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This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

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Suppose $n_1$, $n_2$, $\ldots$, $n_k$ are positive [[integer]]s that are pairwise co-prime.

Then, for any given sequence of integers $a_1$, $a_2$, $\dots$, $a_k$, there exists an integer $x$ solving the following system of simultaneous congruences:

::: \begin\left\{align\right\} x &\equiv a_1 \pmod\left\{n_1\right\} \ x &\equiv a_2 \pmod\left\{n_2\right\} \ &\left\{\right\}\ \ \vdots \ x &\equiv a_k \pmod\left\{n_k\right\} \end\left\{align\right\}

Furthermore, all solutions $x$ of this system are congruent modulo the product, $N=n_1n_2\ldots n_k$.

;Task: Write a program to solve a system of linear congruences by applying the [[wp:Chinese Remainder Theorem|Chinese Remainder Theorem]].

If the system of equations cannot be solved, your program must somehow indicate this.

(It may throw an exception or return a special false value.)

Since there are infinitely many solutions, the program should return the unique solution $s$ where $0 \leq s \leq n_1n_2\ldots n_k$.

''Show the functionality of this program'' by printing the result such that the $n$'s are $\left[3,5,7\right]$ and the $a$'s are $\left[2,3,2\right]$.

'''Algorithm''': The following algorithm only applies if the $n_i$'s are pairwise co-prime.

Suppose, as above, that a solution is required for the system of congruences:

::: $x \equiv a_i \pmod\left\{n_i\right\} \quad\mathrm\left\{for\right\}; i = 1, \ldots, k$

Again, to begin, the product $N = n_1n_2 \ldots n_k$ is defined.

Then a solution $x$ can be found as follows:

For each $i$, the integers $n_i$ and $N/n_i$ are co-prime.

Using the [[wp:Extended Euclidean algorithm|Extended Euclidean algorithm]], we can find integers $r_i$ and $s_i$ such that $r_i n_i + s_i N/n_i = 1$.

Then, one solution to the system of simultaneous congruences is:

::: $x = \sum_\left\{i=1\right\}^k a_i s_i N/n_i$

and the minimal solution,

::: $x \pmod\left\{N\right\}$.

## 11l

{{trans|Python}}

F mul_inv(=a, =b)
V b0 = b
V x0 = 0
V x1 = 1
I b == 1
R 1
L a > 1
V q = a I/ b
(a, b) = (b, a % b)
(x0, x1) = (x1 - q * x0, x0)
I x1 < 0
x1 += b0
R x1

F chinese_remainder(n, a)
V sum = 0
V prod = product(n)
L(n_i, a_i) zip(n, a)
V p = prod I/ n_i
sum += a_i * mul_inv(p, n_i) * p
R sum % prod

V n = [3, 5, 7]
V a = [2, 3, 2]
print(chinese_remainder(n, a))


{{out}}

23


## 360 Assembly

{{trans|REXX}}

*        Chinese remainder theorem 06/09/2015
CHINESE  CSECT
LR     R12,R15
BEGIN    LA     R9,1               m=1
LA     R6,1               j=1
LOOPJ    C      R6,NN              do j=1 to nn
BH     ELOOPJ
LR     R1,R6              j
SLA    R1,2               j*4
M      R8,N-4(R1)         m=m*n(j)
LA     R6,1(R6)           j=j+1
B      LOOPJ
ELOOPJ   LA     R6,1               x=1
LOOPX    CR     R6,R9              do x=1 to m
BH     ELOOPX
LA     R7,1               i=1
LOOPI    C      R7,NN              do i=1 to nn
BH     ELOOPI
LR     R1,R7              i
SLA    R1,2               i*4
LR     R5,R6              x
LA     R4,0
D      R4,N-4(R1)         x//n(i)
C      R4,A-4(R1)         if x//n(i)^=a(i)
BNE    ITERX              then iterate x
LA     R7,1(R7)           i=i+1
B      LOOPI
ELOOPI   MVC    PG(2),=C'x='
XDECO  R6,PG+2            edit x
XPRNT  PG,14              print buffer
B      RETURN
ITERX    LA     R6,1(R6)           x=x+1
B      LOOPX
ELOOPX   XPRNT  NOSOL,17           print
RETURN   XR     R15,R15            rc=0
BR     R14
NN       DC     F'3'
N        DC     F'3',F'5',F'7'
A        DC     F'2',F'3',F'2'
PG       DS     CL80
NOSOL    DC     CL17'no solution found'
YREGS
END    CHINESE


{{out}}


x=          23



Using the package Mod_Inv from [[http://rosettacode.org/wiki/Modular_inverse#Ada]].

with Ada.Text_IO, Mod_Inv;

procedure Chin_Rema is
N: array(Positive range <>) of Positive := (3, 5, 7);
A: array(Positive range <>) of Positive := (2, 3, 2);
Tmp: Positive;
Prod: Positive := 1;
Sum: Natural := 0;

begin
for I in N'Range loop
Prod := Prod * N(I);
end loop;

for I in A'Range loop
Tmp := Prod / N(I);
Sum := Sum + A(I) * Mod_Inv.Inverse(Tmp, N(I)) * Tmp;
end loop;
end Chin_Rema;


## AWK

{{trans|C}} We are using the split-function to create both arrays, thus the indices start at 1. This is the only difference to the C version.

# Usage: GAWK -f CHINESE_REMAINDER_THEOREM.AWK
BEGIN {
len = split("3 5 7", n)
len = split("2 3 2", a)
printf("%d\n", chineseremainder(n, a, len))
}
function chineseremainder(n, a, len,    p, i, prod, sum) {
prod = 1
sum = 0
for (i = 1; i <= len; i++)
prod *= n[i]
for (i = 1; i <= len; i++) {
p = prod / n[i]
sum += a[i] * mulinv(p, n[i]) * p
}
return sum % prod
}
function mulinv(a, b,    b0, t, q, x0, x1) {
# returns x where (a * x) % b == 1
b0 = b
x0 = 0
x1 = 1
if (b == 1)
return 1
while (a > 1) {
q = int(a / b)
t = b
b = a % b
a = t
t = x0
x0 = x1 - q * x0
x1 = t
}
if (x1 < 0)
x1 += b0
return x1
}


{{out}}

23


## Bracmat

{{trans|C}}

( ( mul-inv
=   a b b0 q x0 x1
.   !arg:(?a.?b:?b0)
& ( !b:1
|   0:?x0
& 1:?x1
&   whl
' ( !a:>1
&   (!b.mod$(!a.!b):?q.!x1+-1*!q*!x0.!x0) : (?a.?b.?x0.?x1) ) & ( !x1:<0&!b0+!x1 | !x1 ) ) ) & ( chinese-remainder = n a as p ns ni prod sum . !arg:(?n.?a) & 1:?prod & 0:?sum & !n:?ns & whl'(!ns:%?ni ?ns&!prod*!ni:?prod) & !n:?ns & !a:?as & whl ' ( !ns:%?ni ?ns & !as:%?ai ?as & div$(!prod.!ni):?p
& !sum+!ai*mul-inv$(!p.!ni)*!p:?sum ) & mod$(!sum.!prod):?arg
& !arg
)
& 3 5 7:?n
& 2 3 2:?a
& put$(str$(chinese-remainder$(!n.!a) \n)) );  Output: 23  ## C When n are not pairwise coprime, the program crashes due to division by zero, which is one way to convey error. #include <stdio.h> // returns x where (a * x) % b == 1 int mul_inv(int a, int b) { int b0 = b, t, q; int x0 = 0, x1 = 1; if (b == 1) return 1; while (a > 1) { q = a / b; t = b, b = a % b, a = t; t = x0, x0 = x1 - q * x0, x1 = t; } if (x1 < 0) x1 += b0; return x1; } int chinese_remainder(int *n, int *a, int len) { int p, i, prod = 1, sum = 0; for (i = 0; i < len; i++) prod *= n[i]; for (i = 0; i < len; i++) { p = prod / n[i]; sum += a[i] * mul_inv(p, n[i]) * p; } return sum % prod; } int main(void) { int n[] = { 3, 5, 7 }; int a[] = { 2, 3, 2 }; printf("%d\n", chinese_remainder(n, a, sizeof(n)/sizeof(n[0]))); return 0; }  ## C++ #include <iostream> #include <numeric> #include <vector> #include <execution> using namespace std; int mulInv(int a, int b) { int b0 = b; int x0 = 0; int x1 = 1; if (b == 1) { return 1; } while (a > 1) { int q = a / b; int amb = a % b; a = b; b = amb; int xqx = x1 - q * x0; x1 = x0; x0 = xqx; } if (x1 < 0) { x1 += b0; } return x1; } int chineseRemainder(vector<int> n, vector<int> a) { int prod = std::reduce(std::execution::seq, n.begin(), n.end(), 1, [](int a, int b) { return a * b; }); int sm = 0; for (int i = 0; i < n.size(); i++) { int p = prod / n[i]; sm += a[i] * mulInv(p, n[i])*p; } return sm % prod; } int main() { vector<int> n = { 3, 5, 7 }; vector<int> a = { 2, 3, 2 }; cout << chineseRemainder(n,a) << endl; return 0; }  {{out}} 23  ## C# using System; using System.Linq; namespace ChineseRemainderTheorem { class Program { static void Main(string[] args) { int[] n = { 3, 5, 7 }; int[] a = { 2, 3, 2 }; int result = ChineseRemainderTheorem.Solve(n, a); int counter = 0; int maxCount = n.Length - 1; while (counter <= maxCount) { Console.WriteLine($"{result} ≡ {a[counter]} (mod {n[counter]})");
counter++;
}
}
}

