Is binomial(n, k)=0 (mod n) when gcd(n, k)=1? When gcd(n, k)>1?
All variables are understood to be positive integers. Let $n\mid m$ mean $n$ is a factor of $m$, a.k.a. $m$ is divisible by $n$ or $m=0\pmod n$.


  Is it correct that if $\gcd(n, k)=1$, then $n\mid\binom nk$?


Yes. 

Since $k\binom nk=n\binom{n-1}{k-1}$, we know $k\binom nk$ is divisible by $n$. Since $\gcd(n,k)=1$,  we must have $\binom nk$ is divisible by $n$. 




  Is it correct that if $\gcd(n, k)&gt;1$, then $\binom nk&gt;0\pmod n$?


The relation of "greater than" or "less than" modulo $n$ does not make sense since, for example, $-1=1=2019\pmod 2$.

What you have in mind should be the following question.


  Is it correct that if $\gcd(n, k)&gt;1$, then $n\not\mid\binom nk$?


Not necessarily. 

The smallest counterexample is $\gcd(10,4)=2\neq1$ but $10\mid\binom{10}4=210$. In fact, if $n=2\pmod8$, then $n\mid\binom n4$.



Kummer's theorem provides a way to compute the exponent of the highest power of a given prime number dividing a binomial coefficient. Peter Taylor points out that it can be used to determine whether $n\mid\binom nk$ in polynomial time if we know the prime factorisation of $k$ or $n$. However, no algorithm is known to be able to obtain the prime factorisation of an integer in polynomial-time. 

Here is a related interesting article, divisibility properties of binomial coefficients by K. R. Mclean.



Exercise 1. Show that if $m\mid n$ and $\gcd(m,k)=1$, then  $m\mid\binom nk$.

Exercise 2. Show that $2i\mid\binom{2i}4$ iff $i=1\pmod4$.

Exercise 3. Find infinitely many pairs of $n$ and $k$ such that $1\lt k \mid n$ and $n\mid\binom nk$. (Hint, one such family can be found by setting $k=6$.)