Consider the following question.
Let be $m \in \mathbb{N}$. Can one find $n \in \mathbb{N}$ such that $m$ divides $\binom{n}{k}$ for all $1 \leq k \leq n-1$?
If not, does it hold for any particular $m \in \mathbb{N}$?
I actually do not know how to start.
Clearly, given that $m = 1,$ we have that $m$ divides $\binom n k$ for all integers $1 \leq k \leq n - 1.$
Given that $m$ is a prime, one can prove that $m$ divides $\binom n k$ for all $1 \leq k \leq n - 1$ so long as $n = m.$
Proof. Given that $m$ is prime and $n = m,$ we have that $$\binom m k = \frac{m!}{(m - k)! k!}$$ is an integer that is divisible by $m.$ Explicitly, the integers $(m - k)!$ and $k!$ do not divide $m$ by hypothesis that $m$ is prime and $1 \leq k \leq m - 1,$ hence we can factor out $m$ to obtain $$\binom m k = m \cdot \frac{(m - 1)!}{(m - k)! k!}.$$ But this says precisely that $m$ divides $\binom m k = \binom n k.$ QED.
Upon inpsection, if $m$ is prime, it seems that this could be true for the case that $n = m^i$ for some integer $i \geq 1.$ (Edit: Jyrki Lahtonen has kindly directed my attention to this answer from Andreas Caranti that confirms my hunch.) Beyond these few cases, I suspect that it is not true that $m$ divides $\binom n k$ for all $1 \leq k \leq n - 1,$ but again, I cannot confirm my suspicions with a proof.
$$v_p\binom{n}{k}=v_p(n!)-v_p((n-k)!)-v_p(k!)$$ For $m|\binom{n}{k}$, we have $v_p(m)\le v_p\binom{n}{k}$ for all the prime divisors $p$ of $m$.
By legendre's formula, we have $$v_p\binom{n}{k}=\sum_{i=1}^{\infty} \left(\left\lfloor \frac{n}{p^i}\right\rfloor -\left(\left\lfloor \frac{n-k}{p^i}\right\rfloor+ \left\lfloor \frac{k}{p^i}\right\rfloor\right)\right)$$ Does that help?