public static class ChineseRemainderTheorem
{
public static int Solve(int[] n, int[] a)
{
int prod = n.Aggregate(1, (i, j) => i * j);
int p;
int sm = 0;
for (int i = 0; i < n.Length; i++)
{
p = prod / n[i];
sm += a[i] * ModularMultiplicativeInverse(p, n[i]) * p;
}
return sm % prod;
}

private static int ModularMultiplicativeInverse(int a, int mod)
{
int b = a % mod;
for (int x = 1; x < mod; x++)
{
if ((b * x) % mod == 1)
{
return x;
}
}
return 1;
}
}
}


## Clojure

Modeled after the Python version http://rosettacode.org/wiki/Category:Python

(ns test-p.core
(:require [clojure.math.numeric-tower :as math]))

(defn extended-gcd
"The extended Euclidean algorithm
Returns a list containing the GCD and the Bézout coefficients
corresponding to the inputs. "
[a b]
(cond (zero? a) [(math/abs b) 0 1]
(zero? b) [(math/abs a) 1 0]
:else (loop [s 0
s0 1
t 1
t0 0
r (math/abs b)
r0 (math/abs a)]
(if (zero? r)
[r0 s0 t0]
(let [q (quot r0 r)]
(recur (- s0 (* q s)) s
(- t0 (* q t)) t
(- r0 (* q r)) r))))))

(defn chinese_remainder
" Main routine to return the chinese remainder "
[n a]
(let [prod (apply * n)
reducer (fn [sum [n_i a_i]]
(let [p (quot prod n_i)           ; p = prod / n_i
egcd (extended-gcd p n_i)   ; Extended gcd
inv_p (second egcd)]        ; Second item is the inverse
(+ sum (* a_i inv_p p))))
sum-prod (reduce reducer 0 (map vector n a))] ; Replaces the Python for loop to sum
; (map vector n a) is same as
;                                             ; Python's version Zip (n, a)
(mod sum-prod prod)))                             ; Result line

(def n [3 5 7])
(def a [2 3 2])

(println (chinese_remainder n a))



'''Output:'''


23



## Coffeescript

crt = (n,a) ->
sum = 0
prod = n.reduce (a,c) -> a*c
for [ni,ai] in _.zip n,a
p = prod // ni
sum += ai * p * mulInv p,ni
sum % prod

mulInv = (a,b) ->
b0 = b
[x0,x1] = [0,1]
if b==1 then return 1
while a > 1
q = a // b
[a,b] = [b, a % b]
[x0,x1] = [x1-q*x0, x0]
if x1 < 0 then x1 += b0
x1

print crt [3,5,7], [2,3,2]


'''Output:'''


23



## Common Lisp

Using function ''invmod'' from [[http://rosettacode.org/wiki/Modular_inverse#Common_Lisp]].


(defun chinese-remainder (am)
"Calculates the Chinese Remainder for the given set of integer modulo pairs.
Note: All the ni and the N must be coprimes."
(loop :for (a . m) :in am
:with mtot = (reduce #'* (mapcar #'(lambda(X) (cdr X)) am))
:with sum  = 0
:finally (return (mod sum mtot))
:do
(incf sum (* a (invmod (/ mtot m) m) (/ mtot m)))))



'''Output:'''


* (chinese-remainder '((2 . 3) (3 . 5) (2 . 7)))

23
* (chinese-remainder '((10 . 11) (4 . 12) (12 . 13)))

1000
* (chinese-remainder '((19 . 100) (0 . 23)))

1219
* (chinese-remainder '((10 . 11) (4 . 22) (9 . 19)))

debugger invoked on a SIMPLE-ERROR in thread
invmod: Values 418 and 11 are not coprimes.

Type HELP for debugger help, or (SB-EXT:EXIT) to exit from SBCL.

restarts (invokable by number or by possibly-abbreviated name):
0: [ABORT] Exit debugger, returning to top level.

(INVMOD 418 11)
0]



## D

{{trans|Python}}

import std.stdio, std.algorithm;

T chineseRemainder(T)(in T[] n, in T[] a) pure nothrow @safe @nogc
in {
assert(n.length == a.length);
} body {
static T mulInv(T)(T a, T b) pure nothrow @safe @nogc {
auto b0 = b;
T x0 = 0, x1 = 1;
if (b == 1)
return T(1);
while (a > 1) {
immutable q = a / b;
immutable amb = a % b;
a = b;
b = amb;
immutable xqx = x1 - q * x0;
x1 = x0;
x0 = xqx;
}
if (x1 < 0)
x1 += b0;
return x1;
}

immutable prod = reduce!q{a * b}(T(1), n);

T p = 1, sm = 0;
foreach (immutable i, immutable ni; n) {
p = prod / ni;
sm += a[i] * mulInv(p, ni) * p;
}
return sm % prod;
}

void main() {
immutable n = [3, 5, 7],
a = [2, 3, 2];
chineseRemainder(n, a).writeln;
}


{{out}}

23


## EasyLang

{{trans|C}} func mul_inv a b . x1 . b0 = b x1 = 1 if b <> 1 while a > 1 q = a / b t = b b = a mod b a = t t = x0 x0 = x1 - q * x0 x1 = t . if x1 < 0 x1 += b0 . . . func remainder . n[] a[] r . prod = 1 sum = 0 for i range len n[] prod *= n[i] . for i range len n[] p = prod / n[i] call mul_inv p n[i] h sum += a[i] * h * p r = sum mod prod . . n[] = [ 3 5 7 ] a[] = [ 2 3 2 ] call remainder n[] a[] h print h


{{out}}

txt
23


## EchoLisp

'''egcd''' - extended gcd - and '''crt-solve''' - chinese remainder theorem solve - are included in math.lib.


