Divisibility of multinomial by a prime number What is the condition for divisibility of multinomial $ \dbinom {n}{x_1, x_2, \dots, x_k} $ by a prime $p$?
Update:
I tried to solve using a generalisation of Lucas Theorem by representing the $n$ as well as $x_1, x_2,\dots,x_k$ in base $p$ representation. 
I am following this site - 
Editorial to a problem
I am stuck on the part where it says that "..to be non-zero modulo p, it has to satisfy $n_j$ = $\Sigma{x_{i_j}}$ for all j..."
 A: A multinomial coefficient can be expressed using the binomial ones:
$${{n}\choose{x_1,x_2,\ldots,x_k}}={{n}\choose{x_1}}{{n-x_1}\choose{x_2}}{{n-x_1-x_2}\choose{x_3}}\ldots{{n-x_1-x_2-\ldots-x_{k-2}}\choose{x_{k-1}}}$$
Since a product is divisible by $p$ if and only if at least one of the factors is divisible by $p$, one can apply any of the divisibility criteria for binomials to obtain the answer for multinomials too.
The value of the multinomial doesn't depend on the order of $x_1, \ldots, x_k$ values in its "denominator", allowing them to be rearranged arbitrarily. Thus, it's actually sufficient to check the divisibility of ${{n}\choose{x_i}}$ for $1\leq i\leq k$. 

Alternatively, one can take the definition of a multinomial coefficient
$${{n}\choose{x_1,x_2,\ldots,x_k}} = \frac{n!}{x_1!x_2!\ldots x_k!}$$
and calculate the $p$-adic valuation of numerator and denominator directly using the equality:
$$v_p(n!)=\sum_{i\geq 1} \left\lfloor \frac{n}{p^i}\right\rfloor$$
The coefficient will then be divisible by $p$ if and only if $v_p(n!)>v_p(x_1!)+v_p(x_2!)+\ldots+v_p(x_k!)$
