Here is the problem: Let $\displaystyle n$ be a positive integer. Show that in any set of $\displaystyle n$ consecutive integers there is exactly one divisible by $\displaystyle n$.

Here is the solution:

Let $\displaystyle a, a+1,...,a+n-1$ be the integers in the sequence. The integers $\displaystyle (a+i) \mod n$, $\displaystyle i=0,1,2...,n-1$, are distinct because $\displaystyle 0< (a+j)-(a+k)<n$ whenever $\displaystyle 0 \le k < j < n-1$. Because there are n possible values for $\displaystyle (a+i) \mod n$ and there are n different integers integers in the set, each of these values is taken exactly once. It follows that there is exactly one integer in the sequence that is divisible by n.

Something I don't quite understand: How does $\displaystyle 0< (a+j)-(a+k)<n$

make the values calculated from the modulus distinct?