(lib 'math)
math.lib v1.10 ® EchoLisp

(crt-solve '(2 3 2) '(3 5 7))
→ 23
(crt-solve '(2 3 2) '(7 1005 15))
💥 error: mod[i] must be co-primes : assertion failed : 1005



## Elixir

{{trans|Ruby}} {{works with|Elixir|1.2}} Brute-force:

defmodule Chinese do
def remainder(mods, remainders) do
max = Enum.reduce(mods, fn x,acc -> x*acc end)
Enum.zip(mods, remainders)
|> Enum.map(fn {m,r} -> Enum.take_every(r..max, m) |> MapSet.new end)
|> Enum.reduce(fn set,acc -> MapSet.intersection(set, acc) end)
|> MapSet.to_list
end
end

IO.inspect Chinese.remainder([3,5,7], [2,3,2])
IO.inspect Chinese.remainder([10,4,9], [11,22,19])
IO.inspect Chinese.remainder([11,12,13], [10,4,12])


{{out}}


[23]
[]
[1000]



## Erlang

{{trans|OCaml}}

-module(crt).
-import(lists, [zip/2, unzip/1, foldl/3, sum/1]).
-export([egcd/2, mod/2, mod_inv/2, chinese_remainder/1]).

egcd(_, 0) -> {1, 0};
egcd(A, B) ->
{S, T} = egcd(B, A rem B),
{T, S - (A div B)*T}.

mod_inv(A, B) ->
{X, Y} = egcd(A, B),
if
A*X + B*Y =:= 1 -> X;
true -> undefined
end.

mod(A, M) ->
X = A rem M,
if
X < 0 -> X + M;
true -> X
end.

calc_inverses([], []) -> [];
calc_inverses([N | Ns], [M | Ms]) ->
case mod_inv(N, M) of
undefined -> undefined;
Inv -> [Inv | calc_inverses(Ns, Ms)]
end.

chinese_remainder(Congruences) ->
{Residues, Modulii} = unzip(Congruences),
ModPI = foldl(fun(A, B) -> A*B end, 1, Modulii),
CRT_Modulii = [ModPI div M || M <- Modulii],
case calc_inverses(CRT_Modulii, Modulii) of
undefined -> undefined;
Inverses ->
Solution = sum([A*B || {A,B} <- zip(CRT_Modulii,
[A*B || {A,B} <- zip(Residues, Inverses)])]),
mod(Solution, ModPI)
end.


{{out}}


16> crt:chinese_remainder([{10,11}, {4,12}, {12,13}]).
1000
17> crt:chinese_remainder([{10,11}, {4,22}, {9,19}]).
undefined
18> crt:chinese_remainder([{2,3}, {3,5}, {2,7}]).
23



let rec sieve cs x N =
match cs with
| [] -> Some(x)
| (a,n)::rest ->
let arrProgress = Seq.unfold (fun x -> Some(x, x+N)) x
let firstXmodNequalA = Seq.tryFind (fun x -> a = x % n)
match firstXmodNequalA (Seq.take n arrProgress) with
| None -> None
| Some(x) -> sieve rest x (N*n)

[ [(2,3);(3,5);(2,7)];
[(10,11); (4,22); (9,19)];
[(10,11); (4,12); (12,13)] ]
|> List.iter (fun congruences ->
let cs =
congruences
|> List.map (fun (a,n) -> (a % n, n))
|> List.sortBy (snd>>(~-))
match sieve (List.tail cs) (fst an) (snd an) with
| None    -> printfn "no solution"
| Some(x) -> printfn "result = %i" x
)


{{out}}

result = 23
no solution
result = 1000


### Or for those who prefer unsieved

This uses [[Greatest_common_divisor#F.23]] to verify valid input, can be simplified if you know input has a solution.

This uses [[Modular_inverse#F.23]]


//Chinese Division Theorem: Nigel Galloway: April 3rd., 2017
let CD n g =
match Seq.fold(fun n g->if (gcd n g)=1 then n*g else 0) 1 g with
|0 -> None
|fN-> Some ((Seq.fold2(fun n i g -> n+i*(fN/g)*(MI g ((fN/g)%g))) 0 n g)%fN)



{{out}}


CD [10;4;12] [11;12;13] -> Some 1000
CD [10;4;9] [11;22;19]  -> None
CD [2;3;2] [3;5;7]      -> Some 23



## Factor

USING: math.algebra prettyprint ;
{ 2 3 2 } { 3 5 7 } chinese-remainder .


{{out}}


23



## Forth

Tested with GNU FORTH

: egcd ( a b -- a b )
dup 0= IF
2drop 1 0
ELSE
dup -rot /mod               \ -- b r=a%b q=a/b
-rot recurse                \ -- q (s,t) = egcd(b, r)
>r swap r@ * - r> swap      \ -- t (s - q*t)
THEN ;

: egcd>gcd  ( a b x y -- n )  \ calculate gcd from egcd
rot * -rot * + ;

: mod-inv  ( a m -- a' )     \ modular inverse with coprime check
2dup egcd over >r egcd>gcd r> swap 1 <> -24 and throw ;

: array-product ( adr count -- n )
1 -rot  cells bounds ?DO  i @ *  cell +LOOP ;

2dup array-product   locals| M count m[] a[] |
0  \ result
count 0 DO
m[] i cells + @
dup M swap /
dup rot mod-inv *
a[] i cells + @ * +
LOOP  M mod ;

create crt-residues[]  10 cells allot
create crt-moduli[]    10 cells allot

: crt ( .... n -- n )  \ takes pairs of "n (mod m)" from stack.
10 min  locals| n |
n 0 DO
crt-moduli[] i cells + !
crt-residues[] i cells + !
LOOP
crt-residues[] crt-moduli[] n crt-from-array ;



{{out}}


Gforth 0.7.2, Copyright (C) 1995-2008 Free Software Foundation, Inc.
Gforth comes with ABSOLUTELY NO WARRANTY; for details type license'
Type bye' to exit
10 11  4 12  12 13  3 crt . 1000  ok
10 11  4 22   9 19  3 crt .
:2: Invalid numeric argument
10 11  4 22   9 19  3 >>>crt<<< .



## FunL

import integers.modinv

def crt( congruences ) =
N = product( n | (_, n) <- congruences )
sum( a*modinv(N/n, n)*N/n | (a, n) <- congruences ) mod N

println( crt([(2, 3), (3, 5), (2, 7)]) )


{{out}}


23



## Go

Go has the Extended Euclidean algorithm in the GCD function for big integers in the standard library. GCD will return 1 only if numbers are coprime, so a result != 1 indicates the error condition.

package main

import (
"fmt"
"math/big"
)

var one = big.NewInt(1)

func crt(a, n []*big.Int) (*big.Int, error) {
p := new(big.Int).Set(n[0])
for _, n1 := range n[1:] {
p.Mul(p, n1)
}
var x, q, s, z big.Int
for i, n1 := range n {
q.Div(p, n1)
z.GCD(nil, &s, n1, &q)
if z.Cmp(one) != 0 {
return nil, fmt.Errorf("%d not coprime", n1)
}
}
return x.Mod(&x, p), nil
}

func main() {
n := []*big.Int{
big.NewInt(3),
big.NewInt(5),
big.NewInt(7),
}
a := []*big.Int{
big.NewInt(2),
big.NewInt(3),
big.NewInt(2),
}
fmt.Println(crt(a, n))
}


{{out}} Two values, the solution x and an error value.


23 <nil>



{{trans|Erlang}}

import Control.Monad (zipWithM)

egcd :: Int -> Int -> (Int, Int)
egcd _ 0 = (1, 0)
egcd a b = (t, s - q * t)
where
(s, t) = egcd b r
(q, r) = a quotRem b

modInv :: Int -> Int -> Either String Int
modInv a b =
case egcd a b of
(x, y)
| a * x + b * y == 1 -> Right x
| otherwise ->
Left $"No modular inverse for " ++ show a ++ " and " ++ show b chineseRemainder :: [Int] -> [Int] -> Either String Int chineseRemainder residues modulii = zipWithM modInv crtModulii modulii >>= (Right . (mod modPI) . sum . zipWith (*) crtModulii . zipWith (*) residues) where modPI = product modulii crtModulii = (modPI div) <$> modulii

main :: IO ()
main =
mapM_ (putStrLn . either id show) $uncurry chineseRemainder <$>
[ ([10, 4, 12], [11, 12, 13])
, ([10, 4, 9], [11, 22, 19])
, ([2, 3, 2], [3, 5, 7])
]


{{Out}}

1000
No modular inverse for 418 and 11
23


{{trans|Python}} with error check added. Works in both languages:

link numbers   # for gcd()

procedure main()
write(cr([3,5,7],[2,3,2]) | "No solution!")
write(cr([10,4,9],[11,22,19]) | "No solution!")
end

procedure cr(n,a)
if 1 ~= gcd(n[i := !*n],a[i]) then fail  # Not pairwise coprime
(prod := 1, sm := 0)
every prod *:= !n
every p := prod/(ni := n[i := !*n]) do sm +:= a[i] * mul_inv(p,ni) * p
return sm%prod
end

procedure mul_inv(a,b)
if b = 1 then return 1
(b0 := b, x0 := 0, x1 := 1)
while q := (1 < a)/b do {
(t := a, a := b, b := t%b)
(t := x0, x0 := x1-q*t, x1 := t)
}
return if x1 < 0 then x1+b0 else x1
end


Output:


->crt
23
No solution!
->



## J

'''Solution''' (''brute force''):

   crt =: (1 + ] - {:@:[ -: {.@:[ | ])^:_&0@:,:


'''Example''':

   3 5 7 crt 2 3 2
23
11 12 13 crt 10 4 12
1000


'''Notes''': This is a brute force approach and does not meet the requirement for explicit notification of an an unsolvable set of equations (it just spins forever). A much more thorough and educational approach can be found on the [[j:Essays/Chinese%20Remainder%20Theorem|J wiki's Essay on the Chinese Remainder Thereom]].

