I am aware of proof of divisibility of binomial coefficients of a prime $p$. I've seen it is easy to show that when $0<k<p$ $$\binom{p}{k}\equiv 0 \mod p$$
Can there be anything stronger. Suppose I have an integer $n=pq$. Can we prove that $p|\binom{pq}{k}$. I feel like this is an impossibility since, looking at a case like $n=15$, $3|\binom{15}{4}$, but $3\not|\binom{15}{3}$. $5|\binom{15}{3}$ but $5\not|\binom{15}{5}$. Yet there ARE binomial coefficients where that 15 divide evenly.
Is there a way to determine which coefficients will successfully be divided by $n$ and which will be divisible by $p$ or $q$? And are there stronger divisibility laws for powers of a prime?
If $n = pq$, then $p$ divides $\left( \begin{array}{c} n \\ k \end{array} \right)$ if $p$ does not divide $k$, and sometimes in some other special cases. For example $3$ divides $\left( \begin{array}{c} 12 \\ 6 \end{array} \right)$.
The full answer: write the base $p$ expansions of $n$ and $k$:
\begin{align*} n & = n_mp^m + \cdots + n_1p^1 + n_0 \\ k & = k_mp^m + \cdots + k_1p^1 + k_0 \\ \end{align*}
Then $p$ divides $\left( \begin{array}{c} n \\ k \end{array} \right)$ if and only if there exists some index $i$ with $k_i > n_i$. This follows from Lucas' Theorem. In particular, since $p$ divides $n$, $n_0 = 0$, and so $p$ will always divide $\left( \begin{array}{c} n \\ k \end{array} \right)$ as long as $p$ does not divide $k$.