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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 ``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

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## Javascript

 ``
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 ``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

 ` 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

 ``
Output:
`A = 2, B = 3`

Time comlplexity : O(n)

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