Let $a_1, a_2, \cdots , a_k$ and $n_1, n_2, \cdots , n_k$ be integers and $n_i$ are pairwise coprime.
Chinese remainder theorem state that the system : $$\begin{matrix} x & = & a_1 \pmod{n_1} \\ & \vdots & \\ x & = & a_k \pmod{n_k} \end{matrix}$$ had a unique solution modulo $N = n_1 \times n_2 \times \cdots \times n_k $.
And in this case we have : $$x = \sum_{j=1}^{k} a_j N_j y_j \pmod{N}$$ With $N_j = N/n_j$ and $y_j = N_j^{-1} \pmod{n_j}$.