Composite numbers dividing binomial coefficients

The answer is $k=0.$ The reason is simple, if $n$ is composite, then there is a prime number $p$ that is a factor of $n$ , and it is easy to see that $n$ is not a factor of $\binom{n}{p}.$

If $n$ is composite and $p$ is a prime dividing $n$ , then $1 < p < n$ and ${n \choose p} \; = \; \frac{n(n-1)(n-2)\cdots(n-p+1)}{p!}$ Of the $p$ consecutive integers being multiplied together in the numerator, only $n$ is a multiple of $p$ . Thus the index of $p$ in the numerator (the number of times $p$ divides the numerator) is equal to the index of $p$ in $n$ . On the other hand, the index of $p$ in the denominator is $1$ . This tells us that the index of $p$ in $\binom{n}{p}$ is $1$ less than the index of $p$ in $n$ , which means that $n$ cannot divide $\binom{n}{p}$ .