# Chinese remainder theorem

(Redirected from Chinese Remainder Theorem)
Chinese remainder theorem
You are encouraged to solve this task according to the task description, using any language you may know.

Suppose   ${\displaystyle n_{1}}$,   ${\displaystyle n_{2}}$,   ${\displaystyle \ldots }$,   ${\displaystyle n_{k}}$   are positive integers that are pairwise co-prime.

Then, for any given sequence of integers   ${\displaystyle a_{1}}$,   ${\displaystyle a_{2}}$,   ${\displaystyle \dots }$,   ${\displaystyle a_{k}}$,   there exists an integer   ${\displaystyle x}$   solving the following system of simultaneous congruences:

{\displaystyle {\begin{aligned}x&\equiv a_{1}{\pmod {n_{1}}}\\x&\equiv a_{2}{\pmod {n_{2}}}\\&{}\ \ \vdots \\x&\equiv a_{k}{\pmod {n_{k}}}\end{aligned}}}

Furthermore, all solutions   ${\displaystyle x}$   of this system are congruent modulo the product,   ${\displaystyle N=n_{1}n_{2}\ldots n_{k}}$.

Write a program to solve a system of linear congruences by applying the   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   ${\displaystyle s}$   where   ${\displaystyle 0\leq s\leq n_{1}n_{2}\ldots n_{k}}$.

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

Algorithm:   The following algorithm only applies if the   ${\displaystyle n_{i}}$'s   are pairwise co-prime.

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

${\displaystyle x\equiv a_{i}{\pmod {n_{i}}}\quad \mathrm {for} \;i=1,\ldots ,k}$

Again, to begin, the product   ${\displaystyle N=n_{1}n_{2}\ldots n_{k}}$   is defined.

Then a solution   ${\displaystyle x}$   can be found as follows:

For each   ${\displaystyle i}$,   the integers   ${\displaystyle n_{i}}$   and   ${\displaystyle N/n_{i}}$   are co-prime.

Using the   Extended Euclidean algorithm,   we can find integers   ${\displaystyle r_{i}}$   and   ${\displaystyle s_{i}}$   such that   ${\displaystyle r_{i}n_{i}+s_{i}N/n_{i}=1}$.

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

${\displaystyle x=\sum _{i=1}^{k}a_{i}s_{i}N/n_{i}}$

and the minimal solution,

${\displaystyle x{\pmod {N}}}$.

## 11l

Translation of: 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))
Output:
23

## 360 Assembly

Translation of: 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
Output:
x=          23


## AArch64 Assembly

Works with: as version Raspberry Pi 3B version Buster 64 bits
or android 64 bits with application Termux
/* ARM assembly AARCH64 Raspberry PI 3B */
/*  program chineserem64.s   */

/************************************/
/* Constantes                       */
/************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"

/*********************************/
/* Initialized data              */
/*********************************/
.data
szMessResult:       .asciz "Result = "
szCarriageReturn:   .asciz "\n"
.align 2
.equ ARRAYSIZE,  (. - arrayA)/8
/*********************************/
/* UnInitialized data            */
/*********************************/
.bss
sZoneConv:           .skip 24
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main
main:
mov x2,#ARRAYSIZE            //  array size
bl chineseremainder

bl conversion10             // call décimal conversion
mov x0,#3
ldr x2,qAdrsZoneConv        // insert conversion in message
bl displayStrings           // display message

100:                            // standard end of the program
mov x0, #0                  // return code
mov x8,EXIT
svc #0                      // perform the system call

/******************************************************************/
/*     compute chinese remainder                                  */
/******************************************************************/
/* x0 contains n array address */
/* x1 contains a array address */
/* x2 contains array size      */
chineseremainder:
stp x1,lr,[sp,-16]!       // save  registers
stp x2,x3,[sp,-16]!       // save  registers
stp x4,x5,[sp,-16]!       // save  registers
stp x6,x7,[sp,-16]!       // save  registers
stp x8,x9,[sp,-16]!       // save  registers
mov x4,#1                 // product
mov x5,#0                 // sum
mov x6,#0                 // indice
1:
ldr x3,[x0,x6,lsl #3]     // load a value
mul x4,x3,x4              // compute product
cmp x6,x2
blt 1b
mov x6,#0
mov x7,x0                 // save entry
mov x8,x1
mov x9,x2
2:
mov x0,x4                 // product
ldr x1,[x7,x6,lsl #3]     // value of n
sdiv x2,x0,x1
mov x0,x2                 //  p
bl inverseModulo
mul x0,x2,x0              // = product / n * invmod
ldr x3,[x8,x6,lsl #3]     //  value a
madd x5,x0,x3,x5          // sum = sum + (result1 * a)
cmp x6,x9
blt 2b
sdiv x1,x5,x4             // divide sum by produc
msub x0,x1,x4,x5          // compute remainder
100:
ldp x8,x9,[sp],16        // restaur  registers
ldp x6,x7,[sp],16        // restaur  registers
ldp x4,x5,[sp],16        // restaur  registers
ldp x2,x3,[sp],16        // restaur  registers
ldp x1,lr,[sp],16            // restaur  registers
ret
/***************************************************/
/*   Calcul modulo inverse                            */
/***************************************************/
/* x0 cont.quad number, x1 modulo              */
/* x0 return result               */
inverseModulo:
stp x1,lr,[sp,-16]!          // save  registers
stp x2,x3,[sp,-16]!          // save  registers
stp x4,x5,[sp,-16]!          // save  registers
stp x6,x7,[sp,-16]!          // save  registers
mov x7,x1            // save Modulo
mov x6,x1            // A   x0=B
mov x4,#1            // X
mov x5,#0            // Y
1:   //
cmp x0,#0            // B = 0
beq 2f
mov x1,x0            // T = B
mov x0,x6            // A
sdiv x2,x0,x1        // A / T
msub x0,x2,x1,x0     // B and x2=Q
mov x6,x1            // A=T
mov x1,x4            // T=X
msub x4,x2,x1,x5     // X=Y-(Q*T)
mov x5,x1            // Y=T
b 1b
2:
add x7,x7,x5         // = Y + N
cmp x5,#0            // Y > 0
bge 3f
mov x0,x7
b 100f
3:
mov x0,x5
100:
ldp x6,x7,[sp],16        // restaur  registers
ldp x4,x5,[sp],16        // restaur  registers
ldp x2,x3,[sp],16        // restaur  registers
ldp x1,lr,[sp],16            // restaur  registers
ret
/***************************************************/
/*   display multi strings                    */
/***************************************************/
/* x0  contains number strings address */
/* other address on the stack */
displayStrings:            // INFO:  affichageStrings
stp x1,lr,[sp,-16]!    // save  registers
stp x2,x3,[sp,-16]!    // save  registers
stp x4,x5,[sp,-16]!    // save  registers
add fp,sp,#48          // save paraméters address (6 registers saved * 8 bytes)
mov x4,x0              // save strings number
cmp x4,#0              // 0 string -> end
ble 100f
mov x0,x1              // string 1
bl affichageMess
cmp x4,#1              // number > 1
ble 100f
mov x0,x2
bl affichageMess
cmp x4,#2
ble 100f
mov x0,x3
bl affichageMess
cmp x4,#3
ble 100f
mov x3,#3
sub x2,x4,#4
1:                          // loop extract address string on stack
ldr x0,[fp,x2,lsl #3]
bl affichageMess
subs x2,x2,#1
bge 1b
100:
ldp x4,x5,[sp],16        // restaur  registers
ldp x2,x3,[sp],16        // restaur  registers
ldp x1,lr,[sp],16        // restaur  registers
ret
/***************************************************/
/*      ROUTINES INCLUDE                           */
/***************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
Output:
Result = 23


