Tool to compute congruences with the chinese remainder theorem. The Chinese Remainder Theorem helps to solve congruence equation systems in modular arithmetic.

Chinese Remainder - dCode

Tag(s) : Arithmetics

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Tool to compute congruences with the chinese remainder theorem. The Chinese Remainder Theorem helps to solve congruence equation systems in modular arithmetic.

The **Chinese remainder** theorem is the name given to a system of congruances (modular equations). The original problem is to calculate a number of elements which remainders (of their Euclidean division) are known.

Example: If they are arranged by 3 there remains 2. If they are arranged by 5, there remain 3 and if they are arranged by 7, there remain 2. How many objects are there?

Take a list of \( k \) coprimes integers \( n_1, ..., n_k \) and their product \( n = \prod_{i=1}^k n_i \). For all integers \( a_1, ... , a_k \), it exists another integer \( x \) which is unique modulo \( n \), such as :

$$ \begin{matrix} x \equiv a_1\pmod{n_1} \\ \ldots \\ x \equiv a_k\pmod{n_k} \end{matrix} $$

To find a solution of the congruence system, take the numbers \( \hat{n}_i = \frac n{n_i} = n_1 \ldots n_{i-1}n_{i+1}\ldots n_k \) which are also coprimes. To find the modular inverses, use the Bezout theorem to find integers \( u_i \) and \( v_i \) such as \( u_i n_i + v_i \hat{n}_i = 1 \). Here, \( v_i \) is the modular inverse of \( \hat{n}_i \) modulo \( n_i \).

Take then the numbers \( e_i = v_i \hat{n}_i \equiv 1 \mod{n_i} \). A particular solution of the **Chinese remainders** theorem is $$ x = \sum_{i=1}^k a_i e_i~ $$

dCode accepts numbers as pairs (remainder, modulo) or written x = A mod B

Example: \( (2,3),(3,5),(2,7) \iff \left\{ \begin{array}{ll} x = 2 \mod 3 \\ x = 3 \mod 5 \\ x = 2 \mod 7 \end{array} \right. \Rightarrow x = 23 \)

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Source : https://www.dcode.fr/chinese-remainder

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