Find two numbers with the given LCM and minimum possible difference - GeeksforGeeks

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++
Java
Python3
C#
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Function to return the LCM of a and bint lcm(int a, int b){    return (a / __gcd(a, b) * b);} // Function to find and print the two numbersvoid 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 codeint main(){    int x = 12;     findNums(x);     return 0;}
// Java implementation of the approachclass 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 implementation of the approachfrom math import gcd as __gcd, sqrt, ceil # Function to return the LCM of a and bdef lcm(a, b):    return (a // __gcd(a, b) * b) # Function to find and print the two numbersdef 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 codex = 12 findNums(x) # This code is contributed by mohit kumar 29
// C# implementation of the approachusing 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
<script>// Javascript implementation of the approach // Function to return the LCM of a and bfunction lcm(a,b){     return (a / __gcd(a, b) * b);} function __gcd(a,b){     return b == 0 ? a : __gcd(b, a % b);} // Function to find and print the two numbersfunction findNums(x){    let ans = -1;           // To find the factors        for (let 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, Math.floor(x / i)) == x)            {                   ans = i;            }        }        document.write(ans + " " + Math.floor(x / ans));} // Driver codelet x = 12;findNums(x); // This code is contributed by patel2127</script>
3 4
 
Time Complexity: O(n1/2 * log(max(a, b)))
Auxiliary Space: O(log(max(a, b)))
mohit kumar 29
29AjayKumar
Code_Mech
patel2127
subham348
LCM
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Mathematical
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