Maximum possible GCD for a pair of integers with sum N

Last Updated : 16 Dec, 2022

Given an integer N, the task is to find the maximum possible GCD of a pair of integers such that their sum is N.

Examples :

Input: N = 30
Output: 15
Explanation: GCD of (15, 15) is 15, which is the maximum possible GCD

Input: N = 33
Output: 11
Explanation: GCD of (11, 22) is 11, which is the maximum possible GCD

Naive Approach:
The simplest approach to solve this problem is to calculate GCD for all pair of integers with sum N and find the maximum possible GCD among them.
Time complexity: O(N²logN)
Auxiliary Space: O(1)

Efficient Approach:
Follow the steps given below to optimize the above approach:

Iterate up to ?N and find the largest proper factor of N.
If N is prime, i.e. no factor could be obtained, print 1, as all pairs are co-prime.
Otherwise, print the largest possible factor as the answer.

Below is the implementation of the above approach:

C++

// C++ Program to find the maximum 
// possible GCD of any pair with 
// sum N 
#include <bits/stdc++.h> 
using namespace std; 

// Function to find the required 
// GCD value 
int maxGCD(int N) 
{ 
    for (int i = 2; i * i <= N; i++) { 

        // If i is a factor of N 
        if (N % i == 0) { 

            // Return the largest 
            // factor possible 
            return N / i; 
        } 
    } 

    // If N is a prime number 
    return 1; 
} 

// Driver Code 
int main() 
{ 
    int N = 33; 
    cout << "Maximum Possible GCD value is : "
        << maxGCD(N) << endl; 
    return 0; 
}

Java

// Java program to find the maximum
// possible GCD of any pair with
// sum N
import java.io.*;
public class GFG{

// Function to find the required
// GCD value
static int maxGCD(int N)
{
    for(int i = 2; i * i <= N; i++)
    {

        // If i is a factor of N
        if (N % i == 0) 
        {

            // Return the largest
            // factor possible
            return N / i;
        }
    }

    // If N is a prime number
    return 1;
}
    
// Driver Code
public static void main(String[] args)
{
    int N = 33;
    System.out.println("Maximum Possible GCD " + 
                       "value is : " + maxGCD(N));
}
}

// This code is conhtributed by rutvik_56

Python3

# Python3 program to find the maximum 
# possible GCD of any pair with 
# sum N 

# Function to find the required 
# GCD value 
def maxGCD(N): 
    
    i = 2
    while(i * i <= N): 

        # If i is a factor of N 
        if (N % i == 0): 

            # Return the largest 
            # factor possible 
            return N // i 
        
        i += 1

    # If N is a prime number 
    return 1

# Driver Code 
N = 33

print("Maximum Possible GCD value is : ",
       maxGCD(N))

# This code is contributed by code_hunt

// C# program to find the maximum
// possible GCD of any pair with
// sum N
using System;

class GFG{

// Function to find the required
// GCD value
static int maxGCD(int N)
{
    for(int i = 2; i * i <= N; i++)
    {

        // If i is a factor of N
        if (N % i == 0) 
        {

            // Return the largest
            // factor possible
            return N / i;
        }
    }

    // If N is a prime number
    return 1;
}
    
// Driver Code
public static void Main(String[] args)
{
    int N = 33;
    Console.WriteLine("Maximum Possible GCD " + 
                      "value is : " + maxGCD(N));
}
}

// This code is contributed by Princi Singh

JavaScript

<script>

// Javascript program to find the maximum
// possible GCD of any pair with
// sum N

// Function to find the required
// GCD value
function maxGCD(N)
{
    for(var i = 2; i * i <= N; i++)
    {
        
        // If i is a factor of N
        if (N % i == 0) 
        {
            
            // Return the largest
            // factor possible
            return N / i;
        }
    }

    // If N is a prime number
    return 1;
}

// Driver code
var N = 33; 
document.write("Maximum Possible GCD " + 
               "value is :" + maxGCD(N));

// This code is contributed by Ankita saini.
   
</script>

Output:

Maximum Possible GCD value is : 11

Time Complexity: O(?N)
Auxiliary Space: O(1)

Maximize GCD of all possible pairs from 1 to N

codeku

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Maximum possible GCD for a pair of integers with sum N

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