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Find two numbers B and C such that their product is A and their GCD is maximum

  • Last Updated : 19 Jan, 2022

Given a positive integer, A. The task is to find two numbers B and C such that their product is A and their GCD should be maximum.

Examples: 

Input: A = 72
Output: 12 6
Explanation: The product of 12 and 6 is 72 and GCD(12, 6) is 6 which is maximum possible.  

Input: A = 54
Output: 6 9
Explanation: The product of 6 and 9 is 54, gcd(6, 9) is 3 which is maximum possible. 

 

Approach: This problem can be solved by generating Prime factors of A. To maximize GCD the only possible way is to choose different prime factors so that product of them would give the maximum GCD. Follow the steps given below to solve the problem. 

  • Create a function say, primeFactors() to find all the prime factors of a number.
  • Firstly call primeFactors() and pass A and an array is passed by reference to store all prime factors in a sorted manner.
  • Iterate over the prime factor array, and distribute all the factors of A to B and C alternatively such that,
    • B = primefactor[0] * primefactor[2] * primefactor[4] – – – and so on.
    • C = primefactor[1] * primefactor[3] * primefactor[5] – – – and so on.
  • Output the numbers B and C separated by space.

For Example: N = 72
Prime Factorization of 72 = 2 * 2 * 2 * 3 * 3. 
primefactor[] = {2, 2, 2, 3, 3}
B = primefactor[0] * primefactor[2] * primefactor[4] => 2 * 2 * 3 = 12.
C = primefactor[1] * primefactor[3] => 2 * 3 = 6.
Hence, B = 12 and C = 6.

Below is the implementation of the above approach.

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to stores prime factors
// of a number in a vector.
void primeFactors(int A, vector<int>& prime)
{
    while (A % 2 == 0) {
        prime.push_back(2);
        A = A / 2;
    }
 
    for (int i = 3; i <= sqrt(A); i += 2) {
        while (A % i == 0) {
            prime.push_back(i);
            A = A / i;
        }
    }
 
    if (A > 2)
        prime.push_back(A);
}
 
// Find the numbers B and C
// such that gcd(B, C) is max
// and product is A.
void maxPairGCD(int A)
{
    // Vector to store prime factors
    vector<int> prime;
 
    primeFactors(A, prime);
    int B = 1, C = 1;
    int temp = 0;
 
    for (int i = 0; i < prime.size(); i++) {
        if (temp == 0)
            B *= prime[i];
        else
            C *= prime[i];
        temp = temp ^ 1;
    }
    cout << B << " " << C;
}
 
// Driver code
int main()
{
    int A = 72;
 
    // Function Call
    maxPairGCD(A);
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.ArrayList;
 
class GFG
{
 
  // Function to stores prime factors
  // of a number in a vector.
  static void primeFactors(int A, ArrayList<Integer> prime)
  {
    while (A % 2 == 0)
    {
      prime.add(2);
      A = A / 2;
    }
 
    for (int i = 3; i <= Math.sqrt(A); i += 2)
    {
      while (A % i == 0)
      {
        prime.add(i);
        A = A / i;
      }
    }
 
    if (A > 2)
      prime.add(A);
  }
 
  // Find the numbers B and C
  // such that gcd(B, C) is max
  // and product is A.
  static void maxPairGCD(int A)
  {
 
    // Vector to store prime factors
    ArrayList<Integer> prime = new ArrayList<Integer>();
 
    primeFactors(A, prime);
    int B = 1, C = 1;
    int temp = 0;
 
    for (int i = 0; i < prime.size(); i++) {
      if (temp == 0)
        B *= prime.get(i);
      else
        C *= prime.get(i);
      temp = temp ^ 1;
    }
    System.out.println(B + " " + C);
  }
 
  // Driver code
  public static void main(String args[])
  {
    int A = 72;
 
    // Function Call
    maxPairGCD(A);
  }
}
 
// This code is contributed by Saurabh Jaiswal

Python3




# Python 3 program for above approach
import math
 
# Function to stores prime factors
# of a number in a vector.
def primeFactors(A,  prime):
 
    while (A % 2 == 0):
        prime.append(2)
        A = A / 2
 
    for i in range(3, int(math.sqrt(A)) + 12):
        while (A % i == 0):
            prime.append(i)
            A = A / i
 
    if (A > 2):
        prime.append(A)
 
# Find the numbers B and C
# such that gcd(B, C) is max
# and product is A.
def maxPairGCD(A):
 
    # Vector to store prime factors
    prime = []
 
    primeFactors(A, prime)
    B = 1
    C = 1
    temp = 0
 
    for i in range(len(prime)):
        if (temp == 0):
            B *= prime[i]
 
        else:
            C *= prime[i]
        temp = temp ^ 1
 
    print(B, C)
 
# Driver code
if __name__ == "__main__":
 
    A = 72
 
    # Function Call
    maxPairGCD(A)
 
    # This code is contributed by ukasp.

C#




// C# program to implement
// the above approach
using System;
using System.Collections;
 
class GFG
{
 
  // Function to stores prime factors
  // of a number in a vector.
  static void primeFactors(int A, ArrayList prime)
  {
    while (A % 2 == 0) {
      prime.Add(2);
      A = A / 2;
    }
 
    for (int i = 3; i <= Math.Sqrt(A); i += 2) {
      while (A % i == 0) {
        prime.Add(i);
        A = A / i;
      }
    }
 
    if (A > 2)
      prime.Add(A);
  }
 
  // Find the numbers B and C
  // such that gcd(B, C) is max
  // and product is A.
  static void maxPairGCD(int A)
  {
    // Vector to store prime factors
    ArrayList prime = new ArrayList();
 
    primeFactors(A, prime);
    int B = 1, C = 1;
    int temp = 0;
 
    for (int i = 0; i < prime.Count; i++) {
      if (temp == 0)
        B *= (int)prime[i];
      else
        C *= (int)prime[i];
      temp = temp ^ 1;
    }
    Console.Write(B + " " + C);
  }
 
  // Driver code
  public static void Main()
  {
    int A = 72;
 
    // Function Call
    maxPairGCD(A);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
        // JavaScript code for the above approach
 
        // Function to stores prime factors
        // of a number in a vector.
        function primeFactors(A, prime)
        {
         
            while (A % 2 == 0) {
                prime.push(2);
                A = A / 2;
            }
 
            for (let i = 3; i <= Math.sqrt(A); i += 2) {
                while (A % i == 0) {
                    prime.push(i);
                    A = A / i;
                }
            }
 
            if (A > 2)
                prime.push(A);
 
            return prime;
        }
 
        // Find the numbers B and C
        // such that gcd(B, C) is max
        // and product is A.
        function maxPairGCD(A)
        {
         
            // Vector to store prime factors
            let prime = [];
 
            prime = primeFactors(A, prime);
            let B = 1, C = 1;
            let temp = 0;
 
            for (let i = 0; i < prime.length; i++) {
                if (temp == 0)
                    B *= prime[i];
                else
                    C *= prime[i];
                temp = temp ^ 1;
            }
            document.write(B + " " + C)
        }
 
        // Driver code
        let A = 72;
 
        // Function Call
        maxPairGCD(A);
 
  // This code is contributed by Potta Lokesh
    </script>
Output
12 6

 

Time Complexity: O(Sqrt(A) )
Auxiliary Space: O(log(A))

 


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