# Maximum possible GCD for a pair of integers with sum N

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   ` `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 ` `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#

 `// 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 `

Output:

```Maximum Possible GCD value is : 11
```

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

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