## Action!

INT FUNC MulInv(INT a,b)
INT b0,x0,x1,q,tmp

IF b=1 THEN RETURN (1) FI

b0=b x0=0 x1=1
WHILE a>1
DO
q=a/b

tmp=b
b=a MOD b
a=tmp

tmp=x0
x0=x1-q*x0
x1=tmp
OD
IF x1<0 THEN
x1==+b0
FI
RETURN (x1)

INT FUNC ChineseRemainder(BYTE ARRAY n,a BYTE len)
INT prod,sum,p,m
BYTE i

prod=1 sum=0
FOR i=0 TO len-1
DO
prod==*n(i)
OD
FOR i=0 TO len-1
DO
p=prod/n(i)
m=MulInv(p,n(i))
sum==+a(i)*m*p
OD
RETURN (sum MOD prod)

PROC Main()
BYTE ARRAY n=[3 5 7],a=[2 3 2]
INT res

res=ChineseRemainder(n,a,3)
PrintI(res)
RETURN
Output:
23


Using the package Mod_Inv from [[1]].

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;


## ALGOL 68

Translation of: C – with some error messages
BEGIN # Chinese remainder theorewm - translated from the C sample             #

PROC mul inv = ( INT a in, b in )INT:
IF b in = 1
THEN 1
ELSE
INT b0 = b in;
INT a := a in, b := b in, x0 := 0, x1 := 1;
WHILE a > 1 DO
IF b = 0 THEN
print( ( "Numbers not pairwise coprime", newline ) ); stop
FI;
INT q = a OVER b;
INT t;
t := b; b := a MOD b; a := t;
t := x0; x0 := x1 - q * x0; x1 := t
OD;
IF x1 < 0 THEN x1 + b0 ELSE x1 FI
FI # mul inv # ;

PROC chinese remainder = ( []INT n, a )INT:
IF LWB n /= LWB a OR UPB n /= UPB a OR ( UPB a - LWB a ) + 1 < 1
THEN print( ( "Array bounds mismatch or empty arrays", newline ) ); stop
ELSE
INT prod := 1, sum := 0;
FOR i FROM LWB n TO UPB n DO prod *:= n[ i ] OD;
IF prod = 0 THEN
print( ( "Numbers not pairwise coprime", newline ) ); stop
FI;
FOR i FROM LWB n TO UPB n DO
INT p = prod OVER n[ i ];
sum +:= a[ i ] * mul inv( p, n[ i ] ) * p
OD;
sum MOD prod
FI # chinese remainder # ;

print( ( whole( chinese remainder( ( 3, 5, 7 ), ( 2, 3, 2 ) ), 0 ), newline ) )

END
Output:
23


## ARM Assembly

Works with: as version Raspberry Pi
or android 32 bits with application Termux
/* ARM assembly Raspberry PI or android with termux */
/*  program chineserem.s   */

/* REMARK 1 : this program use routines in a include file
see task Include a file language arm assembly
for the routine affichageMess conversion10
see at end of this program the instruction include */
/* for constantes see task include a file in arm assembly */
/************************************/
/* Constantes                       */
/************************************/
.include "../constantes.inc"

/*********************************/
/* Initialized data              */
/*********************************/
.data
szMessResult:       .asciz "Result = "
szCarriageReturn:   .asciz "\n"
.align 2
arrayN:           .int 3,5,7
arrayA:           .int 2,3,2
.equ ARRAYSIZE,  (. - arrayA)/4
/*********************************/
/* UnInitialized data            */
/*********************************/
.bss
sZoneConv:           .skip 24
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main
main:
mov r2,#ARRAYSIZE            @  array size
bl chineseremainder

bl conversion10             @ call décimal conversion
mov r0,#3
ldr r2,iAdrsZoneConv        @ insert conversion in message
bl displayStrings           @ display message

100:                            @ standard end of the program
mov r0, #0                  @ return code
mov r7, #EXIT               @ request to exit program
svc #0                      @ perform the system call

