# 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

**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 <bits/stdc++.h> ` `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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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.. ` |

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## PHP

`<?php ` `// PHP implementation of the approach ` ` ` `// Function to return the smallest number ` `// which on dividing with any element from ` `// the range [2, N] leaves a remainder of 1 ` `function` `getMinNum(` `$N` `) ` `{ ` ` ` `// Find the LCM of the elements ` ` ` `// from the range [2, N] ` ` ` `$lcm` `= 1; ` ` ` `for` `(` `$i` `= 2; ` `$i` `<= ` `$N` `; ` `$i` `++) ` ` ` `$lcm` `= ((` `$i` `* ` `$lcm` `) / (__gcd(` `$i` `, ` `$lcm` `))); ` ` ` ` ` `// Return the required number ` ` ` `return` `(` `$lcm` `+ 1); ` `} ` ` ` `function` `__gcd(` `$a` `, ` `$b` `) ` `{ ` ` ` `if` `(` `$b` `== 0) ` ` ` `return` `$a` `; ` ` ` `return` `__gcd(` `$b` `, ` `$a` `% ` `$b` `); ` `} ` ` ` `// Driver code ` ` ` `$N` `= 5; ` `echo` `(getMinNum(` `$N` `)); ` ` ` `// This code has been contributed by ajit.... ` `?> ` |

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**Output:**

61

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