Search for a tool
Chinese Remainder

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

Results

Chinese Remainder -

Tag(s) : Arithmetics

Share
Share
dCode and more

dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!
A suggestion ? a feedback ? a bug ? an idea ? Write to dCode!


Please, check our dCode Discord community for help requests!
NB: for encrypted messages, test our automatic cipher identifier!


Feedback and suggestions are welcome so that dCode offers the best 'Chinese Remainder' tool for free! Thank you!

Chinese Remainder

Chinese Remainder Calculator


Loading...
(if this message do not disappear, try to refresh this page)


Answers to Questions (FAQ)

What is the Chinese Remainder Theorem? (Definition)

The Chinese remainder theorem is the name given to a system of congruences (multiple simultaneous 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? This exercise implies to calculate $ x $ such that $ x \equiv 2 \mod 3 $ and $ x \equiv 3 \mod 5 $ and $ x \equiv 2 \mod 7 $

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:

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

How to calculate Chinese remainder?

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 A, modulo B) in equations of the form 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 $

When does the Chinese Remainder Theorem have no solution?

The system of equations with remainders $ r_i $ and modulos $ m_i $ has solutions only if the following modular equation is true: $$ r_1 \mod d = r_2 \mod d = \cdots r_n \mod d $$ with $ d $ the GCD of all modulos $ m_i $.

Source code

dCode retains ownership of the "Chinese Remainder" source code. Except explicit open source licence (indicated Creative Commons / free), the "Chinese Remainder" algorithm, the applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, breaker, translator), or the "Chinese Remainder" functions (calculate, convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (Python, Java, PHP, C#, Javascript, Matlab, etc.) and all data download, script, or API access for "Chinese Remainder" are not public, same for offline use on PC, mobile, tablet, iPhone or Android app!
Reminder : dCode is free to use.

Cite dCode

The copy-paste of the page "Chinese Remainder" or any of its results, is allowed (even for commercial purposes) as long as you credit dCode!
Exporting results as a .csv or .txt file is free by clicking on the export icon
Cite as source (bibliography):
Chinese Remainder on dCode.fr [online website], retrieved on 2024-12-03, https://www.dcode.fr/chinese-remainder

Need Help ?

Please, check our dCode Discord community for help requests!
NB: for encrypted messages, test our automatic cipher identifier!

Questions / Comments

Feedback and suggestions are welcome so that dCode offers the best 'Chinese Remainder' tool for free! Thank you!


https://www.dcode.fr/chinese-remainder
© 2024 dCode — The ultimate 'toolkit' to solve every games / riddles / geocaching / CTF.
 
Feedback