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Find two co-prime integers such that the first divides A and the second divides B

  • Difficulty Level : Hard
  • Last Updated : 11 Jun, 2021

Given two integers A and B, the task is to find two co-prime numbers C1 and C2 such that C1 divides A and C2 divides B.
Examples: 
 

Input: A = 12, B = 16 
Output: 3 4 
12 % 3 = 0 
16 % 4 = 0 
gcd(3, 4) = 1
Input: A = 542, B = 762 
Output: 271 381 
 

 

Naive approach: A simple solution is to store all of the divisors of A and B then iterate over all the divisors of A and B pairwise to find the pair of elements which are co-prime.
Efficient approach: If an integer d divides gcd(a, b) then gcd(a / d, b / d) = gcd(a, b) / d. More formally, if num = gcd(a, b) then gcd(a / num, b / num) = 1 i.e. (a / num) and (b / num) are relatively co-prime. 
So in order to find the required numbers, find gcd(a, b) and store it in a variable gcd. Now the required numbers will be (a / gcd) and (b / gcd).
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the required numbers
void findNumbers(int a, int b)
{
 
    // GCD of the given numbers
    int gcd = __gcd(a, b);
 
    // Printing the required numbers
    cout << (a / gcd) << " " << (b / gcd);
}
 
// Driver code
int main()
{
    int a = 12, b = 16;
 
    findNumbers(a, b);
 
    return 0;
}

Java




// Java implementation of the approach
import java.math.*;
 
class GFG
{
    public static int findGCD(int a, int b)
    {
        if(b == 0)
            return a;
        else
            return findGCD(b, a % b);
    }
 
    // Function to find the required numbers
    static void findNumbers(int a, int b)
    {
     
        // GCD of the given numbers
        int gcd = findGCD(a, b);
         
        // Printing the required numbers
        System.out.println((a / gcd) + " " +
                           (b / gcd));
         
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a = 12, b = 16;
     
        findNumbers(a, b);
    }
}
 
// This code is contributed by Naman_Garg

Python3




# Python3 implementation of the approach
 
# import gcd function from math module
from math import gcd
 
# Function to find the required numbers
def findNumbers(a, b) :
 
    # GCD of the given numbers
    __gcd = gcd(a, b);
 
    # Printing the required numbers
    print((a // __gcd), (b // __gcd));
 
# Driver code
if __name__ == "__main__" :
 
    a = 12; b = 16;
 
    findNumbers(a, b);
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
    public static int findGCD(int a, int b)
    {
        if(b == 0)
            return a;
        else
            return findGCD(b, a % b);
    }
 
    // Function to find the required numbers
    static void findNumbers(int a, int b)
    {
     
        // GCD of the given numbers
        int gcd = findGCD(a, b);
         
        // Printing the required numbers
        Console.Write((a / gcd) + " " +
                      (b / gcd));
         
    }
 
    // Driver code
    static public void Main ()
    {
        int a = 12, b = 16;
     
        findNumbers(a, b);
    }
}
 
// This code is contributed by ajit

Javascript




<script>
 
// Javascript implementation of the approach
 
function findGCD(a, b)
{
    if (b == 0)
        return a;
    else
        return findGCD(b, a % b);
}
 
// Function to find the required numbers
function findNumbers(a, b)
{
     
    // GCD of the given numbers
    var gcd = findGCD(a, b);
     
    // Printing the required numbers
    document.write((a / gcd) + " " +
                   (b / gcd));
}
 
// Driver Code
var a = 12, b = 16;
     
findNumbers(a, b);
 
// This code is contributed by Khushboogoyal499
     
</script>
Output: 
3 4

 




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