The CRT states that, given a set of congruences (where all of the \(m\)s are coprime):

\[ x \equiv a_1\ \text{mod}\ m_1 \\ x \equiv a_2\ \text{mod}\ m_2 \\ \vdots \\ x \equiv a_n\ \text{mod}\ m_n \\ \]

We can solve for \(x\) with the equation:

\[ x \equiv (\sum_{i=1}^n a_i \cdot M_i \cdot M_{i}^{-1})\ \text{mod}\ N \]

Where \(M_i\) is \(\cfrac{m_1m_2 \dots m_n}{m_i}\) and \(M_{i}^{-1}\) is \(M_i\)’s multiplicative inverse modulo \(m_i\).

**Example**

We will start with the congruences

\[ x \equiv 1\ \text{mod}\ 3 \\ x \equiv 2\ \text{mod}\ 7 \\ \]

This gives us the equation:

\[ (1 \times 7 \times 1) + (2 \times 3 \times 5) \]

Because, for \(M_1\), \(\cfrac{3 \cdot 7}{3} = 7\) and \(7^{-1}\ \text{mod}\ 3 = 1\) and for \(M_2\), \(\cfrac{3 \cdot 7}{7} = 3\) and \(3^{-1}\ \text{mod}\ 7 = 5\)

This results in \(7 + 30 = 37\ \text{mod}\ 21\), or \(16\), which we can check: \(16\ \text{mod}\ 3 \equiv 1\) and \(16\ \text{mod}\ 7 \equiv 2\).