## Java

Translation of [[Chinese_remainder_theorem#Python|Python]] via [[Chinese_remainder_theorem#D|D]] {{works with|Java|8}}

import static java.util.Arrays.stream;

public class ChineseRemainderTheorem {

public static int chineseRemainder(int[] n, int[] a) {

int prod = stream(n).reduce(1, (i, j) -> i * j);

int p, sm = 0;
for (int i = 0; i < n.length; i++) {
p = prod / n[i];
sm += a[i] * mulInv(p, n[i]) * p;
}
return sm % prod;
}

private static int mulInv(int a, int b) {
int b0 = b;
int x0 = 0;
int x1 = 1;

if (b == 1)
return 1;

while (a > 1) {
int q = a / b;
int amb = a % b;
a = b;
b = amb;
int xqx = x1 - q * x0;
x1 = x0;
x0 = xqx;
}

if (x1 < 0)
x1 += b0;

return x1;
}

public static void main(String[] args) {
int[] n = {3, 5, 7};
int[] a = {2, 3, 2};
System.out.println(chineseRemainder(n, a));
}
}

23


## jq

This implementation is similar to the one in C, but raises an error if there is no solution, as illustrated in the last example.

# mul_inv(a;b) returns x where (a * x) % b == 1, or else null
def mul_inv(a; b):

# state: [a, b, x0, x1]
def iterate:
.[0] as $a | .[1] as$b
| if $a > 1 then if$b == 0 then null
else ($a /$b | floor) as $q | [$b, ($a %$b), (.[3] - ($q * .[2])), .[2]] | iterate end else . end ; if (b == 1) then 1 else [a,b,0,1] | iterate | if . == null then . else .[3] | if . < 0 then . + b else . end end end; def chinese_remainder(mods; remainders): (reduce mods[] as$i (1; . * $i)) as$prod
| reduce range(0; mods|length) as $i (0; ($prod/mods[$i]) as$p
| mul_inv($p; mods[$i]) as $mi | if$mi == null then error("nogo: p=\($p) mods[\($i)]=\(mods[$i])") else . + (remainders[$i] * $mi *$p)
end )
| . % $prod ;  '''Examples''': chinese_remainder([3,5,7]; [2,3,2]) # => 23 chinese_remainder([100,23]; [19,0]) # => 1219 chinese_remainder([10,4,9]; [11,22,19]) # jq: error: nogo: p=36 mods[0]=10 ## Julia {{works with|Julia|1.2}} function chineseremainder(n::Array, a::Array) Π = prod(n) mod(sum(ai * invmod(Π ÷ ni, ni) * Π ÷ ni for (ni, ai) in zip(n, a)), Π) end @show chineseremainder([3, 5, 7], [2, 3, 2])  {{out}} chineseremainder([3, 5, 7], [2, 3, 2]) = 23  ## Kotlin {{trans|C}} // version 1.1.2 /* returns x where (a * x) % b == 1 */ fun multInv(a: Int, b: Int): Int { if (b == 1) return 1 var aa = a var bb = b var x0 = 0 var x1 = 1 while (aa > 1) { val q = aa / bb var t = bb bb = aa % bb aa = t t = x0 x0 = x1 - q * x0 x1 = t } if (x1 < 0) x1 += b return x1 } fun chineseRemainder(n: IntArray, a: IntArray): Int { val prod = n.fold(1) { acc, i -> acc * i } var sum = 0 for (i in 0 until n.size) { val p = prod / n[i] sum += a[i] * multInv(p, n[i]) * p } return sum % prod } fun main(args: Array<String>) { val n = intArrayOf(3, 5, 7) val a = intArrayOf(2, 3, 2) println(chineseRemainder(n, a)) }  {{out}}  23  ## Lua -- Taken from https://www.rosettacode.org/wiki/Sum_and_product_of_an_array#Lua function prodf(a, ...) return a and a * prodf(...) or 1 end function prodt(t) return prodf(unpack(t)) end function mulInv(a, b) local b0 = b local x0 = 0 local x1 = 1 if b == 1 then return 1 end while a > 1 do local q = math.floor(a / b) local amb = math.fmod(a, b) a = b b = amb local xqx = x1 - q * x0 x1 = x0 x0 = xqx end if x1 < 0 then x1 = x1 + b0 end return x1 end function chineseRemainder(n, a) local prod = prodt(n) local p local sm = 0 for i=1,#n do p = prod / n[i] sm = sm + a[i] * mulInv(p, n[i]) * p end return math.fmod(sm, prod) end n = {3, 5, 7} a = {2, 3, 2} io.write(chineseRemainder(n, a))  {{out}} 23  ## Maple This is a Maple built-in procedure, so it is trivial:  chrem( [2, 3, 2], [3, 5, 7] ); 23  =={{header|Mathematica}} / {{header|Wolfram Language}}== Very easy, because it is a built-in function: ChineseRemainder[{2, 3, 2}, {3, 5, 7}] 23  =={{header|MATLAB}} / {{header|Octave}}== function f = chineseRemainder(r, m) s = prod(m) ./ m; [~, t] = gcd(s, m); f = s .* t * r';  {{out}}  chineseRemainder([2 3 2], [3 5 7]) ans = 23  =={{header|Modula-2}}== MODULE CRT; FROM FormatString IMPORT FormatString; FROM Terminal IMPORT WriteString,WriteLn,ReadChar; PROCEDURE WriteInt(n : INTEGER); VAR buf : ARRAY[0..15] OF CHAR; BEGIN FormatString("%i", buf, n); WriteString(buf) END WriteInt; PROCEDURE MulInv(a,b : INTEGER) : INTEGER; VAR b0,x0,x1,q,amb,xqx : INTEGER; BEGIN b0 := b; x0 := 0; x1 := 1; IF b=1 THEN RETURN 1 END; WHILE a>1 DO q := a DIV b; amb := a MOD b; a := b; b := amb; xqx := x1 - q * x0; x1 := x0; x0 := xqx END; IF x1<0 THEN x1 := x1 + b0 END; RETURN x1 END MulInv; PROCEDURE ChineseRemainder(n,a : ARRAY OF INTEGER) : INTEGER; VAR i : CARDINAL; prod,p,sm : INTEGER; BEGIN prod := n[0]; FOR i:=1 TO HIGH(n) DO prod := prod * n[i] END; sm := 0; FOR i:=0 TO HIGH(n) DO p := prod DIV n[i]; sm := sm + a[i] * MulInv(p, n[i]) * p END; RETURN sm MOD prod END ChineseRemainder; TYPE TA = ARRAY[0..2] OF INTEGER; VAR n,a : TA; BEGIN n := TA{3, 5, 7}; a := TA{2, 3, 2}; WriteInt(ChineseRemainder(n, a)); WriteLn; ReadChar END CRT.  {{out}} 23  ## Nim {{trans|C}} proc mulInv(a0, b0): int = var (a, b, x0) = (a0, b0, 0) result = 1 if b == 1: return while a > 1: let q = a div b a = a mod b swap a, b result = result - q * x0 swap x0, result if result < 0: result += b0 proc chineseRemainder[T](n, a: T): int = var prod = 1 var sum = 0 for x in n: prod *= x for i in 0 .. <n.len: let p = prod div n[i] sum += a[i] * mulInv(p, n[i]) * p sum mod prod echo chineseRemainder([3,5,7], [2,3,2])  Output: 23  ## OCaml This is using the Jane Street Ocaml Core library. open Core.Std open Option.Monad_infix let rec egcd a b = if b = 0 then (1, 0) else let q = a/b and r = a mod b in let (s, t) = egcd b r in (t, s - q*t) let mod_inv a b = let (x, y) = egcd a b in if a*x + b*y = 1 then Some x else None let calc_inverses ns ms = let rec list_inverses ns ms l = match (ns, ms) with | ([], []) -> Some l | ([], _) | (_, []) -> assert false | (n::ns, m::ms) -> let inv = mod_inv n m in match inv with | None -> None | Some v -> list_inverses ns ms (v::l) in list_inverses ns ms [] >>= fun l -> Some (List.rev l) let chinese_remainder congruences = let (residues, modulii) = List.unzip congruences in let mod_pi = List.reduce_exn modulii ~f:( * ) in let crt_modulii = List.map modulii ~f:(fun m -> mod_pi / m) in calc_inverses crt_modulii modulii >>= fun inverses -> Some (List.map3_exn residues inverses crt_modulii ~f:(fun a b c -> a*b*c) |> List.reduce_exn ~f:(+) |> fun n -> let n' = n mod mod_pi in if n' < 0 then n' + mod_pi else n')  {{out}}  utop # chinese_remainder [(10, 11); (4, 12); (12, 13)];; - : int option = Some 1000 utop # chinese_remainder [(10, 11); (4, 22); (9, 19)];; - : int option = None  ## PARI/GP chivec(residues, moduli)={ my(m=Mod(0,1)); for(i=1,#residues, m=chinese(Mod(residues[i],moduli[i]),m) ); lift(m) }; chivec([2,3,2], [3,5,7])  {{out}} 23  Pari's chinese function takes a vector in the form [Mod(a1,n1), Mod(a2, n2), ...], so we can do this directly: lift( chinese([Mod(2,3),Mod(3,5),Mod(2,7)]) )  or to take the residue/moduli array as above: chivec(residues,moduli)={ lift(chinese(vector(#residues,i,Mod(residues[i],moduli[i])))) }  ## Perl There are at least three CPAN modules for this: ntheory (Math::Prime::Util), Math::ModInt, and Math::Pari. All three handle bigints. {{libheader|ntheory}} use ntheory qw/chinese/; say chinese([2,3], [3,5], [2,7]);  {{out}} 23  The function returns undef if no common residue class exists. The combined modulus can be obtained using the lcm function applied to the moduli (e.g. lcm(3,5,7) = 105 in the example above). use Math::ModInt qw(mod); use Math::ModInt::ChineseRemainder qw(cr_combine); say cr_combine(mod(2,3),mod(3,5),mod(2,7));  {{out}} mod(23, 105)  This returns a Math::ModInt object, which if no common residue class exists will be a special undefined object. The modulus and residue methods may be used to extract the integer components. === Non-pairwise-coprime === All three modules will also handle cases where the moduli are not pairwise co-prime but a solution exists, e.g.: use ntheory qw/chinese lcm/; say chinese( [2328,16256], [410,5418] ), " mod ", lcm(16256,5418);  {{out}} 28450328 mod 44037504  ## Perl 6 {{trans|C}} {{works with|Rakudo|2015.12}} # returns x where (a * x) % b == 1 sub mul-inv($a is copy, $b is copy) { return 1 if$b == 1;
my ($b0, @x) =$b, 0, 1;
($a,$b, @x) = (
$b,$a % $b, @x[1] - ($a div $b)*@x[0], @x[0] ) while$a > 1;
@x[1] += $b0 if @x[1] < 0; return @x[1]; } sub chinese-remainder(*@n) { my \N = [*] @n; -> *@a { N R% [+] map { my \p = N div @n[$_];
@a[$_] * mul-inv(p, @n[$_]) * p
}, ^@n
}
}

