Given two integers **L** and **R**, the task is to find the greatest divisor that divides all the natural numbers in the range **[L, R]**.

**Examples:**

Input:L = 3, R = 12

Output:1

Input:L = 24, R = 24

Output:24

**Approach:** For a range of consecutive integer elements, there are two cases:

- If
**L = R**then the answer will**L**. - If
**L < R**then all consecutive natural numbers in this range are co-primes. So,**1**is the only number that will be able to divide all the elements of the range.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the greatest divisor that ` `// divides all the natural numbers in the range [l, r] ` `int` `find_greatest_divisor(` `int` `l, ` `int` `r) ` `{ ` ` ` `if` `(l == r) ` ` ` `return` `l; ` ` ` ` ` `return` `1; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `l = 2, r = 12; ` ` ` ` ` `cout << find_greatest_divisor(l, r); ` `} ` |

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## Java

`// Java implementation of the approach ` ` ` `class` `GFG { ` ` ` `// Function to return the greatest divisor that ` `// divides all the natural numbers in the range [l, r] ` ` ` `static` `int` `find_greatest_divisor(` `int` `l, ` `int` `r) { ` ` ` `if` `(l == r) { ` ` ` `return` `l; ` ` ` `} ` ` ` ` ` `return` `1` `; ` ` ` `} ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String[] args) { ` ` ` `int` `l = ` `2` `, r = ` `12` `; ` ` ` ` ` `System.out.println(find_greatest_divisor(l, r)); ` ` ` `} ` `} ` `// This code is contributed by PrinciRaj1992 ` |

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## Python3

`# Python3 implementation of the approach ` ` ` `# Function to return the greatest divisor that ` `# divides all the natural numbers in the range [l, r] ` `def` `find_greatest_divisor(l, r): ` ` ` ` ` `if` `(l ` `=` `=` `r): ` ` ` `return` `l; ` ` ` ` ` `return` `1` `; ` ` ` ` ` `# Driver Code ` ` ` `l ` `=` `2` `; ` `r ` `=` `12` `; ` ` ` `print` `(find_greatest_divisor(l, r)); ` ` ` `#This code is contributed by Shivi_Aggarwal ` |

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## C#

`// C# implementation of the approach ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `// Function to return the greatest divisor that ` ` ` `// divides all the natural numbers in the range [l, r] ` ` ` `static` `int` `find_greatest_divisor(` `int` `l, ` `int` `r) { ` ` ` `if` `(l == r) { ` ` ` `return` `l; ` ` ` `} ` ` ` ` ` `return` `1; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main() { ` ` ` `int` `l = 2, r = 12 ; ` ` ` ` ` `Console.WriteLine(find_greatest_divisor(l, r)); ` ` ` `} ` ` ` `// This code is contributed by Ryuga ` ` ` `} ` |

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## PHP

`<?php ` `// PHP implementation of the approach ` ` ` `// Function to return the greatest ` `// divisor that divides all the natural ` `// numbers in the range [l, r] ` `function` `find_greatest_divisor(` `$l` `, ` `$r` `) ` `{ ` ` ` `if` `(` `$l` `== ` `$r` `) ` ` ` `return` `$l` `; ` ` ` ` ` `return` `1; ` `} ` ` ` `// Driver Code ` `$l` `= 2; ` `$r` `= 12; ` ` ` `echo` `find_greatest_divisor(` `$l` `, ` `$r` `); ` ` ` `// This code is contributed by jit_t ` `?> ` |

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**Output:**

1

**Time Complexity:** O(1)

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