/******************************************************************/
/*     compute chinese remainder                                  */
/******************************************************************/
/* r0 contains n array address */
/* r1 contains a array address */
/* r2 contains array size      */
chineseremainder:
push {r1-r9,lr}           @ save  registers
mov r4,#1                 @ product
mov r5,#0                 @ sum
mov r6,#0                 @ indice
1:
ldr r3,[r0,r6,lsl #2]     @ load a value
mul r4,r3,r4              @ compute product
cmp r6,r2
blt 1b
mov r6,#0
mov r7,r0                 @ save entry
mov r8,r1
mov r9,r2
2:
mov r0,r4                 @ product
ldr r1,[r7,r6,lsl #2]     @ value of n
bl division
mov r0,r2                 @  p
bl inverseModulo
mul r0,r2,r0              @ = product / n * invmod
ldr r3,[r8,r6,lsl #2]     @  value a
mla r5,r0,r3,r5           @ sum = sum + (result1 * a)
cmp r6,r9
blt 2b
mov r0,r5                  @ sum
mov r1,r4                  @ product
bl division
mov r0,r3

100:
pop {r1-r9,pc}              @ restaur registers
/***************************************************/
/*   Calcul modulo inverse                            */
/***************************************************/
/* r0 containt number, r1 modulo              */
/* x0 return result               */
inverseModulo:
push {r1-r7,lr}       @ save  registers
mov r7,r1            // save Modulo
mov r6,r1            // A   r0=B
mov r4,#1            // X
mov r5,#0            // Y
1:   //
cmp r0,#0            // B = 0
beq 2f
mov r1,r0            // T = B
mov r0,r6            // A
bl division          // A / T
mov r0,r3            // B and r2=Q
mov r6,r1            // A=T
mov r1,r4            // T=X
mls r4,r2,r1,r5      // X=Y-(Q*T)
mov r5,r1            // Y=T
b 1b
2:
add r7,r7,r5         // = Y + N
cmp r5,#0            // Y > 0
bge 3f
mov r0,r7
b 100f
3:
mov r0,r5
100:
pop {r1-r7,pc}
/***************************************************/
/*   display multi strings                    */
/***************************************************/
/* r0  contains number strings address */
/* other address on the stack */
displayStrings:            @ INFO:  affichageStrings
push {r1-r4,fp,lr}     @ save des registres
add fp,sp,#24          @ save paraméters address (6 registers saved * 4 bytes)
mov r4,r0              @ save strings number
cmp r4,#0              @ 0 string -> end
ble 100f
mov r0,r1              @ string 1
bl affichageMess
cmp r4,#1              @ number > 1
ble 100f
mov r0,r2
bl affichageMess
cmp r4,#2
ble 100f
mov r0,r3
bl affichageMess
cmp r4,#3
ble 100f
mov r3,#3
sub r2,r4,#4
1:                            @ loop extract address string on stack
ldr r0,[fp,r2,lsl #2]
bl affichageMess
subs r2,#1
bge 1b
100:
pop {r1-r4,fp,pc}
/***************************************************/
/*      ROUTINES INCLUDE                           */
/***************************************************/
.include "../affichage.inc"
Output:
Result = 23


## Arturo

mulInv: function [a0, b0][
[a b x0]: @[a0 b0 0]
result: 1
if b = 1 -> return result
while [a > 1][
q: a / b
a: a % b
tmp: a
a: b
b: tmp
result: result - q * x0
tmp: x0
x0: result
result: tmp
]
if result < 0 -> result: result + b0
return result
]

chineseRemainder: function [N, A][
prod: 1
s: 0

loop N 'x -> prod: prod * x

loop.with:'i N 'x [
p: prod / x
s: s + (mulInv p x) * p * A\[i]
]
return s % prod
]

print chineseRemainder [3 5 7] [2 3 2]

Output:
23

## AWK

Translation of: 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
}

Output:
23

## BQN

Multiplicative Modular inverse function taken from BQNcrate.

MulInv←⊣|·⊑{0=𝕨?1‿0;(⌽-(0⋈𝕩⌊∘÷𝕨)⊸×)𝕨𝕊˜𝕨|𝕩}
ChRem←{
num 𝕊 rem:
prod←×´num
prod|+´rem×(⊢×num⊸(MulInv¨))prod⌊∘÷num
}

•Show 3‿5‿7 ChRem 2‿3‿2
•Show 10‿4‿9 ChRem 11‿22‿19

23
172


## Bracmat

Translation of: 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$(!n.!a) \n)) ); Output: 23 ## C When n are not pairwise coprime, the program crashes due to division by zero, which is one way to convey error. #include <stdio.h> // returns x where (a * x) % b == 1 int mul_inv(int a, int b) { int b0 = b, t, q; int x0 = 0, x1 = 1; if (b == 1) return 1; while (a > 1) { q = a / b; t = b, b = a % b, a = t; t = x0, x0 = x1 - q * x0, x1 = t; } if (x1 < 0) x1 += b0; return x1; } int chinese_remainder(int *n, int *a, int len) { int p, i, prod = 1, sum = 0; for (i = 0; i < len; i++) prod *= n[i]; for (i = 0; i < len; i++) { p = prod / n[i]; sum += a[i] * mul_inv(p, n[i]) * p; } return sum % prod; } int main(void) { int n[] = { 3, 5, 7 }; int a[] = { 2, 3, 2 }; printf("%d\n", chinese_remainder(n, a, sizeof(n)/sizeof(n[0]))); return 0; }  ## C# using System; using System.Linq; namespace ChineseRemainderTheorem { class Program { static void Main(string[] args) { int[] n = { 3, 5, 7 }; int[] a = { 2, 3, 2 }; int result = ChineseRemainderTheorem.Solve(n, a); int counter = 0; int maxCount = n.Length - 1; while (counter <= maxCount) { Console.WriteLine($"{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;
}
}
}


## C++

// Requires C++17
#include <iostream>
#include <numeric>
#include <vector>
#include <execution>

template<typename _Ty> _Ty mulInv(_Ty a, _Ty b) {
_Ty b0 = b;
_Ty x0 = 0;
_Ty x1 = 1;

if (b == 1) {
return 1;
}

while (a > 1) {
_Ty q = a / b;
_Ty amb = a % b;
a = b;
b = amb;

_Ty xqx = x1 - q * x0;
x1 = x0;
x0 = xqx;
}

if (x1 < 0) {
x1 += b0;
}

return x1;
}

template<typename _Ty> _Ty chineseRemainder(std::vector<_Ty> n, std::vector<_Ty> a) {
_Ty prod = std::reduce(std::execution::seq, n.begin(), n.end(), (_Ty)1, [](_Ty a, _Ty b) { return a * b; });