say chinese-remainder(3, 5, 7)(2, 3, 2);


{{out}}

23


## Phix

{{trans|C}} Uses the function mul_inv() from [[Modular_inverse#Phix]]

function chinese_remainder(sequence n, a)
integer p, prod = 1, tot = 0;
for i=1 to length(n) do prod *= n[i] end for
for i=1 to length(n) do
p = prod / n[i];
object m = mul_inv(p, n[i])
if string(m) then return "fail" end if
tot += a[i] * m * p;
end for
return mod(tot,prod)
end function

?chinese_remainder({3,5,7},{2,3,2})
?chinese_remainder({11,12,13},{10,4,12})
?chinese_remainder({11,22,19},{10,4,9})
?chinese_remainder({100,23},{19,0})


{{out}}


23
1000
"fail"
1219



## PicoLisp

(de modinv (A B)
(let (B0 B  X0 0  X1 1  Q 0  T1 0)
(while (< 1 A)
(setq
Q (/ A B)
T1 B
B (% A B)
A T1
T1 X0
X0 (- X1 (* Q X0))
X1 T1 ) )
(if (lt0 X1) (+ X1 B0) X1) ) )

(de chinrem (N A)
(let P (apply * N)
(%
(sum
'((N A)
(setq T1 (/ P N))
(* A (modinv T1 N) T1) )
N
A )
P ) ) )

(println
(chinrem (3 5 7) (2 3 2))
(chinrem (11 12 13) (10 4 12)) )

(bye)


## PureBasic

EnableExplicit
DisableDebugger
DataSection
LBL_n1:
Data.i 3,5,7
LBL_a1:
Data.i 2,3,2
LBL_n2:
Data.i 11,12,13
LBL_a2:
Data.i 10,4,12
LBL_n3:
Data.i 10,4,9
LBL_a3:
Data.i 11,22,19
EndDataSection

Procedure ErrorHdl()
Print(ErrorMessage())
Input()
EndProcedure

Macro PrintData(n,a)
Define Idx.i=0
Print("[")
While n+SizeOf(Integer)*Idx<a
Print("( ")
Print(Str(PeekI(a+SizeOf(Integer)*Idx)))
Print(" . ")
Print(Str(PeekI(n+SizeOf(Integer)*Idx)))
Print(" )")
Idx+1
Wend
Print(~"]\nx = ")
EndMacro

Define p.i=1
Wend
ProcedureReturn p
EndProcedure

Procedure.i Eval_x1(a.i,b.i)
Define b0.i=b, x0.i=0, x1.i=1, q.i, t.i
If b=1 : ProcedureReturn x1 : EndIf
While a>1
q=Int(a/b)
t=b : b=a%b : a=t
t=x0 : x0=x1-q*x0 : x1=t
Wend
If x1<0 : ProcedureReturn x1+b0 : EndIf
ProcedureReturn x1
EndProcedure

p=Int(prod/b) : a=p
Idx+1
Wend
ProcedureReturn sum%prod
EndProcedure

OnErrorCall(@ErrorHdl())
OpenConsole("Chinese remainder theorem")
PrintData(?LBL_n1,?LBL_a1)
PrintN(Str(ChineseRem(?LBL_n1,?LBL_a1)))
PrintData(?LBL_n2,?LBL_a2)
PrintN(Str(ChineseRem(?LBL_n2,?LBL_a2)))
PrintData(?LBL_n3,?LBL_a3)
PrintN(Str(ChineseRem(?LBL_n3,?LBL_a3)))
Input()


{{out}}

[( 2 . 3 )( 3 . 5 )( 2 . 7 )]
x = 23
[( 10 . 11 )( 4 . 12 )( 12 . 13 )]
x = 1000
[( 11 . 10 )( 22 . 4 )( 19 . 9 )]
x = Division by zero


## Python

### =Python 2.7=

# Python 2.7
def chinese_remainder(n, a):
sum = 0
prod = reduce(lambda a, b: a*b, n)

for n_i, a_i in zip(n, a):
p = prod / n_i
sum += a_i * mul_inv(p, n_i) * p
return sum % prod

def mul_inv(a, b):
b0 = b
x0, x1 = 0, 1
if b == 1: return 1
while a > 1:
q = a / b
a, b = b, a%b
x0, x1 = x1 - q * x0, x0
if x1 < 0: x1 += b0
return x1

if __name__ == '__main__':
n = [3, 5, 7]
a = [2, 3, 2]
print chinese_remainder(n, a)


{{out}}

23


### =Python 3.6=

# Python 3.6
from functools import reduce
def chinese_remainder(n, a):
sum = 0
prod = reduce(lambda a, b: a*b, n)
for n_i, a_i in zip(n, a):
p = prod // n_i
sum += a_i * mul_inv(p, n_i) * p
return sum % prod

def mul_inv(a, b):
b0 = b
x0, x1 = 0, 1
if b == 1: return 1
while a > 1:
q = a // b
a, b = b, a%b
x0, x1 = x1 - q * x0, x0
if x1 < 0: x1 += b0
return x1

if __name__ == '__main__':
n = [3, 5, 7]
a = [2, 3, 2]
print(chinese_remainder(n, a))


{{out}}

23


### Functional

Using an option type to represent the possibility that there may or may not be a solution for any given pair of input lists.

