Find two numbers with the given LCM and minimum possible difference

• Difficulty Level : Hard
• Last Updated : 14 Jun, 2021

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 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

 // 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

 # 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#

 // 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

Javascript


Output:
3 4

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