_Ty sm = 0;
for (int i = 0; i < n.size(); i++) {
_Ty 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;
}

Output:
23

## 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 [[2]].

(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]


## Crystal

Translation of: Ruby
def extended_gcd(a, b)
last_remainder, remainder = a.abs, b.abs
x, last_x = 0, 1

until remainder == 0
tmp = remainder
quotient, remainder = last_remainder.divmod(remainder)
last_remainder = tmp
x, last_x = last_x - quotient * x, x
end

return last_remainder, last_x * (a < 0 ? -1 : 1)
end

def invmod(e, et)
g, x = extended_gcd(e, et)
unless g == 1
raise "Multiplicative inverse modulo does not exist"
end
return x % et
end

def chinese_remainder(mods, remainders)
max = mods.product
series = remainders.zip(mods).map { |r, m| r * max * invmod(max // m, m) // m }
return series.sum % max
end

puts chinese_remainder([3, 5, 7], [2, 3, 2])
puts chinese_remainder([5, 7, 9, 11], [1, 2, 3, 4])


Output:

23
1731


## D

Translation of: 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;
}

Output:
23

## Delphi

Translation of: Java
program ChineseRemainderTheorem;

uses
System.SysUtils, Velthuis.BigIntegers;

function mulInv(a, b: BigInteger): BigInteger;
var
b0, x0, x1, q, amb, xqx: BigInteger;
begin
b0 := b;
x0 := 0;
x1 := 1;

if (b = 1) then
exit(1);

while (a > 1) do
begin
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;

Result := x1;
end;

function chineseRemainder(n: TArray<BigInteger>; a: TArray<BigInteger>)
: BigInteger;
var
i: Integer;
prod, p, sm: BigInteger;
begin
prod := 1;

for i := 0 to High(n) do
prod := prod * n[i];

p := 0;
sm := 0;

for i := 0 to High(n) do
begin
p := prod div n[i];
sm := sm + a[i] * mulInv(p, n[i]) * p;
end;
Result := sm mod prod;
end;

var
n, a: TArray<BigInteger>;

begin
n := [3, 5, 7];
a := [2, 3, 2];

Writeln(chineseRemainder(n, a).ToString);
end.

Output:
23

## EasyLang

Translation of: C
func mul_inv a b .
b0 = b
x1 = 1
if b <> 1
while a > 1
q = a div b
t = b
b = a mod b
a = t
t = x0
x0 = x1 - q * x0
x1 = t
.
if x1 < 0
x1 += b0
.
.
return x1
.
proc remainder . n[] a[] r .
prod = 1
sum = 0
for i = 1 to len n[]
prod *= n[i]
.
for i = 1 to len n[]
p = prod / n[i]
sum += a[i] * (mul_inv p n[i]) * p
r = sum mod prod
.
.
n[] = [ 3 5 7 ]
a[] = [ 2 3 2 ]
remainder n[] a[] h
print h

Output:
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

Translation of: Ruby
Works with: Elixir version 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])

Output:
[23]
[]
[1000]


## Erlang

Translation of: 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.

Output:
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


## F#

### sieving

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
)

Output:
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)

Output:
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 .

Output:
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 ;

Output:
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<<< .


## Fortran

Written in Fortran-77 style (fixed form, GO TO), directly translated from the problem description.

* RC task: use the Chinese Remainder Theorem to solve a system of congruences.

FUNCTION crt(n, residues, moduli)
IMPLICIT INTEGER (A-Z)
DIMENSION residues(n), moduli(n)

p = product(moduli)
crt = 0
DO 10 i = 1, n
m = p/moduli(i)
CALL egcd(moduli(i), m, r, s, gcd)
IF (gcd .ne. 1) GO TO 20 ! error exit
10       crt = crt + residues(i)*s*m
crt = modulo(crt, p)
RETURN

20    crt = -1 ! will never be negative, so flag an error
END

* Compute egcd(a, b), returning x, y, g s.t.
*   g = gcd(a, b) and a*x + b*y = g
*
SUBROUTINE egcd(a, b, x, y, g)
IMPLICIT INTEGER (A-Z)

g = a
u = 0
v = 1
w = b
x = 1
y = 0

1     IF (w .eq. 0) RETURN
q = g/w
u next = x - q*u
v next = y - q*v
w next = g - q*w
x = u
y = v
g = w
u = u next
v = v next
w = w next
GO TO 1
END

PROGRAM Chinese Remainder
IMPLICIT INTEGER (A-Z)

PRINT *, crt(3, [2, 3, 2], [3, 5, 7])
PRINT *, crt(3, [2, 3, 2], [3, 6, 7]) ! no solution

END

Output:
          23
-1


## FreeBASIC

Partial

Translation of: C

. Uses the code from Greatest_common_divisor#Recursive_solution as an include.

#include "gcd.bas"
function mul_inv( a as integer, b as integer ) as integer
if b = 1 then return 1
for i as integer = 1 to b
if a*i mod b = 1 then return i
next i
return 0
end function

function chinese_remainder(n() as integer, a() as integer) as integer
dim as integer p, i, prod = 1, sum = 0, ln = ubound(n)
for p = 0 to ln-1
for i = p+1 to ln
if gcd(n(i), n(p))>1 then
print "N not coprime"
end
end if
next i
next p
for i = 0 to ln
prod *= n(i)
next i
for i = 0 to ln
p = prod/n(i)
sum += a(i) * mul_inv(p, n(i))*p
next i
return sum mod prod
end function

dim as integer n(0 to 2) = { 3, 5, 7 }
dim as integer a(0 to 2) = { 2, 3, 2 }

print chinese_remainder(n(), a())
Output:
23

## Frink

This example solves an extended version of the Chinese Remainder theorem by allowing an optional third parameter d which defaults to 0 and is an integer. The solution returned is the smallest solution >= d. (This optional parameter is common in many/most real-world applications of the Chinese Remainder Theorem.)

