# Chinese Remainder Theorem (CRT)

Given $$N = \prod\_{i=1}^{k}n\_i$$ where each moduli(or divisor) $$n\_i$$ is a **pairwise** [**coprime**](https://en.wikipedia.org/wiki/Coprime_integers), 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](mailto:undefined) of Fractalyze


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