Given an integer **N** (? 2), the task is to find the maximum GCD among all pairs possible by the integers in the range **[1, N]**.

**Example:**

Input:N = 5Output:2Explanation :

GCD(1, 2) : 1

GCD(1, 3) : 1

GCD(1, 4) : 1

GCD(1, 5) : 1

GCD(2, 3) : 1

GCD(2, 4) :2

GCD(2, 5) : 1

GCD(3, 4) : 1

GCD(3, 5) : 1

GCD(4, 5) : 1

Input:N = 6Output:3Explanation:GCD of pair (3, 6) is the maximum.

**Naive Approach:**

The simplest approach to solve the problem is to generate all possible pairs from **[1, N]** and calculate **GCD** of each pair. Finally, print the maximum GCD obtained. **Time Complexity:** O(N^{2}logN) **Auxiliary Space:** O(1)

**Efficient Approach: **

Follow the steps below to solve the problem:

- Since all the pairs are distinct, then, for any pair
**{a, b}**with GCD**g**, either of**a**or**b**is greater than**g**. - Considering
**b**to be the greater number,**b ? 2g**, since**2g**is the smallest multiple of**g**greater than it. - Since
**b**cannot exceed**N**, and**2g ? N**. - Therefore,
**g ? floor(n/2)**. - Therefore, the maximum GCD that can be obtained is floor(n/2), when pair chosen is
**(floor(n/2), 2*floor(n/2)).**

Illustration:

N = 6

Maximum GCD = 6/2 = 3, occurs for the pair (3, 6)

Below is the implementation of the above approach:

## C++

`// C++ Program to implement` `// the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to obtain the maximum` `// gcd of all pairs from 1 to n` `void` `find(` `int` `n)` `{` ` ` `// Print the answer` ` ` `cout << n / 2 << endl;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 5;` ` ` `// Function call` ` ` `find(n);` ` ` `return` `0;` `}` |

## Java

`// Java Program to implement` `// the approach` `class` `GFG{` ` ` `// Function to obtain the maximum` `// gcd of all pairs from 1 to n` `static` `void` `find(` `int` `n)` `{` ` ` `// Print the answer` ` ` `System.out.println(n / ` `2` `);` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `5` `;` ` ` `// Function call` ` ` `find(n);` `}` `}` `// This code is contributed by Ritik Bansal` |

## Python3

`# Python3 program to implement` `# the approach` `# Function to obtain the maximum` `# gcd of all pairs from 1 to n` `def` `find(n):` ` ` `# Print the answer` ` ` `print` `(n ` `/` `/` `2` `)` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `# Given n` ` ` `n ` `=` `5` ` ` `# Function call` ` ` `find(n)` `# This code is contributed by Shivam Singh` |

## C#

`// C# Program to implement` `// the approach` `using` `System;` `class` `GFG{` ` ` `// Function to obtain the maximum` `// gcd of all pairs from 1 to n` `static` `void` `find(` `int` `n)` `{` ` ` `// Print the answer` ` ` `Console.Write(n / 2);` `}` ` ` `// Driver code` `public` `static` `void` `Main(` `string` `[] args)` `{` ` ` `int` `n = 5;` ` ` `// Function call` ` ` `find(n);` `}` `}` ` ` `// This code is contributed by rock_cool` |

## Javascript

`<script>` `// Javascript program to implement` `// the approach` `// Function to obtain the maximum` `// gcd of all pairs from 1 to n` `function` `find(n)` `{` ` ` ` ` `// Print the answer` ` ` `document.write(parseInt(n / 2, 10) + ` `"</br>"` `);` `}` `// Driver code` `let n = 5;` `// Function call` `find(n);` `// This code is contributed by divyeshrabadiya07` `</script>` |

**Output:**

2

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

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