There are many ways to solve the problem, using, for instance, Euler-Fermat theorem. But one student, André Reys Leal, obtained a clever, simple solution: consider all numbers of the form $1999\dots 91$ with more than two nines. If one of them is multiple of $1991$ we are done. If not, since there are infinite of them there are two that are the same modulo $1991$. By subtracting them, we obtain $1999\dots 98000\dots 0$, which is a multiple of $1991$. We may ignore all rightmost zeros, but we do keep three of them, obtaining $1999\dots 98000$, which is still a multiple of $1991$. Sum $1991$ to it and we are done.