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) : Mathematics, 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 consider a number of elements which we know the remainder of their Euclidean division.

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?

Consider 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, consider 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 \( n_i \).

Consider 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), but the simplest is to write 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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