Chinese Remainder Theorem (CRT)

Given $N = \prod_{i=1}^{k}n_i$ where each moduli(or divisor) $n_i$ is a pairwise coprime, and the set of $a_i$ is given as below:

$$x \equiv a_1 \pmod {n_1} \\ \vdots \\ x \equiv a_k \pmod {n_k}$$

Then the system has a unique solution $x$ under modulo $N$ such that:

$$x \equiv s_1 \cdot \prod_{i \ne 1} n_i + s_2 \cdot \prod_{i \ne 2} n_i + \cdots + s_k \cdot \prod_{i \ne k} n_i \pmod N$$

where each $s_i$ satisfies:

$$s_i \cdot \prod_{j \ne i} n_j = a_i \pmod {n_i}$$

In other words, a set of modulus statements can be reduced to a single statement.

Take the example below:

$$x = \begin{cases} 2 \pmod 3 \\ 3 \pmod 5 \\ 2 \pmod 7 \end{cases}$$

$$35 \cdot s_1 \equiv 2 \cdot s_1 \equiv 2 \pmod 3 \\ 21 \cdot s_2 \equiv s_2 \equiv 3 \pmod 5 \\ 15 \cdot s_3 \equiv s_3 \equiv 2 \pmod 7 \\$$

Solving these gives $s_1 = 1, s_2 = 3$ and $s_3 = 2$. Then the solution is

$$x \equiv 1 \cdot 35 + 3 \cdot 21 + 2 \cdot 15 \equiv 128 \equiv 23 \pmod {105}$$

Written by ryan Kim from A41