# 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 = 5Output:61

61 % 2 = 1

61 % 3 = 1

61 % 4 = 1

61 % 5 = 1Input:N = 2Output:3Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the

DSA Self Paced Courseat a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please referComplete Interview Preparation Course.In case you wish to attend

live classeswith experts, please referDSA Live Classes for Working ProfessionalsandCompetitive Programming Live for Students.

**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;` `}` |

## 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

`<?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....` `?>` |

## Javascript

`<script>` `// Javascript 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]` ` ` `var` `lcm = 1;` ` ` `for` `(` `var` `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` `var` `N = 5;` `document.write( getMinNum(N));` `</script>` |

**Output:**

61

**Time Complexity: **O(N * log(N) )

**Auxiliary Space: **O(1)