(Note that the procedural versions above both fail with a '''ZeroDivisionError''' on inputs for which no solution is found).

'''Chinese remainder theorem'''

from functools import reduce

# cnRemainder :: [Int] -> [Int] -> Either String Int
def cnRemainder(ms):
'''Chinese remainder theorem.
(moduli, residues) -> Either explanation or solution
'''
def go(ms, rs):
mp = numericProduct(ms)
cms = [(mp // x) for x in ms]

def possibleSoln(invs):
return Right(
sum(map(
mul,
cms, map(mul, rs, invs)
)) % mp
)
return bindLR(
zipWithEither(modMultInv)(cms)(ms)
)(possibleSoln)

return lambda rs: go(ms, rs)

# modMultInv :: Int -> Int -> Either String Int
def modMultInv(a, b):
'''Modular multiplicative inverse.'''
x, y = eGcd(a, b)
return Right(x) if 1 == (a * x + b * y) else (
Left('no modular inverse for ' + str(a) + ' and ' + str(b))
)

# egcd :: Int -> Int -> (Int, Int)
def eGcd(a, b):
'''Extended greatest common divisor.'''
def go(a, b):
if 0 == b:
return (1, 0)
else:
q, r = divmod(a, b)
(s, t) = go(b, r)
return (t, s - q * t)
return go(a, b)

# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''Tests of soluble and insoluble cases.'''

print(
fTable(
__doc__ + ':\n\n         (moduli, residues) -> ' + (
'Either solution or explanation\n'
)
)(repr)(
compose(quoted(' '))(repr)
)
)(uncurry(cnRemainder))([
([10, 4, 12], [11, 12, 13]),
([11, 12, 13], [10, 4, 12]),
([10, 4, 9], [11, 22, 19]),
([3, 5, 7], [2, 3, 2]),
([2, 3, 2], [3, 5, 7])
])
)

# GENERIC -------------------------------------------------

# Left :: a -> Either a b
def Left(x):
'''Constructor for an empty Either (option type) value
with an associated string.'''
return {'type': 'Either', 'Right': None, 'Left': x}

# Right :: b -> Either a b
def Right(x):
'''Constructor for a populated Either (option type) value'''
return {'type': 'Either', 'Left': None, 'Right': x}
# any :: (a -> Bool) -> [a] -> Bool

def any_(p):
'''True if p(x) holds for at least
one item in xs.'''
def go(xs):
for x in xs:
if p(x):
return True
return False
return lambda xs: go(xs)

# bindLR (>>=) :: Either a -> (a -> Either b) -> Either b
def bindLR(m):
Two computations sequentially composed,
with any value produced by the first
passed as an argument to the second.'''
return lambda mf: (
mf(m.get('Right')) if None is m.get('Left') else m
)

# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
'''Right to left function composition.'''
return lambda f: lambda x: g(f(x))

# curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
'''A curried function derived
from an uncurried function.'''
return lambda a: lambda b: f(a, b)

# either :: (a -> c) -> (b -> c) -> Either a b -> c
def either(fl):
'''The application of fl to e if e is a Left value,
or the application of fr to e if e is a Right value.'''
return lambda fr: lambda e: fl(e['Left']) if (
None is e['Right']
) else fr(e['Right'])

# fTable :: String -> (a -> String) ->
#                     (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function ->
fx display function ->
f -> value list -> tabular string.'''
def go(xShow, fxShow, f, xs):
w = max(map(compose(len)(xShow), xs))
return s + '\n' + '\n'.join([
xShow(x).rjust(w, ' ') + (' -> ') + fxShow(f(x))
for x in xs
])
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)

# numericProduct :: [Num] -> Num
def numericProduct(xs):
'''The arithmetic product of all numbers in xs.'''
return reduce(mul, xs, 1)

# partitionEithers :: [Either a b] -> ([a],[b])
def partitionEithers(lrs):
'''A list of Either values partitioned into a tuple
of two lists, with all Left elements extracted
into the first list, and Right elements
extracted into the second list.
'''
def go(a, x):
ls, rs = a
r = x.get('Right')
return (ls + [x.get('Left')], rs) if None is r else (
ls, rs + [r]
)
return reduce(go, lrs, ([], []))

# quoted :: Char -> String -> String
def quoted(c):
'''A string flanked on both sides
by a specified quote character.
'''
return lambda s: c + s + c

# uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
'''A function over a tuple,
derived from a curried function.'''
return lambda xy: f(xy[0])(xy[1])

# zipWithEither :: (a -> b -> Either String  c)
#            -> [a] -> [b] -> Either String [c]
def zipWithEither(f):
'''Either a list of results if f succeeds with every pair
in the zip of xs and ys, or an explanatory string
if any application of f returns no result.
'''
def go(xs, ys):
ls, rs = partitionEithers(map(f, xs, ys))
return Left(ls[0]) if ls else Right(rs)
return lambda xs: lambda ys: go(xs, ys)

# MAIN ---
if __name__ == '__main__':
main()


{{Out}}

Chinese remainder theorem:

(moduli, residues) -> Either solution or explanation

([10, 4, 12], [11, 12, 13]) -> 'No solution: no modular inverse for 48 and 10'
([11, 12, 13], [10, 4, 12]) ->  1000
([10, 4, 9], [11, 22, 19]) -> 'No solution: no modular inverse for 36 and 10'
([3, 5, 7], [2, 3, 2]) ->  23
([2, 3, 2], [3, 5, 7]) -> 'No solution: no modular inverse for 6 and 2'


## R

{{trans|C}}

mul_inv <- function(a, b)
{
b0 <- b
x0 <- 0L
x1 <- 1L

if (b == 1) return(1L)
while(a > 1){
q <- as.integer(a/b)

t <- b
b <- a %% b
a <- t

t <- x0
x0 <- x1 - q*x0
x1 <- t
}

if (x1 < 0) x1 <- x1 + b0
return(x1)
}

chinese_remainder <- function(n, a)
{
len <- length(n)

prod <- 1L
sum <- 0L

for (i in 1:len) prod <- prod * n[i]

for (i in 1:len){
p <- as.integer(prod / n[i])
sum <- sum + a[i] * mul_inv(p, n[i]) * p
}

return(sum %% prod)
}

n <- c(3L, 5L, 7L)
a <- c(2L, 3L, 2L)

chinese_remainder(n, a)


{{out}}

23


## Racket

This is more of a demonstration of the built-in function "solve-chinese", than anything. A bit cheeky, I know... but if you've got a dog, why bark yourself?

Take a look in the "math/number-theory" package it's full of goodies! URL removed -- I can't be doing the Dutch recaptchas I'm getting.

