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Divide the two given numbers by their common divisors
  • Difficulty Level : Medium
  • Last Updated : 27 Apr, 2021

Given two numbers A and B, the task is to Divide the two numbers A and B by their common divisors. The numbers A and B is less than 10^8.
Examples: 
 

Input: A = 10, B = 15
Output: A = 2, B = 3
The common factors are 1, 5

Input: A = 100, B = 150
Output: A = 2, B = 3

 

Naive Approach: Iterate from i = 1 to minimum of A and B and check whether i is a factor of both A and B. If i is a factor of A and B then Divide the two numbers A and B by i.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// print the numbers after dividing
// them by their common factors
void divide(int a, int b)
{
    // iterate from 1 to minimum of a and b
    for (int i = 2; i <= min(a, b); i++) {
 
        // if i is the common factor
        // of both the numbers
        while (a % i == 0 && b % i == 0) {
            a = a / i;
            b = b / i;
        }
    }
 
    cout << "A = " << a << ", B = " << b << endl;
}
 
// Driver code
int main()
{
    int A = 10, B = 15;
 
    // divide A and B by their common factors
    divide(A, B);
 
    return 0;
}

Java




// Java implementation of above approach
import java.util.*;
 
class solution
{
 
// print the numbers after dividing
// them by their common factors
static void divide(int a, int b)
{
    // iterate from 1 to minimum of a and b
    for (int i = 2; i <= Math.min(a, b); i++) {
 
        // if i is the common factor
        // of both the numbers
        while (a % i == 0 && b % i == 0) {
            a = a / i;
            b = b / i;
        }
    }
 
    System.out.println("A = "+a+", B = "+b);
}
 
// Driver code
public static void main(String args[])
{
    int A = 10, B = 15;
 
    // divide A and B by their common factors
    divide(A, B);
 
}
}
 
// This code is contributed by
// Surendra_Gangwar

Python3




# Python3 implementation of above approach
 
# print the numbers after dividing
# them by their common factors
def divide(a, b) :
     
    # iterate from 1 to minimum of a and b
    for i in range(2, min(a, b) + 1) :
 
        # if i is the common factor
        # of both the numbers
        while (a % i == 0 and b % i == 0) :
            a = a // i
            b = b // i
 
    print("A =", a, ", B =", b)
 
# Driver code
if __name__ == "__main__" :
 
    A, B = 10, 15
 
    # divide A and B by their
    # common factors
    divide(A, B)
     
# This code is contributed by Ryuga

C#




// C# implementation of above approach
using System;
 
class GFG
{
     
// print the numbers after dividing
// them by their common factors
static void divide(int a, int b)
{
    // iterate from 1 to minimum of a and b
    for (int i = 2; i <= Math.Min(a, b); i++)
    {
 
        // if i is the common factor
        // of both the numbers
        while (a % i == 0 && b % i == 0)
        {
            a = a / i;
            b = b / i;
        }
    }
    Console.WriteLine("A = "+a+", B = "+b);
}
 
// Driver code
static public void Main ()
{
    int A = 10, B = 15;
 
    // divide A and B by their common factors
    divide(A, B);
}
}
 
// This code is contributed by ajit.

PHP




<?php
// PHP implementation of above approach
// print the numbers after dividing
// them by their common factors
 
function divide($a, $b)
{
    // iterate from 1 to minimum of a and b
    for ($i = 2; $i <= min($a, $b); $i++)
    {
 
        // if i is the common factor
        // of both the numbers
        while ($a % $i == 0 &&
                $b % $i == 0)
        {
            $a = $a / $i;
            $b = $b / $i;
        }
    }
    echo "A = ", $a, ", B = ", $b, "\n";
}
 
// Driver code
$A = 10;
$B = 15;
 
// divide A and B by their common factors
divide($A, $B);
 
