We are given two arrays num[0..k-1] and rem[0..k-1]. In num[0..k-1], every pair is coprime (gcd for every pair is 1). We need to find minimum positive number x such that:

x % num[0] = rem[0], x % num[1] = rem[1], ....................... x % num[k-1] = rem[k-1]

Basically, we are given k numbers which are pairwise coprime, and given remainders of these numbers when an unknown number x is divided by them. We need to find the minimum possible value of x that produces given remainders.

**Examples :**

Input: num[] = {5, 7}, rem[] = {1, 3} Output: 31 Explanation: 31 is the smallest number such that: (1) When we divide it by 5, we get remainder 1. (2) When we divide it by 7, we get remainder 3. Input: num[] = {3, 4, 5}, rem[] = {2, 3, 1} Output: 11 Explanation: 11 is the smallest number such that: (1) When we divide it by 3, we get remainder 2. (2) When we divide it by 4, we get remainder 3. (3) When we divide it by 5, we get remainder 1.

* Chinise Remainder Theorem states that there always exists an x that satisfies given congruences.* Below is theorem statement adapted from wikipedia.

Let num[0], num[1], …num[k-1] be positive integers that are pairwise coprime. Then, for any given sequence of integers rem[0], rem[1], … rem[k-1], there exists an integer x solving the following system of simultaneous congruences.

The first part is clear that there exist an x. The second part basically states that all solutions (including the minimum one) produce the same remainder when divided by product of n[0], num[1], .. num[k-1]. In above example, product is 3*4*5 = 60. And 11 is one solution, other solutions are 71, 131, .. etc. All these solutions produce same remainder when divide by 60, i.e., they are of the form 11 + m*60 where m >= 0.

A **Naive Approach to find x** is to start with 1 and one by one increment it and check if dividing it with given elements in num[] produces corresponding remainders in rem[]. Once we find such a x, we return it.

Below is implementation of Naive Approach.

## C

// A C++ program to demonstrate working of Chinise remainder // Theorem #include<bits/stdc++.h> using namespace std; // k is size of num[] and rem[]. Returns the smallest // number x such that: // x % num[0] = rem[0], // x % num[1] = rem[1], // .................. // x % num[k-2] = rem[k-1] // Assumption: Numbers in num[] are pairwise coprime // (gcd for every pair is 1) int findMinX(int num[], int rem[], int k) { int x = 1; // Initialize result // As per the Chinise remainder theorem, // this loop will always break. while (true) { // Check if remainder of x % num[j] is // rem[j] or not (for all j from 0 to k-1) int j; for (j=0; j<k; j++ ) if (x%num[j] != rem[j]) break; // If all remainders matched, we found x if (j == k) return x; // Else try next numner x++; } return x; } // Driver method int main(void) { int num[] = {3, 4, 5}; int rem[] = {2, 3, 1}; int k = sizeof(num)/sizeof(num[0]); cout << "x is " << findMinX(num, rem, k); return 0; }

## Java

// A Java program to demonstrate working of Chinise remainder // Theorem import java.io.*; class GFG { // k is size of num[] and rem[]. Returns the smallest // number x such that: // x % num[0] = rem[0], // x % num[1] = rem[1], // .................. // x % num[k-2] = rem[k-1] // Assumption: Numbers in num[] are pairwise coprime // (gcd for every pair is 1) static int findMinX(int num[], int rem[], int k) { int x = 1; // Initialize result // As per the Chinise remainder theorem, // this loop will always break. while (true) { // Check if remainder of x % num[j] is // rem[j] or not (for all j from 0 to k-1) int j; for (j=0; j<k; j++ ) if (x%num[j] != rem[j]) break; // If all remainders matched, we found x if (j == k) return x; // Else try next numner x++; } } // Driver method public static void main(String args[]) { int num[] = {3, 4, 5}; int rem[] = {2, 3, 1}; int k = num.length; System.out.println("x is " + findMinX(num, rem, k)); } } /*This code is contributed by Nikita Tiwari.*/

## C#

// C# program to demonstrate working // of Chinise remainder Theorem using System; class GFG { // k is size of num[] and rem[]. // Returns the smallest // number x such that: // x % num[0] = rem[0], // x % num[1] = rem[1], // .................. // x % num[k-2] = rem[k-1] // Assumption: Numbers in num[] // are pairwise coprime // (gcd for every pair is 1) static int findMinX(int []num, int []rem, int k) { // Initialize result int x = 1; // As per the Chinise remainder theorem, // this loop will always break. while (true) { // Check if remainder of x % num[j] is // rem[j] or not (for all j from 0 to k-1) int j; for (j = 0; j < k; j++ ) if (x % num[j] != rem[j]) break; // If all remainders matched, we found x if (j == k) return x; // Else try next numner x++; } } // Driver code public static void Main() { int []num = {3, 4, 5}; int []rem = {2, 3, 1}; int k = num.Length; Console.WriteLine("x is " + findMinX(num, rem, k)); } } // This code is contributed by Sam007.

## PHP

<?php // A PHP program to demonstrate // working of Chinise remainder Theorem // k is size of num[] and rem[]. // Returns the smallest number x // such that: // x % num[0] = rem[0], // x % num[1] = rem[1], // .................. // x % num[k-2] = rem[k-1] // Assumption: Numbers in num[] // are pairwise coprime (gcd for // every pair is 1) function findMinX($num, $rem, $k) { $x = 1; // Initialize result // As per the Chinise remainder // theorem, this loop will // always break. while (true) { // Check if remainder of // x % num[j] is rem[j] // or not (for all j from // 0 to k-1) $j; for ($j = 0; $j < $k; $j++ ) if ($x % $num[$j] != $rem[$j]) break; // If all remainders // matched, we found x if ($j == $k) return $x; // Else try next numner $x++; } return $x; } // Driver Code $num = array(3, 4, 5); $rem = array(2, 3, 1); $k = sizeof($num); echo "x is " , findMinX($num, $rem, $k); // This code is contributed by ajit ?>

**Output :**

x is 11

See below link for efficient method to find x.

**Chinese Remainder Theorem | Set 2 (Inverse Modulo based Implementation)**

This article is contributed by **Ruchir Garg**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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