#lang racket
(require (only-in math/number-theory solve-chinese))
(define as '(2 3 2))
(define ns '(3 5 7))
(solve-chinese as ns)


{{out}}

23


## REXX

### algebraic

/*REXX program demonstrates  Sun Tzu's  (or Sunzi's)  Chinese Remainder  Theorem.       */
parse arg Ns As .                                /*get optional arguments from the C.L. */
if Ns=='' | Ns==","  then Ns = '3,5,7'           /*Ns not specified?   Then use default.*/
if As=='' | As==","  then As = '2,3,2'           /*As  "      "          "   "      "   */
say 'Ns: ' Ns
say 'As: ' As;                   say
Ns=space(translate(Ns, , ','));  #=words(Ns)     /*elide any superfluous blanks from N's*/
As=space(translate(As, , ','));  _=words(As)     /*  "    "       "        "      "  A's*/
if #\==_   then do;  say  "size of number sets don't match.";   exit 131;    end
if #==0    then do;  say  "size of the  N  set isn't valid.";   exit 132;    end
if _==0    then do;  say  "size of the  A  set isn't valid.";   exit 133;    end
N=1                                              /*the product─to─be  for  prod(n.j).   */
do j=1  for #                              /*process each number for  As  and Ns. */
n.j=word(Ns,j);  N=N*n.j                   /*get an  N.j  and calculate product.  */
a.j=word(As,j)                             /* "   "  A.j  from the  As  list.     */
end   /*j*/

do    x=1  for N                           /*use a simple algebraic method.       */
do i=1  for #                           /*process each   N.i  and  A.i  number.*/
if x//n.i\==a.i  then iterate x         /*is modulus correct for the number X ?*/
end   /*i*/                             /* [↑]  limit solution to the product. */
say 'found a solution with X='   x         /*display one possible solution.       */
exit                                       /*stick a fork in it,  we're all done. */
end      /*x*/

say 'no solution found.'                         /*oops, announce that solution ¬ found.*/


{{out|output|text= when using the default inputs:}}


Ns:  3,5,7
As:  2,3,2

found a solution with X= 23



### congruences sets

/*REXX program demonstrates  Sun Tzu's  (or Sunzi's)  Chinese Remainder  Theorem.       */
parse arg Ns As .                                /*get optional arguments from the C.L. */
if Ns=='' | Ns==","  then Ns = '3,5,7'           /*Ns not specified?   Then use default.*/
if As=='' | As==","  then As = '2,3,2'           /*As  "      "          "   "      "   */
say 'Ns: ' Ns
say 'As: ' As;                   say
Ns=space(translate(Ns, , ','));  #=words(Ns)     /*elide any superfluous blanks from N's*/
As=space(translate(As, , ','));  _=words(As)     /*  "    "       "        "      "  A's*/
if #\==_   then do;  say  "size of number sets don't match.";   exit 131;    end
if #==0    then do;  say  "size of the  N  set isn't valid.";   exit 132;    end
if _==0    then do;  say  "size of the  A  set isn't valid.";   exit 133;    end
N=1                                              /*the product─to─be  for  prod(n.j).   */
do j=1  for #                              /*process each number for  As  and Ns. */
n.j=word(Ns,j);  N=N*n.j                   /*get an  N.j  and calculate product.  */
a.j=word(As,j)                             /* "   "  A.j  from the  As  list.     */
end   /*j*/
@.=                                              /* [↓]  converts congruences ───► sets.*/
do i=1  for #;  _=a.i;  @.i._=a.i;  p=a.i
do N; p=p+n.i;  @.i.p=p;  end            /*build a (array) list of modulo values*/
end   /*i*/
/* [↓]  find common number in the sets.*/
do   x=1  for N;  if @.1.x==''  then iterate                       /*locate a number. */
do v=2  to #;   if @.v.x==''  then iterate x;  end               /*Is in all sets ? */
say 'found a solution with X='    x            /*display one possible solution.       */
exit                                           /*stick a fork in it,  we're all done. */
end   /*x*/

say 'no solution found.'                         /*oops, announce that solution ¬ found.*/


{{out|output|text= is identical to the 1st REXX version.}}

## Ruby

Brute-force.


def chinese_remainder(mods, remainders)
max = mods.inject( :* )
series = remainders.zip( mods ).map{|r,m| r.step( max, m ).to_a }
series.inject( :& ).first #returns nil when empty
end

p chinese_remainder([3,5,7], [2,3,2])     #=> 23
p chinese_remainder([10,4,9], [11,22,19]) #=> nil



Similar to above, but working with large(r) numbers.


def extended_gcd(a, b)
last_remainder, remainder = a.abs, b.abs
x, last_x, y, last_y = 0, 1, 1, 0
while remainder != 0
last_remainder, (quotient, remainder) = remainder, last_remainder.divmod(remainder)
x, last_x = last_x - quotient*x, x
y, last_y = last_y - quotient*y, y
end
return last_remainder, last_x * (a < 0 ? -1 : 1)
end

def invmod(e, et)
g, x = extended_gcd(e, et)
if g != 1
raise 'Multiplicative inverse modulo does not exist!'
end
x % et
end

def chinese_remainder(mods, remainders)
max = mods.inject( :* )  # product of all moduli
series = remainders.zip(mods).map{ |r,m| (r * max * invmod(max/m, m) / m) }
series.inject( :+ ) % max
end

p chinese_remainder([3,5,7], [2,3,2])     #=> 23
p chinese_remainder([17353461355013928499, 3882485124428619605195281, 13563122655762143587], [7631415079307304117, 1248561880341424820456626, 2756437267211517231]) #=> 937307771161836294247413550632295202816
p chinese_remainder([10,4,9], [11,22,19]) #=> nil



## Rust

fn egcd(a: i64, b: i64) -> (i64, i64, i64) {
if a == 0 {
(b, 0, 1)
} else {
let (g, x, y) = egcd(b % a, a);
(g, y - (b / a) * x, x)
}
}

fn mod_inv(x: i64, n: i64) -> Option<i64> {
let (g, x, _) = egcd(x, n);
if g == 1 {
Some((x % n + n) % n)
} else {
None
}
}

fn chinese_remainder(residues: &[i64], modulii: &[i64]) -> Option<i64> {
let prod = modulii.iter().product::<i64>();

let mut sum = 0;

for (&residue, &modulus) in residues.iter().zip(modulii) {
let p = prod / modulus;
sum += residue * mod_inv(p, modulus)? * p
}

Some(sum % prod)
}

fn main() {
let modulii = [3,5,7];
let residues = [2,3,2];

match chinese_remainder(&residues, &modulii) {
Some(sol) => println!("{}", sol),
None      => println!("modulii not pairwise coprime")
}

}


## Scala

{{Out}}Best seen running in your browser either by [https://scalafiddle.io/sf/9QZvFht/0 ScalaFiddle (ES aka JavaScript, non JVM)] or [https://scastie.scala-lang.org/njcaS3BFT6GtaWT2cHiwXg Scastie (remote JVM)].

import scala.util.{Success, Try}

object ChineseRemainderTheorem extends App {

def chineseRemainder(n: List[Int], a: List[Int]): Option[Int] = {
require(n.size == a.size)
val prod = n.product

def iter(n: List[Int], a: List[Int], sm: Int): Int = {
def mulInv(a: Int, b: Int): Int = {
def loop(a: Int, b: Int, x0: Int, x1: Int): Int = {
if (a > 1) loop(b, a % b, x1 - (a / b) * x0, x0) else x1
}

if (b == 1) 1
else {
val x1 = loop(a, b, 0, 1)
if (x1 < 0) x1 + b else x1
}
}

if (n.nonEmpty) {
val p = prod / n.head

} else sm
}

Try {
iter(n, a, 0) % prod
} match {
case Success(v) => Some(v)
case _          => None
}
}

println(chineseRemainder(List(3, 5, 7), List(2, 3, 2)))
println(chineseRemainder(List(11, 12, 13), List(10, 4, 12)))
println(chineseRemainder(List(11, 22, 19), List(10, 4, 9)))

