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Find two numbers with the given LCM and minimum possible difference
• Last Updated : 24 Jun, 2020

Given an integer X, the task is to find two integers A and B such that LCM(A, B) = X and the difference between the A and B is minimum possible.

Examples:

Input: X = 6
Output: 2 3
LCM(2, 3) = 6 and (3 – 2) = 1
which is the minimum possible.

Input X = 7
Output: 1 7

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: An approach to solve this problem is to find all the factors of the given number using the approach discussed in this article and then find the pair (A, B) that satisfies the given conditions and has the minimum possible difference.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the LCM of a and b``int` `lcm(``int` `a, ``int` `b)``{``    ``return` `(a / __gcd(a, b) * b);``}`` ` `// Function to find and print the two numbers``void` `findNums(``int` `x)``{`` ` `    ``int` `ans;`` ` `    ``// To find the factors``    ``for` `(``int` `i = 1; i <= ``sqrt``(x); i++) {`` ` `        ``// To check if i is a factor of x and``        ``// the minimum possible number``        ``// satisfying the given conditions``        ``if` `(x % i == 0 && lcm(i, x / i) == x) {`` ` `            ``ans = i;``        ``}``    ``}``    ``cout << ans << ``" "` `<< (x / ans);``}`` ` `// Driver code``int` `main()``{``    ``int` `x = 12;`` ` `    ``findNums(x);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{`` ` `    ``// Function to return the LCM of a and b``    ``static` `int` `lcm(``int` `a, ``int` `b)``    ``{``        ``return` `(a / __gcd(a, b) * b);``    ``}`` ` `    ``static` `int` `__gcd(``int` `a, ``int` `b) ``    ``{``        ``return` `b == ``0` `? a : __gcd(b, a % b);``    ``}`` ` `    ``// Function to find and print the two numbers``    ``static` `void` `findNums(``int` `x)``    ``{`` ` `        ``int` `ans = -``1``;`` ` `        ``// To find the factors``        ``for` `(``int` `i = ``1``; i <= Math.sqrt(x); i++)``        ``{`` ` `            ``// To check if i is a factor of x and``            ``// the minimum possible number``            ``// satisfying the given conditions``            ``if` `(x % i == ``0` `&& lcm(i, x / i) == x)``            ``{`` ` `                ``ans = i;``            ``}``        ``}``        ``System.out.print(ans + ``" "` `+ (x / ans));``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args) ``    ``{``        ``int` `x = ``12``;`` ` `        ``findNums(x);``    ``}``}`` ` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach``from` `math ``import` `gcd as __gcd, sqrt, ceil`` ` `# Function to return the LCM of a and b``def` `lcm(a, b):``    ``return` `(a ``/``/` `__gcd(a, b) ``*` `b)`` ` `# Function to find and print the two numbers``def` `findNums(x):`` ` `    ``ans ``=` `0`` ` `    ``# To find the factors``    ``for` `i ``in` `range``(``1``, ceil(sqrt(x))):`` ` `        ``# To check if i is a factor of x and``        ``# the minimum possible number``        ``# satisfying the given conditions``        ``if` `(x ``%` `i ``=``=` `0` `and` `lcm(i, x ``/``/` `i) ``=``=` `x):`` ` `            ``ans ``=` `i`` ` `    ``print``(ans, (x``/``/``ans))`` ` `# Driver code``x ``=` `12`` ` `findNums(x)`` ` `# This code is contributed by mohit kumar 29`

## C#

 `// C# implementation of the approach``using` `System;`` ` `class` `GFG``{`` ` `    ``// Function to return the LCM of a and b``    ``static` `int` `lcm(``int` `a, ``int` `b)``    ``{``        ``return` `(a / __gcd(a, b) * b);``    ``}`` ` `    ``static` `int` `__gcd(``int` `a, ``int` `b) ``    ``{``        ``return` `b == 0 ? a : __gcd(b, a % b);``    ``}`` ` `    ``// Function to find and print the two numbers``    ``static` `void` `findNums(``int` `x)``    ``{`` ` `        ``int` `ans = -1;`` ` `        ``// To find the factors``        ``for` `(``int` `i = 1; i <= Math.Sqrt(x); i++)``        ``{`` ` `            ``// To check if i is a factor of x and``            ``// the minimum possible number``            ``// satisfying the given conditions``            ``if` `(x % i == 0 && lcm(i, x / i) == x)``            ``{`` ` `                ``ans = i;``            ``}``        ``}``        ``Console.Write(ans + ``" "` `+ (x / ans));``    ``}`` ` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args) ``    ``{``        ``int` `x = 12;`` ` `        ``findNums(x);``    ``}``}`` ` `// This code is contributed by 29AjayKumar`
Output:
```3 4
```

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