This program also works with arbitrarily-large integers and peforms efficiently due to Frink's built-in modInverse function.

Input is validated and useful error messages are emitted if the input data is invalid. If a solution cannot be found, this returns the special value undef.

/** arguments:
[r, m, d=0]  where r and m are arrays of the remainder terms r and the
modulus terms m respectively.  These must be of the same length.
returns
x, the unique solution mod N where N is the product of all the M terms where x &gt;= d.
*/
ChineseRemainder[r, m, d=0] :=
{
if length[r] != length[m]
{
println["ChineseRemainder:  r and m must be arrays of the same length."]
return undef
}

N = product[m]

y = new array
z = new array
x = 0
for i = rangeOf[m]
{
y@i = N / m@i
z@i = modInverse[y@i, m@i]
if z@i == undef
{
println["ChineseRemainder:  modInverse returned undef for modInverse[" + y@i + ", " + m@i + "]"]
return undef
}

x = x + r@i y@i z@i
}

xp = x mod N
f = d div N
r = f * N + xp
if r < d
r = r + N

return r
}

println[ChineseRemainder[[2,3,2],[3,5,7]] ]
Output:
23


## 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)]) )
Output:
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))
}

Output:

Two values, the solution x and an error value.

23 <nil>


## Groovy

Translation of: Java
class ChineseRemainderTheorem {
static int chineseRemainder(int[] n, int[] a) {
int prod = 1
for (int i = 0; i < n.length; i++) {
prod *= n[i]
}

int p, sm = 0
for (int i = 0; i < n.length; i++) {
p = prod.intdiv(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.intdiv(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
}

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

Output:
23

Translation of: 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])
]

Output:
1000
No modular inverse for 418 and 11
23

## Icon and Unicon

Translation of: Python

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 wiki's Essay on the Chinese Remainder Thereom.

## Java

Translation of Python via D

Works with: Java version 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

## JavaScript

function crt(num, rem) {
let sum = 0;
const prod = num.reduce((a, c) => a * c, 1);

for (let i = 0; i < num.length; i++) {
const [ni, ri] = [num[i], rem[i]];
const p = Math.floor(prod / ni);
sum += ri * p * mulInv(p, ni);
}
return sum % prod;
}

function mulInv(a, b) {
const b0 = b;
let [x0, x1] = [0, 1];

if (b === 1) {
return 1;
}
while (a > 1) {
const q = Math.floor(a / b);
[a, b] = [b, a % b];
[x0, x1] = [x1 - q * x0, x0];
}
if (x1 < 0) {
x1 += b0;
}
return x1;
}

console.log(crt([3,5,7], [2,3,2]))


Output:

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 version 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])  Output: chineseremainder([3, 5, 7], [2, 3, 2]) = 23 ## Kotlin Translation of: 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)) }  Output: 23  ## Lobster import std def extended_gcd(a, b): var s = 0 var old_s = 1 var t = 1 var old_t = 0 var r = b var old_r = a while r != 0: let quotient = old_r / r old_r, r = r, old_r - quotient * r old_s, s = s, old_s - quotient * s old_t, t = t, old_t - quotient * t return old_r, old_s, old_t, t, s def for2(xs, ys, fun): return for xs.length: fun(xs[_], ys[_]) def crt(xs, ys): let p = reduce(xs): _a * _b var r = 0 for2(xs,ys) x, y: let q = p / x let z,s,_t,_qt,_qs = q.extended_gcd(x) if z != 1: return "ng " + x + " not coprime", 0 if s < 0: r += y * (s + x) * q else: r += y * s * q return "ok", r % p def print_crt(xs, ys): let msg, res = crt(xs, ys) print(msg + " " + res) print_crt([3,5,7],[2,3,2]) print_crt([11,12,13],[10,4,12]) print_crt([11,22,19],[10,4,9]) print_crt([100,23],[19,0]) Output: ok 23 ok 1000 ng 11 not coprime 0 ok 1219  ## 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))  Output: 23 ## M2000 Interpreter Translation of: C Function ChineseRemainder(n(), a()) { Function mul_inv(a, b) { if b==1 then =1 : exit b0=b x1=1 : x0=0 while a>1 q=a div b t=b : b=a mod b: a=t t=x0: x0=x1-q*x0:x1=t end while if x1<0 then x1+=b0 =x1 } def p, i, prod=1, sum for i=0 to len(n())-1 {prod*=n(i)} for i=0 to len(a())-1 p=prod div n(i) sum+=a(i)*mul_inv(p, n(i))*p next =sum mod prod } Print ChineseRemainder((3,5,7), (2,3,2)) Output: 23  ## Maple This is a Maple built-in procedure, so it is trivial: > chrem( [2, 3, 2], [3, 5, 7] ); 23 ## Mathematica / Wolfram Language Very easy, because it is a built-in function: ChineseRemainder[{2, 3, 2}, {3, 5, 7}] 23  ## MATLAB / Octave function f = chineseRemainder(r, m) s = prod(m) ./ m; [~, t] = gcd(s, m); f = s .* t * r';  Output: >> chineseRemainder([2 3 2], [3 5 7]) ans = 23  ## Maxima /* Function that checks pairwise coprimality */ check_pwc(lst):=block( sublist(cartesian_product_list(makelist(i,i,length(lst)),makelist(i,i,length(lst))),lambda([x],x[1]#x[2])), makelist([lst[%%[i][1]],lst[%%[i][2]]],i,length(%%)), makelist(apply('gcd,%%[i]),i,length(%%)), if length(unique(%%))=1 and first(unique(%%))=1 then true)$

/* Chinese remainder function */
c_remainder(A,N):=if check_pwc(N)=false then "chinese remainder theorem not applicable" else block(
cn:apply("*",N),
makelist(gcdex(cn/N[i],N[i]),i,1,length(N)),
makelist(A[i]*%%[i][1]*cn/N[i],i,1,length(N)),
apply("+",%%),
mod(%%,cn));
Alis:[2,3,2]$Nlis:[3,5,7]$
c_remainder(Alis,Nlis);

Output:
23


## Modula-2

MODULE CRT;
FROM FormatString IMPORT FormatString;

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;

END CRT.

