$\gcd(m,n)$ divides $\binom{m}{n} ?$ One of my students asked me this question.

Can anyone please explain the necessary and sufficient conditions such that

$\gcd(m,n)$ divides $\binom{m}{n} ?$

Is it true that $\gcd(m,n) | \binom{m}{n} \iff \gcd(m,n)=1$ or $m =kn$ ?

I'm sure of it in backward direction. I need help about the forward case?
For primes $p$, Kummer's theorem tells you the $p$-adic order of $m \choose n$.  What you want is that this is at least the $p$-adic order of $g = \gcd(m,n)$ for each prime dividing $g$.  In particular, note that the $p$-adic order of ${m \choose n}$ is the same as that of ${mp \choose np}$.  

EDIT: Thus one way to find examples where $1 &lt; \gcd(m,n) \mid {m \choose n}$ is to start with $m_0, n_0$ coprime, find a prime power $p^k$ dividing ${m_0 \choose n_0}$, and take $m = p^k m_0$, $n = p^k n_0$.
$\gcd(m,n) | \binom{m}{n} \iff gcd(m,n)=1$ or if $gcd(m,n)\neq1$ then the prime factorization of $n$ and $m$ must complain with the following:

$n = p^{\alpha_{1}}q^{\alpha_{2}}\dots w^{\alpha_{n}}$

$m=p^{\alpha_{1}+i}q^{\alpha_{2}+j}\dots w^{\alpha_{n}+k}x$ with  $i,j,k,x\geq 1$

Let's explain it:

$\binom{m}{n} = \frac{m!}{n!(m-n)!} = \frac{m(m-1)\dots(m-n+1)(m-n)!}{n!(m-n)!} = \frac{m(m-1)\dots(m-n+1)!}{n!}$

If $gcd(m,n) = n$ and $m=p^{\alpha_{1}+i}q^{\alpha_{2}+j}\dots w^{\alpha_{n}+k}x$ then

$\frac{p^{\alpha_{1}+i-1}q^{\alpha_{2}+j-1}\dots w^{\alpha_{n}+k-1}x(m-1)\dots(m-n+1)!}{(n-1)!}$

then you want to divide this binomial by $gcd(m,n)=n$ so:

$\frac{p^{\alpha_{1}+i-1}q^{\alpha_{2}+j-1}\dots w^{\alpha_{n}+k-1}x(m-1)\dots(m-n+1)!}{p^{\alpha_{1}}q^{\alpha_{2}}\dots w^{\alpha_{n}}(n-1)!} \Rightarrow \frac{p^{\alpha_{1}+i-2}q^{\alpha_{2}+j-2}\dots w^{\alpha_{n}+k-2}x(m-1)\dots(m-n+1)!}{(n-1)!}$

so it's satisfied when $i,j,k \geq 1$