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
Explaination:
Maximum possible GCD between all possible pairs is 5 which occurs for the pair (10, 5).

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 

Output:

5 10

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

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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