Output:
23

## Nim

Translation of: C
proc mulInv(a0, b0: int): 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 without the Jane Street Ocaml Core Library.

exception Modular_inverse
let inverse_mod a = function
| 1 -> 1
| b -> let rec inner a b x0 x1 =
if a <= 1 then x1
else if  b = 0 then raise Modular_inverse
else inner b (a mod b) (x1 - (a / b) * x0) x0 in
let x = inner a b 0 1 in
if x < 0 then x + b else x

let chinese_remainder_exn congruences =
let mtot = congruences
|> List.map (fun (_, x) -> x)
|> List.fold_left ( *) 1 in
(List.fold_left (fun acc (r, n) ->
acc + r * inverse_mod (mtot / n) n * (mtot / n)
) 0 congruences)
mod mtot

let chinese_remainder congruences =
try Some (chinese_remainder_exn congruences)
with modular_inverse -> None


This is using the Jane Street Ocaml Core library.

open Core.Std

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')

Output:
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])
Output:
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]))))
}

## Pascal

A console application in Free Pascal, created with the Lazarus IDE.

Follows the Perl examples: (1) expresses each solution as a residue class; (2) if the moduli are not pairwise coprime, still finds a solution if there is one.

// Rosetta Code task "Chinese remainder theorem".
program ChineseRemThm;
uses SysUtils;
type TIntArray = array of integer;

