Minimum possible value of max(A, B) such that LCM(A, B) = C

Given an integer C, the task is to find the minimum possible value of max(A, B) such that LCM(A, B) = C.

Examples:

Input: C = 6
Output: 3
max(1, 6) = 6
max(2, 3) = 3
and min(6, 3) = 3

Input: C = 9
Output: 9

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 take the overall minimum of the maximum of these pairs.



Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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 return the minimized value
int getMinValue(int c)
{
    int ans = INT_MAX;
  
    // To find the factors
    for (int i = 1; i <= sqrt(c); i++) {
  
        // To check if i is a factor of c and
        // the minimum possible number
        // satisfying the given conditions
        if (c % i == 0 && lcm(i, c / i) == c) {
  
            // Update the answer
            ans = min(ans, max(i, c / i));
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int c = 6;
  
    cout << getMinValue(c);
  
    return 0;
}

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Java

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// Java implementation of the approach
class Solution
{
    // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    
        // Everything divides 0 
        if (a == 0
        return b; 
        if (b == 0
        return a; 
          
        // base case 
        if (a == b) 
            return a; 
          
        // a is greater 
        if (a > b) 
            return __gcd(a - b, b); 
        return __gcd(a, b - a); 
    }
      
    // Function to return the LCM of a and b
    static int lcm(int a, int b)
    {
        return (a / __gcd(a, b) * b);
    }
  
    // Function to return the minimized value
    static int getMinValue(int c)
    
        int ans = Integer.MAX_VALUE;
  
        // To find the factors
        for (int i = 1; i <= Math.sqrt(c); i++) 
        {
  
            // To check if i is a factor of c and
            // the minimum possible number
            // satisfying the given conditions
            if (c % i == 0 && lcm(i, c / i) == c)
            {
  
                // Update the answer
                ans = Math.min(ans, Math.max(i, c / i));
            }
        }
        return ans;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int c = 6;
  
        System.out.println(getMinValue(c));
    }
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Python implementation of the approach
import sys
  
# Recursive function to return gcd of a and b
def __gcd(a, b):
      
    # Everything divides 0
    if (a == 0):
        return b;
    if (b == 0):
        return a;
  
    # base case
    if (a == b):
        return a;
  
    # a is greater
    if (a > b):
        return __gcd(a - b, b);
    return __gcd(a, b - a);
  
# Function to return the LCM of a and b
def lcm(a, b):
    return (a / __gcd(a, b) * b);
  
# Function to return the minimized value
def getMinValue(c):
    ans = sys.maxsize;
  
    # To find the factors
    for i in range(1, int(pow(c, 1/2)) + 1):
  
        # To check if i is a factor of c and
        # the minimum possible number
        # satisfying the given conditions
        if (c % i == 0 and lcm(i, c / i) == c):
  
            # Update the answer
            ans = min(ans, max(i, c / i));
    return int(ans);
  
# Driver code
if __name__ == '__main__':
    c = 6;
  
    print(getMinValue(c));
      
# This code is contributed by 29AjayKumar

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C#

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// C# implementation of the approach 
using System;
  
class GFG
    // Recursive function to return gcd of a and b 
    static int __gcd(int a, int b) 
    
        // Everything divides 0 
        if (a == 0) 
        return b; 
        if (b == 0) 
        return a; 
          
        // base case 
        if (a == b) 
            return a; 
          
        // a is greater 
        if (a > b) 
            return __gcd(a - b, b); 
        return __gcd(a, b - a); 
    
      
    // Function to return the LCM of a and b 
    static int lcm(int a, int b) 
    
        return (a / __gcd(a, b) * b); 
    
  
    // Function to return the minimized value 
    static int getMinValue(int c) 
    
        int ans = int.MaxValue; 
  
        // To find the factors 
        for (int i = 1; i <= Math.Sqrt(c); i++) 
        
  
            // To check if i is a factor of c and 
            // the minimum possible number 
            // satisfying the given conditions 
            if (c % i == 0 && lcm(i, c / i) == c) 
            
  
                // Update the answer 
                ans = Math.Min(ans, Math.Max(i, c / i)); 
            
        
        return ans; 
    
  
    // Driver code 
    public static void Main() 
    
        int c = 6; 
  
        Console.WriteLine(getMinValue(c)); 
    
  
// This code is contributed by AnkitRai01

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Output:

3

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