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).

Approach:
Follow the steps below to solve the problem:



  1. 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}).

  2. 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++

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    // 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;
    }

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    Python3

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    # 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

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    Output:

    5 10
    

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

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