}


$include "seed7_05.s7i"; include "bigint.s7i"; const func integer: modInverse (in integer: a, in integer: b) is return ord(modInverse(bigInteger conv a, bigInteger conv b)); const proc: main is func local const array integer: n is [] (3, 5, 7); const array integer: a is [] (2, 3, 2); var integer: num is 0; var integer: prod is 1; var integer: sum is 0; var integer: index is 0; begin for num range n do prod *:= num; end for; for key index range a do num := prod div n[index]; sum +:= a[index] * modInverse(num, n[index]) * num; end for; writeln(sum mod prod); end func;  {{out}}  23  ## Sidef {{trans|Perl 6}} func chinese_remainder(*n) { var N = n.prod func (*a) { n.range.sum { |i| var p = (N / n[i]) a[i] * p.invmod(n[i]) * p } % N } } say chinese_remainder(3, 5, 7)(2, 3, 2)  {{out}}  23  ## Swift import Darwin /* * Function: euclid * Usage: (r,s) = euclid(m,n) * -------------------------- * The extended Euclidean algorithm subsequently performs * Euclidean divisions till the remainder is zero and then * returns the Bézout coefficients r and s. */ func euclid(_ m:Int, _ n:Int) -> (Int,Int) { if m % n == 0 { return (0,1) } else { let rs = euclid(n % m, m) let r = rs.1 - rs.0 * (n / m) let s = rs.0 return (r,s) } } /* * Function: gcd * Usage: x = gcd(m,n) * ------------------- * The greatest common divisor of two numbers a and b * is expressed by ax + by = gcd(a,b) where x and y are * the Bézout coefficients as determined by the extended * euclidean algorithm. */ func gcd(_ m:Int, _ n:Int) -> Int { let rs = euclid(m, n) return m * rs.0 + n * rs.1 } /* * Function: coprime * Usage: truth = coprime(m,n) * --------------------------- * If two values are coprime, their greatest common * divisor is 1. */ func coprime(_ m:Int, _ n:Int) -> Bool { return gcd(m,n) == 1 ? true : false } coprime(14,26) //coprime(2,4) /* * Function: crt * Usage: x = crt(a,n) * ------------------- * The Chinese Remainder Theorem supposes that given the * integers n_1...n_k that are pairwise co-prime, then for * any sequence of integers a_1...a_k there exists an integer * x that solves the system of linear congruences: * * x === a_1 (mod n_1) * ... * x === a_k (mod n_k) */ func crt(_ a_i:[Int], _ n_i:[Int]) -> Int { // There is no identity operator for elements of [Int]. // The offset of the elements of an enumerated sequence // can be used instead, to determine if two elements of the same // array are the same. let divs = n_i.enumerated() // Check if elements of n_i are pairwise coprime divs.filter{$0.0 < n.0 }
divs.forEach{
n in divs.filter{ $0.0 < n.0 }.forEach{ assert(coprime(n.1,$0.1))
}
}

// Calculate factor N
let N = n_i.map{$0}.reduce(1, *) // Euclidean algorithm determines s_i (and r_i) var s:[Int] = [] // Using euclidean algorithm to calculate r_i, s_i n_i.forEach{ s += [euclid($0, N / $0).1] } // Solve for x var x = 0 a_i.enumerated().forEach{ x +=$0.1 * s[$0.0] * N / n_i[$0.0]
}

// Return minimal solution
return x % N
}

let a = [2,3,2]
let n = [3,5,7]

let x = crt(a,n)

print(x)


{{out}}

23


## Tcl

{{trans|C}}

proc ::tcl::mathfunc::mulinv {a b} {
if {$b == 1} {return 1} set b0$b; set x0 0; set x1 1
while {$a > 1} { set x0 [expr {$x1 - ($a /$b) * [set x1 $x0]}] set b [expr {$a % [set a $b]}] } incr x1 [expr {($x1 < 0) * $b0}] } proc chineseRemainder {nList aList} { set sum 0; set prod [::tcl::mathop::* {*}$nList]
foreach n $nList a$aList {
set p [expr {$prod /$n}]
incr sum [expr {$a * mulinv($p, $n) *$p}]
}
expr {$sum %$prod}
}
puts [chineseRemainder {3 5 7} {2 3 2}]


{{out}}

23


## uBasic/4tH

{{trans|C}} @(000) = 3 : @(001) = 5 : @(002) = 7 @(100) = 2 : @(101) = 3 : @(102) = 2

Print Func (_Chinese_Remainder (3))

' -------------------------------------

@(000) = 11 : @(001) = 12 : @(002) = 13 @(100) = 10 : @(101) = 04 : @(102) = 12

Print Func (_Chinese_Remainder (3))

' -------------------------------------

End

                                   ' returns x where (a * x) % b == 1


_Mul_Inv Param (2) ' ( a b -- n) Local (4)

c@ = b@ d@ = 0 e@ = 1

If b@ = 1 Then Return (1)

Do While a@ > 1 f@ = a@ / b@ Push b@ : b@ = a@ % b@ : a@ = Pop() Push d@ : d@ = e@ - f@ * d@ : e@ = Pop() Loop

If e@ < 0 Then e@ = e@ + c@

Return (e@)

_Chinese_Remainder Param (1) ' ( len -- n) Local (5)

b@ = 1 c@ = 0

For d@ = 0 Step 1 While d@ < a@ b@ = b@ * @(d@) Next

For d@ = 0 Step 1 While d@ < a@ e@ = b@ / @(d@) c@ = c@ + (@(100 + d@) * Func (_Mul_Inv (e@, @(d@))) * e@) Next

Return (c@ % b@)


{{out}}

txt

23
1000

0 OK, 0:1034



## VBA

Uses the function mul_inv() from [[Modular_inverse#VBA]] {{trans|Phix}}

Private Function chinese_remainder(n As Variant, a As Variant) As Variant
Dim p As Long, prod As Long, tot As Long
prod = 1: tot = 0
For i = 1 To UBound(n)
prod = prod * n(i)
Next i
Dim m As Variant
For i = 1 To UBound(n)
p = prod / n(i)
m = mul_inv(p, n(i))
If WorksheetFunction.IsText(m) Then
chinese_remainder = "fail"
Exit Function
End If
tot = tot + a(i) * m * p
Next i
chinese_remainder = tot Mod prod
End Function
Public Sub re()
Debug.Print chinese_remainder([{3,5,7}], [{2,3,2}])
Debug.Print chinese_remainder([{11,12,13}], [{10,4,12}])
Debug.Print chinese_remainder([{11,22,19}], [{10,4,9}])
Debug.Print chinese_remainder([{100,23}], [{19,0}])
End Sub


{{out}}

 23
1000
fail
1219


## Visual Basic .NET

{{trans|C#}}

Module Module1

Function ModularMultiplicativeInverse(a As Integer, m As Integer) As Integer
Dim b = a Mod m
For x = 1 To m - 1
If (b * x) Mod m = 1 Then
Return x
End If
Next
Return 1
End Function

Function Solve(n As Integer(), a As Integer()) As Integer
Dim prod = n.Aggregate(1, Function(i, j) i * j)
Dim sm = 0
Dim p As Integer
For i = 0 To n.Length - 1
p = prod / n(i)
sm = sm + a(i) * ModularMultiplicativeInverse(p, n(i)) * p
Next
Return sm Mod prod
End Function

Sub Main()
Dim n = {3, 5, 7}
Dim a = {2, 3, 2}

Dim result = Solve(n, a)

Dim counter = 0
Dim maxCount = n.Length - 1
While counter <= maxCount
Console.WriteLine(\$"{result} = {a(counter)} (mod {n(counter)})")
counter = counter + 1
End While
End Sub

End Module


{{out}}

23 = 2 (mod 3)
23 = 3 (mod 5)
23 = 2 (mod 7)


## zkl

{{trans|Go}} Using the GMP library, gcdExt is the Extended Euclidean algorithm.

var BN=Import("zklBigNum"), one=BN(1);

fcn crt(xs,ys){
p:=xs.reduce('*,BN(1));
X:=BN(0);
foreach x,y in (xs.zip(ys)){
q:=p/x;
z,s,_:=q.gcdExt(x);
if(z!=one) throw(Exception.ValueError("%d not coprime".fmt(x)));
X+=y*s*q;
}
return(X % p);
}

println(crt(T(3,5,7),  T(2,3,2)));    //-->23
println(crt(T(11,12,13),T(10,4,12))); //-->1000
println(crt(T(11,22,19), T(10,4,9))); //-->ValueError: 11 not coprime


## ZX Spectrum Basic

{{trans|C}}

10 DIM n(3): DIM a(3)
20 FOR i=1 TO 3
40 NEXT i
50 DATA 3,2,5,3,7,2
100 LET prod=1: LET sum=0
110 FOR i=1 TO 3: LET prod=prod*n(i): NEXT i
120 FOR i=1 TO 3
130 LET p=INT (prod/n(i)): LET a=p: LET b=n(i)
140 GO SUB 1000
150 LET sum=sum+a(i)*x1*p
160 NEXT i
170 PRINT FN m(sum,prod)
180 STOP
200 DEF FN m(a,b)=a-INT (a/b)*b: REM Modulus function
1000 LET b0=b: LET x0=0: LET x1=1
1010 IF b=1 THEN RETURN
1020 IF a<=1 THEN GO TO 1100
1030 LET q=INT (a/b)
1040 LET t=b: LET b=FN m(a,b): LET a=t
1050 LET t=x0: LET x0=x1-q*x0: LET x1=t
1060 GO TO 1020
1100 IF x1<0 THEN LET x1=x1+b0
1110 RETURN

23