Annual Harvard-MIT Mathematics Tournament - Team Round: B Division

Solution:

Suppose that $a_{i}=j(i)+i-b_{i} n$, where $b_{i}$ is an integer. Note that $f\left(i-b_{i} n\right)=i-b_{i} n+j(i)=a_{i}$. Since $\{i-b_{i} n \mid i=0,1, \ldots, n-1\}$ contains $n$ distinct integers (as their residue $\bmod n$ are all distinct), and $f$ is a permutation, we see that after applying the map $f$, the resulting set $\{a_{0}, a_{1}, \ldots, a_{n-1}\}$ is a set of $n$ distinct integers. Since $0 \leq a_{i}<n$ from definition, we see that $(a_{0}, a_{1}, \ldots, a_{n-1})$ is a permutation of $(0,1, \ldots, n-1)$.