# Difference between revisions of "Chinese Remainder Theorem"

 This is an AoPSWiki Word of the Week for July 11-July 16

The Chinese Remainder Theorem is a number theoretic result. It is one of the only theorems named for an oriental person or place, due to the closed development of mathematics in the western world.

## Theorem

Let $m$ be relatively prime to $n$. Then each residue class mod $mn$ is equal to the intersection of a unique residue class mod $m$ and a unique residue class mod $n$, and the intersection of each residue class mod $m$ with a residue class mod $n$ is a residue class mod $mn$.

## Proof

If $a \equiv b \pmod{mn}$, then $a$ and $b$ clearly differ by a multiple of $mn$, so $a \equiv b \pmod{m}$ and $a \equiv b \pmod{n}$. This is the first part of the theorem. The converse follows because $a$ and $b$ must differ by a multiple of $m$ and $n$, and $\mbox{lcm}(m,n) = mn$. This is the second part of the theorem.

## Applicability

Much like the Fundamental Theorem of Arithmetic, many people seem to take this theorem for granted before they consciously turn their attention to it. It ubiquity derives from the fact that many results can be easily proven mod (a power of a prime), and can then be generalized to mod $m$ using the Chinese Remainder Theorem. For intance, Fermat's Little Theorem may be generalized to the Fermat-Euler Theorem in this manner. Its application in problem-solving is similar.

General Case: the proof of the general case follows by repeatedly applying to above result (k-1) times.

## Extended version of the theorem

Suppose one tried to divide a group of objects into $2$, $3$ and $4$ parts instead and found $1$, $1$ and $2$ fish left over, respectively. One can show this to be impossible, because any number giving remainder $1$ modulo $2$ must be odd and any number giving remainder $2$ modulo $4$ must be even. Thus, the number of objects must be odd and even simultaneously, which is absurd. Thus, there must be some restrictions on the numbers $a_1,\dots,a_n$ to ensure that at least one solution exists. It turns out (and this is the extended version of the theorem) that

The solution exists if and only if $a_i\equiv a_j\mod \gcd(m_i,m_j)$ for all $i,j$ where $\gcd$ stands for the greatest common divisor. Moreover, in the case when the problem is solvable, any two solutions differ by some common multiple of $m_1,\ldots,m_n$.

## Discussion

• Here is an AoPS thread in which the Chinese Remainder Theorem is discussed and implemented.