Find pair with maximum GCD for integers in range 2 to N

Given a number N, the task is to find a pair of integers in the range [2, N] with maximum GCD.
Examples:

Input: N = 10 
Output: 5 
Explanation:
Maximum possible GCD between all possible pairs is 5 which occurs for the pair (10, 5).
Input: N = 13 
Output: 6 
Explanation:
Maximum possible GCD between all possible pairs is 6 which occurs for the pair (12, 6). 

Approach:
Follow the steps below to solve the problem: 

• If N is even, return the pair {N, N / 2}. 

Illustration:
If N = 10, Maximum possible GCD for any pair is 5( for the pair {5, 10}).
If N = 20, Maximum possible GCD for any pair is 10( for the pair {20, 10}). 

• If N is odd, then return the pair{N – 1, (N – 1) / 2}.  

Illustration:
If N = 11, Maximum possible GCD for any pair is 5( for the pair {5, 10}).
If N = 21, Maximum possible GCD for any pair is 10( for the pair {20, 10}). 

Below is the implementation of the above approach: 

C++

// C++ Program to find a pair of
// integers less than or equal
// to N such that their GCD
// is maximum
#include <bits/stdc++.h>
using namespace std;

// Function to find the required
// pair whose GCD is maximum
void solve(int N)
{
    // If N is even
    if (N % 2 == 0) {

        cout << N / 2 << " "
             << N << endl;
    }
    // If N is odd
    else {
        cout << (N - 1) / 2 << " "
             << (N - 1) << endl;
    }
}

// Driver Code
int main()
{
    int N = 10;
    solve(N);
    return 0;
}

Java

// Java program to find a pair of
// integers less than or equal
// to N such that their GCD
// is maximum

class GFG{

// Function to find the required
// pair whose GCD is maximum
static void solve(int N)
{

    // If N is even
    if (N % 2 == 0)
    {
        System.out.print(N / 2 + " " +
                         N + "\n");
    }

    // If N is odd
    else
    {
        System.out.print((N - 1) / 2 + " " +
                         (N - 1) + "\n");
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 10;

    solve(N);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 Program to find a pair
# of integers less than or equal
# to N such that their GCD
# is maximum

# Function to find the required
# pair whose GCD is maximum
def solve(N):

    # If N is even
    if (N % 2 == 0):
        print(N // 2, N)

    # If N is odd
    else :
        print((N - 1) // 2, (N - 1))

# Driver Code
N = 10
solve(N)

# This code is contributed by divyamohan123

C#

// C# program to find a pair of
// integers less than or equal
// to N such that their GCD
// is maximum
using System;
class GFG{

// Function to find the required
// pair whose GCD is maximum
static void solve(int N)
{

    // If N is even
    if (N % 2 == 0)
    {
        Console.Write(N / 2 + " " +
                      N + "\n");
    }

    // If N is odd
    else
    {
        Console.Write((N - 1) / 2 + " " +
                      (N - 1) + "\n");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 10;

    solve(N);
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

// JavaScript program to find a pair of
// integers less than or equal
// to N such that their GCD
// is maximum

    // Function to find the required
    // pair whose GCD is maximum
    function solve(N) {

        // If N is even
        if (N % 2 == 0) {
            document.write(N / 2 + " " + N + "<br/>");
        }

        // If N is odd
        else {
            document.write((N - 1) / 2 + " " + (N - 1) + "<br/>");
        }
    }

    // Driver Code

        var N = 10;

        solve(N);

// This code is contributed by todaysgaurav

</script>

Output:

5 10

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