# Minimum integer such that it leaves a remainder 1 on dividing with any element from the range [2, N]

Given an integer N, the task is to find the minimum possible integer X such that X % M = 1 for all M from the range [2, N]

Examples:

Input: N = 5
Output: 61
61 % 2 = 1
61 % 3 = 1
61 % 4 = 1
61 % 5 = 1

Input: N = 2
Output: 3

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

Approach: Find the lcm of all the integers from the range [2, N] and store it in a variable lcm. Now we know that lcm is the smallest number which is divisible by all the elements from the range [2, N] and to make it leave a remainder of 1 on every division, just add 1 to it i.e. lcm + 1 is the required answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the smallest number ` `// which on dividing with any element from ` `// the range [2, N] leaves a remainder of 1 ` `long` `getMinNum(``int` `N) ` `{ ` `    ``// Find the LCM of the elements ` `    ``// from the range [2, N] ` `    ``int` `lcm = 1; ` `    ``for` `(``int` `i = 2; i <= N; i++) ` `        ``lcm = ((i * lcm) / (__gcd(i, lcm))); ` ` `  `    ``// Return the required number ` `    ``return` `(lcm + 1); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 5; ` ` `  `    ``cout << getMinNum(N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function to return the smallest number ` `    ``// which on dividing with any element from ` `    ``// the range [2, N] leaves a remainder of 1 ` `    ``static` `long` `getMinNum(``int` `N) ` `    ``{ ` `        ``// Find the LCM of the elements ` `        ``// from the range [2, N] ` `        ``int` `lcm = ``1``; ` `        ``for` `(``int` `i = ``2``; i <= N; i++) ` `            ``lcm = ((i * lcm) / (__gcd(i, lcm))); ` ` `  `        ``// Return the required number ` `        ``return` `(lcm + ``1``); ` `    ``} ` ` `  `    ``static` `int` `__gcd(``int` `a, ``int` `b) ` `    ``{ ` `        ``if` `(b == ``0``) ` `            ``return` `a; ` `        ``return` `__gcd(b, a % b); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``int` `N = ``5``; ` `        ``System.out.println(getMinNum(N)); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` `from` `math ``import` `gcd ` ` `  `# Function to return the smallest number  ` `# which on dividing with any element from  ` `# the range [2, N] leaves a remainder of 1  ` `def` `getMinNum(N) :  ` ` `  `    ``# Find the LCM of the elements  ` `    ``# from the range [2, N]  ` `    ``lcm ``=` `1``;  ` `    ``for` `i ``in` `range``(``2``, N ``+` `1``) : ` `        ``lcm ``=` `((i ``*` `lcm) ``/``/` `(gcd(i, lcm)));  ` ` `  `    ``# Return the required number  ` `    ``return` `(lcm ``+` `1``);  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``N ``=` `5``;  ` ` `  `    ``print``(getMinNum(N));  ` ` `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to return the smallest number ` `    ``// which on dividing with any element from ` `    ``// the range [2, N] leaves a remainder of 1 ` `    ``static` `long` `getMinNum(``int` `N) ` `    ``{ ` `        ``// Find the LCM of the elements ` `        ``// from the range [2, N] ` `        ``int` `lcm = 1; ` `        ``for` `(``int` `i = 2; i <= N; i++) ` `            ``lcm = ((i * lcm) / (__gcd(i, lcm))); ` ` `  `        ``// Return the required number ` `        ``return` `(lcm + 1); ` `    ``} ` ` `  `    ``static` `int` `__gcd(``int` `a, ``int` `b) ` `    ``{ ` `        ``if` `(b == 0) ` `            ``return` `a; ` `        ``return` `__gcd(b, a % b); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `N = 5; ` `        ``Console.WriteLine(getMinNum(N)); ` `    ``} ` `} ` ` `  `// This code has been contributed by anuj_67.. `

## PHP

 ` `

Output:

```61
```

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