// Defining EXTRA adds optional explanatory code
{$DEFINE EXTRA} // Return (if possible) a residue res_out that satifies // res_out = res1 modulo mod1, res_out = res2 modulo mod2. // Return mod_out = LCM( mod1, mod2), or mod_out = 0 if there's no solution. procedure Solve2( const res1, res2, mod1, mod2 : integer; out res_out, mod_out : integer); var a, c, d, k, m, m1, m2, r, temp : integer; p, p_prev : integer; {$IFDEF EXTRA}
q, q_prev : integer;
{$ENDIF} begin if (mod1 = 0) or (mod2 = 0) then raise SysUtils.Exception.Create( 'Solve2: Modulus cannot be 0'); m1 := Abs( mod1); m2 := Abs( mod2); // Extended Euclid's algorithm for HCF( m1, m2), except that only one // of the Bezout coefficients is needed (here p, could have used q) c := m1; d := m2; p :=0; p_prev := 1; {$IFDEF EXTRA}
q := 1; q_prev := 0;
{$ENDIF} a := 0; while (d > 0) do begin temp := p_prev - a*p; p_prev := p; p := temp; {$IFDEF EXTRA}
temp := q_prev - a*q;  q_prev := q;  q := temp;
{$ENDIF} a := c div d; temp := c - a*d; c := d; d := temp; end; // Here with c = HCF( m1, m2) {$IFDEF EXTRA}
Assert( c = p*m2 + q*m1); // p and q are the Bezout coefficients
{$ENDIF} // A soution exists iff c divides (res2 - res1) k := (res2 - res1) div c; if res2 - res1 <> k*c then begin res_out := 0; mod_out := 0; // indicate that there's no xolution end else begin m := (m1 div c) * m2; // m := LCM( m1, m2) r:= res2 - k*p*m2; // r := a solution modulo m {$IFDEF EXTRA}
Assert( r = res1 + k*q*m1); // alternative formula in terms of q
{$ENDIF} // Return the solution in the range 0..(m - 1) // Don't trust the compiler with a negative argument to mod if (r >= 0) then r := r mod m else begin r := (-r) mod m; if (r > 0) then r := m - r; end; res_out := r; mod_out := m; end; end; // Return (if possible) a residue res_out that satifies // res_out = res_array[j] modulo mod_array[j], for j = 0..High(res_array). // Return mod_out = LCM of the moduli, or mod_out = 0 if there's no solution. procedure SolveMulti( const res_array, mod_array : TIntArray; out res_out, mod_out : integer); var count, k, m, r : integer; begin count := Length( mod_array); if count <> Length( res_array) then raise SysUtils.Exception.Create( 'Arrays are different sizes') else if count = 0 then raise SysUtils.Exception.Create( 'Arrays are empty'); k := 1; m := mod_array[0]; r := res_array[0]; while (k < count) and (m > 0) do begin Solve2( r, res_array[k], m, mod_array[k], r, m); inc(k); end; res_out := r; mod_out := m; end; // Cosmetic to turn an integer array into a string for printout. function ArrayToString( a : TIntArray) : string; var j : integer; begin result := '['; for j := 0 to High(a) do begin result := result + SysUtils.IntToStr(a[j]); if j < High(a) then result := result + ', ' else result := result + ']'; end; end; // For the passed-in res_array and mod_array, show the solution // found by SolveMulti (above), or state that there's no solution. procedure ShowSolution( const res_array, mod_array : TIntArray); var mod_out, res_out : integer; begin SolveMulti( res_array, mod_array, res_out, mod_out); Write( ArrayToString( res_array) + ' mod ' + ArrayToString( mod_array) + ' --> '); if mod_out = 0 then WriteLn( 'No solution') else WriteLn( SysUtils.Format( '%d mod %d', [res_out, mod_out])); end; // Main routine. Examples for Rosetta Code task. begin ShowSolution([2, 3, 2], [3, 5, 7]); ShowSolution([3, 5, 7], [2, 3, 2]); ShowSolution([10, 4, 12], [11, 12, 13]); ShowSolution([1, 2, 3, 4], [5, 7, 9, 11]); ShowSolution([11, 22, 19], [10, 4, 9]); ShowSolution([2328, 410], [16256, 5418]); ShowSolution([19, 0], [100, 23]); end.  Output: [2, 3, 2] mod [3, 5, 7] --> 23 mod 105 [3, 5, 7] mod [2, 3, 2] --> 5 mod 6 [10, 4, 12] mod [11, 12, 13] --> 1000 mod 1716 [1, 2, 3, 4] mod [5, 7, 9, 11] --> 1731 mod 3465 [11, 22, 19] mod [10, 4, 9] --> No solution [2328, 410] mod [16256, 5418] --> 28450328 mod 44037504 [19, 0] mod [100, 23] --> 1219 mod 2300  ## Perl There are at least three CPAN modules for this: ntheory (Math::Prime::Util), Math::ModInt, and Math::Pari. All three handle bigints. Library: ntheory use ntheory qw/chinese/; say chinese([2,3], [3,5], [2,7]);  Output: 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));  Output: 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);  Output: 28450328 mod 44037504 ## Phix Translation of: C Uses the function mul_inv() from Modular_inverse#Phix (reproduced below) function mul_inv(integer a, n) if n<0 then n = -n end if if a<0 then a = n - mod(-a,n) end if integer t = 0, nt = 1, r = n, nr = a; while nr!=0 do integer q = floor(r/nr) {t, nt} = {nt, t-q*nt} {r, nr} = {nr, r-q*nr} end while if r>1 then return "a is not invertible" end if if t<0 then t += n end if return t end function 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})  Output: 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) ## Prolog Created with SWI Prolog. product(A, B, C) :- C is A*B. pair(X, Y, X-Y). egcd(_, 0, 1, 0) :- !. egcd(A, B, X, Y) :- divmod(A, B, Q, R), egcd(B, R, S, X), Y is S - Q*X. modinv(A, B, X) :- egcd(A, B, X, Y), A*X + B*Y =:= 1. crt_fold(A, M, P, R0, R1) :- % system of equations of (x = a) (mod m); p = M/m modinv(P, M, Inv), R1 is R0 + A*Inv*P. crt(Pairs, N) :- maplist(pair, As, Ms, Pairs), foldl(product, Ms, 1, M), maplist(divmod(M), Ms, Ps, _), % p(n) <- M/m(n) foldl(crt_fold, As, Ms, Ps, 0, N0), N is N0 mod M.  Output: ?- crt([2-3, 3-5, 2-7], X). X = 23. ?- crt([10-11, 4-12, 12-13], X). X = 1000. ?- crt([2-3, 1-6], X). false.  ### gprolog version Although consult under gprolog didn't complain, I had a problem with "foldl" above. Therefore, I had to write another version of the Chinese Remainder Therorem /* Chinese remainder Theorem: Input chinrest([2,3,2], [3,5,7], R). -----> R == 23 or chinrest([2,3], [5,13], R). ---------> R == 42 Written along the lines of "Introduction to Algorithms" by Thomas Cormen Charles Leiserson Ronald Rivest compiled with gprolog 1.4.5 (64 Bits) */ chinrest(A, N, X) :- sort(N), prime(N,Nn), !, lenok(A, Nn), /* test as to whether the ni are primes */ product(Nn,P), !, /* P is the product of the ni */ milist(P, Nn, Mi), /* The Mi List: mi = n/ni */ cilist(Mi, Nn, Ci), /* The first Ci List: mi-1 mod ni */ mult_lists(Mi, Ci, Ac), /* The ci List :mi*(mi-1 mod ni) */ mult_lists(Ac, A, Ad), /* The ai*ci List */ sum_list(Ad, S), /* Sum of the ai*cis */ X is S mod P, ! . /* a is (a1c1 + ... +akck) mod n */ prime([X|Ys], Zs) :- fd_not_prime(X), !, prime(Ys,Zs). /* sift the primes of [list] */ prime([Y|Ys], [Y|Zs]) :- fd_prime(Y), !, prime(Ys,Zs). prime([],[]). product([], 0). /* n1.n2.n3. ... .ni. ... .nk */ product([H|T], P) :- product_1(T, H, P). product_1([], P, P). product_1([H|T], H0, P) :- product_1(T, H, P0), P is P0 * H0. lenok(A, N) :- length(A, X), length(N, Y), X=:=Y. lenok(_, _) :- write('Please enter equal length prime numbers only'), fail. cilist(Mi, Ni, Ci) :- maplist( (modinv), Mi, Ni, Ci). /* generate the Cis */ mult_lists(Ai, Ci, Ac) :- maplist( (pro), Ai, Ci, Ac). /* The mi*ci */ pro(X, Y, Z) :- Z is X * Y. milist(_, [],[]). milist(P, [H|T],[X|Y]) :- X is truncate(P/H), milist(P, T, Y). modinv(A, B, N) :- eeuclid(A, B, P, _, GCD), GCD =:= 1, N is P mod B. eeuclid(A,B,P,S,GCD) :- A >= B, a_b_p_s_(A,B,P,S,1-0,0-1,GCD), GCD is A*P + B*S. eeuclid(A,B,P,S,GCD) :- A < B, a_b_p_s_(B,A,S,P,1-0,0-1,GCD); GCD is A*P + B*S. a_b_p_s_(A,0,P1,S1,P1-_P2,S1-_S2,A). a_b_p_s_(A,B,P,S,P1-P2,S1-S2,GCD) :- B > 0, A > B, Q is truncate(A/B), B1 is A mod B, P3 is P1-(Q*P2), S3 is S1-(Q*S2), a_b_p_s_(B,B1,P,S,P2-P3,S2-S3,GCD).  Output: ?- chinrest([2,3,2], [3,5,7], X). X = 23 ?- chinrest([2,3], [5,13], X). X = 42  ## 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 Procedure.i Produkt_n(n_Adr.i,a_Adr.i) Define p.i=1 While n_Adr<a_Adr p*PeekI(n_Adr) n_Adr+SizeOf(Integer) 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 Procedure.i ChineseRem(n_Adr.i,a_Adr.i) Define prod.i=Produkt_n(n_Adr,a_Adr), a.i, b.i, p.i, Idx.i=0, sum.i While n_Adr+SizeOf(Integer)*Idx<a_Adr b=PeekI(n_Adr+SizeOf(Integer)*Idx) p=Int(prod/b) : a=p sum+PeekI(a_Adr+SizeOf(Integer)*Idx)*Eval_x1(a,b)*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() Output: [( 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 ### Procedural #### 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)  Output: 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))  Output: 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). Translation of: Haskell Works with: Python version 3.7 '''Chinese remainder theorem''' from operator import (add, mul) 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)( either(compose(quoted("'"))(curry(add)('No solution: ')))( 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): '''Either monad injection operator. 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()  Output: 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 Translation of: 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) Output: 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)  Output: 23 ## Raku (formerly Perl 6) Translation of: C Works with: Rakudo version 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);

Output:
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 0                                     /*stick a fork in it,  we're all done. */
end      /*x*/

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

output   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             /*build a (array) list of modulo values*/
end   /*N*/
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                     /*Is in all sets ? */
end   /*v*/
say 'found a solution with X='    x            /*display one possible solution.       */
exit 0                                         /*stick a fork in it,  we're all done. */
end   /*x*/

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

output   is identical to the 1st REXX version.

