# Find two numbers with the given LCM and minimum possible difference

Last Updated : 20 Feb, 2022

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

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

## Javascript



Output:

3 4

Time Complexity: O(n1/2 * log(max(a, b)))

Auxiliary Space: O(log(max(a, b)))

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