Okay, so we have the Chinese Remainder Theorem:
If $m_1$ and $m_2$ are coprime then the simultaneous congruences $\left( x \equiv a_1 \mod m_1 \right)$, $\left( x \equiv a_2 \mod m \right)$ have a unique solution $\left(\mod m_1 m_2 \right)$.
I want to prove the solution for more than two congruences using the Chinese Remainder Theorem:
$x = \left[\Sigma_{i} {a}_i \frac {\Pi_{i} n_i} {n_i} \left[ \left( \frac {\Pi_{i} n_i} {n_i} \right)^{-1} \right]_{n_i} \right] _N $
I read somewhere that you can do this by induction... But I don't see how that would work...