Irish Mathematical Olympiad

Because $2013 = 3 \cdot 11 \cdot 61$, the requirement $94^n + a \cdot 28^n \equiv 0 \pmod{2013}$ implies $a+1 \equiv 0 \pmod{3}$, $6^n(a+1) \equiv 0 \pmod{11}$ and $28^n((-1)^n + a) \equiv 0 \pmod{61}$. Therefore $a \equiv -1 \pmod{3}$, $a \equiv -1 \pmod{11}$ and $a \equiv (-1)^{n+1} \pmod{61}$. Using the Chinese Remainder Theorem, we see that for $n$ even the smallest positive integer $a$ which satisfies these three congruences is $2012$. For odd $n$, we need to solve $33k-1 \equiv 1 \pmod{61}$. Using Euclid's Algorithm, we obtain $13 \cdot 61 - 24 \cdot 33 = 1$, and we see that $k \equiv -24 \cdot 2 \equiv 13 \pmod{61}$. This yields that the smallest possible $a$ is equal to $33 \cdot 13 - 1 = 428$.