// This code is contributed by Sach_Code
?>

Javascript




<script>
 
// Javascript implementation of above approach
 
// print the numbers after dividing
// them by their common factors
function divide(a, b)
{
    // iterate from 1 to minimum of a and b
    for (let i = 2; i <= Math.min(a, b); i++) {
 
        // if i is the common factor
        // of both the numbers
        while (a % i == 0 && b % i == 0) {
            a = a / i;
            b = b / i;
        }
    }
 
    document.write("A = " + a + ", B = " + b + "<br>");
}
 
// Driver code
 
    let A = 10, B = 15;
 
    // divide A and B by their common factors
    divide(A, B);
 
// This code is contributed by Mayank Tyagi
 
</script>
Output: 
A = 2, B = 3

 

An efficient approach is to use the same concept used in Common divisors of two numbers. Calculate the greatest common divisor (gcd) of given two numbers, and then divide the numbers by their gcd.
 



C++




// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate gcd of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
void commDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);
 
    a = a / n;
    b = b / n;
 
    cout << "A = " << a << ", B = " << b << endl;
}
 
// Driver code
int main()
{
    int a = 10, b = 15;
    commDiv(a, b);
 
    return 0;
}

Java




// Java implementation of above approach
class GFG
{
     
// Function to calculate gcd
// of two numbers
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to calculate all common
// divisors of two given numbers
// a, b --> input integer numbers
static void commDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);
 
    a = a / n;
    b = b / n;
 
    System.out.println("A = " + a +
                       ", B = " + b);
}
 
// Driver code
public static void main(String[] args)
{
    int a = 10, b = 15;
    commDiv(a, b);
}
}
 
// This code is contributed
// by Code_Mech

Python3




# Python3 implementation of above approach
 
# Function to calculate gcd of two numbers
def gcd(a, b):
    if (a == 0):
        return b
    return gcd(b % a, a)
 
# Function to calculate all common
# divisors of two given numbers
# a, b --> input eger numbers
def commDiv(a, b):
     
    # find gcd of a, b
    n = gcd(a, b)
 
    a = a // n
    b = b // n
 
    print("A =", a, ", B =", b)
 
# Driver code
a, b = 10, 15
commDiv(a, b)
 
# This code is contributed
# by mohit kumar

C#




// C# implementation of above approach
using System;
 
class GFG
{
     
// Function to calculate gcd
// of two numbers
static int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
 
// Function to calculate all common
// divisors of two given numbers
// a, b --> input integer numbers
static void commDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);
 
    a = a / n;
    b = b / n;
 
    Console.WriteLine("A = " + a +
                    ", B = " + b);
}
 
// Driver code
public static void Main()
{
    int a = 10, b = 15;
    commDiv(a, b);
}
}
 
// This code is contributed
// by Code_Mech

PHP




<?php
// PHP implementation of above approach
 
// Function to calculate gcd of
// two numbers
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
 
// Function to calculate all common
// divisors of two given numbers
// a, b --> input integer numbers
function commDiv($a, $b)
{
    // find gcd of a, b
    $n = gcd($a, $b);
 
    $a = (int)($a / $n);
    $b = (int)($b / $n);
 
    echo "A = " . $a .
         ", B = " . $b . "\n";
}
 
// Driver code
$a = 10;
$b = 15;
commDiv($a, $b);
 
// This code is contributed by mits
?>

Javascript




<script>
    // Javascript implementation of above approach
     
    // Function to calculate gcd
    // of two numbers
    function gcd(a, b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    // Function to calculate all common
    // divisors of two given numbers
    // a, b --> input integer numbers
    function commDiv(a, b)
    {
        // find gcd of a, b
        let n = gcd(a, b);
 
        a = parseInt(a / n, 10);
        b = parseInt(b / n, 10);
 
        document.write("A = " + a + ", B = " + b);
    }
     
    let a = 10, b = 15;
    commDiv(a, b);
     
</script>
Output: 
A = 2, B = 3

 

Time comlplexity : O(n)
 

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