Let $k,n$ be positive integers such that $k< n$. Show that if $k$ integers are selected from the set $\{1,2,3,\dots,n\}$ then the probability that their sum is divisible by $n$ is $\dfrac1n$. Also note that $\gcd(k,n)=1$.
Not sure how to do this. Looks hard.
Since ${\rm gcd}(k,n)=1$ there are $u$, $v\in{\mathbb Z}$ with $uk+vn=1$, in particular an $u\in[n]$ such that $$ku=1\quad({\rm mod} \ n)\ .$$ Let $P_k:={[n]\choose k}$ be the set of all $k$-subsets of $[n]$, and for any $A\in P_k$ denote by $\sigma(A)$ the sum of its elements, modulo $n$. Consider now the map $$T:\quad P_k\to P_k,\qquad A\mapsto\{x+u\,|\, x\in A\}$$ which adds $u$ to each element of $A$. It follows that $$\sigma\bigl(T(A)\bigr)=\sigma(A)+ku=\sigma(A)+1\ .$$ Together with $T^n={\rm id}$ this allows to conclude that for all $A\in P_k$ the orbit of $A$ under $T$ has $n$ elements $A_i$, and exactly one of these has $\sigma(A_i)=0$.