Difference between revisions of "Chinese Remainder Theorem"

Revision as of 11:08, 6 August 2013

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

Formally stated, the Chinese Remainder Theorem is as follows:

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 $mn$ .

Simply stated:

Suppose you wish to find the least number $x$ which leaves a remainder of:

$\begin{aligned} &y_{1} \text{ when divided by } &d_{1}\\ &y_{2} \text{ when divided by } &d_{2}\\ &\vdots &\vdots\\ &y_{n} \text{ when divided by } & d_{n}\\ \end{aligned}$

such that $d_{1}$ , $d_{2}$ , ... $d_{n}$ are all relatively prime. Let $M = d_{1}d_{2} \cdots d_{n}$ , and $b_{i} = \frac{M}{d_{i}}$ . Now if the numbers $a_{i}$ satisfy:

$a_{i}b_{i} - 1 \equiv 0 \pmod {d_{i}}$

for every $1 \leq i \leq n$ , then a solution for $x$ is:

$x = \sum_{i=1}^n a_{i}b_{i}y_{i} \pmod M$

Proof

If $a \equiv b \pmod{mn}$ , then $a$ and $b$ 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 instance, Fermat's Little Theorem may be generalized to the Fermat-Euler Theorem in this manner.

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

Extended version of the theorem

Suppose one tried to divide a group of fish into $2$ , $3$ and $4$ parts instead and found $1$ , $1$ and $2$ fish left over, respectively. Any number with remainder $1$ mod $2$ must be odd and any number with remainder $2$ mod $4$ must be even. Thus, the number of objects must be odd and even simultaneously, which is a contradiction. Thus, there must be restrictions on the numbers $a_1,\dots,a_n$ to ensure that at least one solution exists. It follows 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 $m_1,\ldots,m_n$ . (the extended version).

Discussion

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

@@ Line 8: / Line 8: @@
-Simply stated,
+Simply stated:
 Suppose you wish to find the least number <math>x</math> which leaves a remainder of:
@@ Line 19: / Line 19: @@
 </center>
-and no <math>d_{i}</math> have any divisors in common for <math>1 \leq i\leq n</math>.
+such that <math>d_{1}</math> ,  <math>d_{2}</math> , ... <math>d_{n}</math> are all relatively prime.
 Let <math>M = d_{1}d_{2} \cdots d_{n}</math>, and <math>b_{i} = \frac{M}{d_{i}}</math>.
 Now if the numbers <math>a_{i}</math> satisfy:
@@ Line 31: / Line 31: @@
 == Proof ==
-If <math>a \equiv b \pmod{mn}</math>, then <math>a</math> and <math>b</math> clearly differ by a multiple of <math>mn</math>, so <math>a \equiv b \pmod{m}</math> and <math>a \equiv b \pmod{n}</math>.  This is the first part of the theorem.  The converse follows because <math>a</math> and <math>b</math> must differ by a multiple of <math>m</math> and <math>n</math>, and <math>\mbox{lcm}(m,n) = mn</math>.  This is the second part of the theorem.
+If <math>a \equiv b \pmod{mn}</math>, then <math>a</math> and <math>b</math> differ by a multiple of <math>mn</math>, so <math>a \equiv b \pmod{m}</math> and <math>a \equiv b \pmod{n}</math>.  This is the first part of the theorem.  The converse follows because <math>a</math> and <math>b</math> must differ by a multiple of <math>m</math> and <math>n</math>, and <math>\mbox{lcm}(m,n) = mn</math>.  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 <math>m</math> using the Chinese Remainder Theorem.  For instance, [[Fermat's Little Theorem]] may be generalized to the [[Fermat-Euler Theorem]] in this manner.  Its application in [[problem-solving]] is similar.
+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 <math>m</math> using the Chinese Remainder Theorem.  For instance, [[Fermat's Little Theorem]] may be generalized to the [[Fermat-Euler Theorem]] in this manner.
-'''General Case''': the proof of the general case follows by repeatedly applying to above result (k-1) times.
+'''General Case''': the proof of the general case follows by induction to the above result (k-1) times.
 ==Extended version of the theorem==
-Suppose one tried to divide a group of fish into <math>2</math>, <math>3</math> and <math>4</math> parts instead and found <math>1</math>, <math>1</math> and <math>2</math> fish left over, respectively. One can show this to be impossible, because any number giving remainder <math>1</math> modulo <math>2</math> must be [[odd integer | odd]] and any number giving remainder <math>2</math> modulo <math>4</math> must be [[even integer | even]]. Thus, the number of objects must be odd and even simultaneously, which is absurd. Thus, there must be some restrictions on the numbers <math>a_1,\dots,a_n</math> to ensure that at least one solution exists.  It turns out (and this is the extended version of the theorem) that
+Suppose one tried to divide a group of fish into <math>2</math>, <math>3</math> and <math>4</math> parts instead and found <math>1</math>, <math>1</math> and <math>2</math> fish left over, respectively. Any number with remainder <math>1</math> mod <math>2</math> must be [[odd integer | odd]] and any number with remainder <math>2</math> mod <math>4</math> must be [[even integer | even]]. Thus, the number of objects must be odd and even simultaneously, which is a contradiction. Thus, there must be restrictions on the numbers <math>a_1,\dots,a_n</math> to ensure that at least one solution exists.  It follows that:
-''The solution exists if and only if <math>a_i\equiv a_j\mod \gcd(m_i,m_j)</math> for all <math>i,j</math> where <math>\gcd</math> stands for the [[greatest common divisor]]. Moreover, in the case when the problem is solvable, any two solutions differ by some [[common multiple]] of <math>m_1,\ldots,m_n</math>.''
+''The solution exists if and only if <math>a_i\equiv a_j\mod \gcd(m_i,m_j)</math> for all <math>i,j</math> where <math>\gcd</math> stands for the [[greatest common divisor]]. Moreover, in the case when the problem is solvable, any two solutions differ by some [[common multiple]] of <math>m_1,\ldots,m_n</math>.'' (the extended version).
 ==See Also==

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