## RPL

Translation of: Python
Works with: Halcyon Calc version 4.2.9
RPL code Comment
  ≪
DUP ROT 1 R→C ROT 0 R→C
WHILE DUP RE REPEAT
OVER RE OVER RE / FLOOR
OVER * NEG ROT +
END
DROP C→R ROT MOD
SWAP 1 == SWAP 0 IFTE
≫ ‘MODINV’ STO

≪ → n a
≪ 0
1 1 n SIZE FOR j n j GET * NEXT
1 n SIZE FOR j
DUP n j GET / FLOOR
ROT OVER n j GET MODINV a j GET * ROT * +
SWAP NEXT
MOD
≫ ≫ ‘NCHREM’ STO

MODINV ( a b → x )  with ax = 1 mod b
3:b   2:(r,u) ← (a,1)   1:(r',u') ← (b,0)
While r' ≠ 0
q ← r // r'
(r - q*r', u - q*u')

Forget (r',u') and calculate u mod b
Test r and return zero if a and b are not co-prime

NCHREM ( n a → remainder )
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
// reorder stack for next iteration
return sum % prod



RPL 2003 version

Works with: HP version 49
≪ → a n
≪ a 1 GET n 1 GET →V2
2 a SIZE FOR j
a j GET n j GET →V2 ICHINREM
NEXT
V→ MOD
≫ ≫ 'NCHREM' STO

{ 2 3 2 } { 3 5 7 } NCHREM

Output:
1: 23.


## 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 = 0, 1
while remainder != 0
last_remainder, (quotient, remainder) = remainder, last_remainder.divmod(remainder)
x, last_x = last_x - quotient*x, x
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

Output:

Best seen running in your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or 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; Output: 23  ## Sidef Translation of: Raku 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)  Output: 23  ## SQL Works with: SQLite version 3.34.0 create temporary table inputs(remainder int, modulus int); insert into inputs values (2, 3), (3, 5), (2, 7); with recursive -- Multiply out the product of moduli multiplication(idx, product) as ( select 1, 1 union all select multiplication.idx+1, multiplication.product * inputs.modulus from multiplication, inputs where inputs.rowid = multiplication.idx ), -- Take the final value from the product table product(final_value) as ( select max(product) from multiplication ), -- Calculate the multiplicative inverse from each equation multiplicative_inverse(id, a, b, x, y) as ( select inputs.modulus, product.final_value / inputs.modulus, inputs.modulus, 0, 1 from inputs, product union all select id, b, a%b, y - (a/b)*x, x from multiplicative_inverse where a>0 ) -- Combine residues into final answer select sum( (y % inputs.modulus) * inputs.remainder * (product.final_value / inputs.modulus) ) % product.final_value from multiplicative_inverse, product, inputs where a=1 and multiplicative_inverse.id = inputs.modulus;  Output: 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)

Output:
23

## Tcl

Translation of: 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}]

Output:
23

## uBasic/4tH

Translation of: 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@)

Output:
23
1000

0 OK, 0:1034


## VBA

Uses the function mul_inv() from Modular_inverse#VBA

Translation of: 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
Output:
 23
1000
fail
1219 

## Visual Basic .NET

Translation of: 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

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

## Wren

Translation of: C
/* returns x where (a * x) % b == 1 */
var mulInv = Fn.new { |a, b|
if (b == 1) return 1
var b0 = b
var x0 = 0
var x1 = 1
while (a > 1) {
var q = (a/b).floor
var t = b
b = a % b
a = t
t = x0
x0 = x1 - q*x0
x1 = t
}
if (x1 < 0) x1 = x1 + b0
return x1
}

var chineseRemainder = Fn.new { |n, a|
var prod = n.reduce { |acc, i| acc * i }
var sum = 0
for (i in 0...n.count) {
var p = (prod/n[i]).floor
sum = sum + a[i]*mulInv.call(p, n[i])*p
}
return sum % prod
}

var n = [3, 5, 7]
var a = [2, 3, 2]
System.print(chineseRemainder.call(n, a))

Output:
23


## XPL0

Translation of: C

When N are not pairwise coprime, the program aborts due to division by zero, which is one way to convey error.

func MulInv(A, B);              \Returns X where rem((A*X) / B) = 1
int  A, B;
int  B0, T, Q;
int  X0, X1;
[B0:= B;  X0:= 0;  X1:= 1;
if B = 1 then return 1;
while A > 1 do
[Q:= A / B;
T:= B;  B:= rem(A/B);  A:= T;
T:= X0;  X0:= X1 - Q*X0;  X1:= T;
];
if X1 < 0 then X1:= X1 + B0;
return X1;
];

func ChineseRem(N, A, Len);
int  N, A, Len;
int  P, I, Prod, Sum;
[Prod:= 1;  Sum:= 0;
for I:= 0 to Len-1 do Prod:= Prod*N(I);
for I:= 0 to Len-1 do
[P:= Prod / N(I);
Sum:= Sum  +  A(I) * MulInv(P,N(I)) * P;
];
return rem(Sum/Prod);
];

int N, A;
[N:= [ 3, 5, 7 ];
A:= [ 2, 3, 2 ];
IntOut(0, ChineseRem(N, A, 3));  CrLf(0);
]
Output:
23


## zkl

Translation of: 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

Translation of: 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