# Smallest number divisible by first n numbers

Given a number **n** find the smallest number evenly divisible by each number 1 to n.**Examples:**

Input : n = 4 Output : 12 Explanation : 12 is the smallest numbers divisible by all numbers from 1 to 4 Input : n = 10 Output : 2520 Input : n = 20 Output : 232792560

If you observe carefully the **ans** must be the **LCM of the numbers 1 to n**.

To find LCM of numbers from 1 to n –

- Initialize ans = 1.

- Iterate over all the numbers from i = 1 to i = n.

At the i’th iteration**ans = LCM(1, 2, …….., i)**. This can be done easily as**LCM(1, 2, …., i) = LCM(ans, i)**.

Thus at i’th iteration we just have to do –

ans = LCM(ans, i) = ans * i / gcd(ans, i) [Using the below property, a*b = gcd(a,b) * lcm(a,b)]

**Note :** In C++ code, the answer quickly exceeds the integer limit, even the long long limit.

Below is the implementation of the logic.

## C++

`// C++ program to find smallest number evenly divisible by` `// all numbers 1 to n` `#include<bits/stdc++.h>` `using` `namespace` `std;` `// Function returns the lcm of first n numbers` `long` `long` `lcm(` `long` `long` `n)` `{` ` ` `long` `long` `ans = 1; ` ` ` `for` `(` `long` `long` `i = 1; i <= n; i++)` ` ` `ans = (ans * i)/(__gcd(ans, i));` ` ` `return` `ans;` `}` `// Driver program to test the above function` `int` `main()` `{` ` ` `long` `long` `n = 20;` ` ` `cout << lcm(n);` ` ` `return` `0;` `}` |

## Java

`// Java program to find the smallest number evenly divisible by` `// all numbers 1 to n` ` ` `class` `GFG{` `static` `long` `gcd(` `long` `a, ` `long` `b)` `{` ` ` `if` `(a%b != ` `0` `)` ` ` `return` `gcd(b,a%b);` ` ` `else` ` ` `return` `b;` `}` `// Function returns the lcm of first n numbers` `static` `long` `lcm(` `long` `n)` `{` ` ` `long` `ans = ` `1` `; ` ` ` `for` `(` `long` `i = ` `1` `; i <= n; i++)` ` ` `ans = (ans * i)/(gcd(ans, i));` ` ` `return` `ans;` `}` ` ` `// Driver program to test the above function` `public` `static` `void` `main(String []args)` `{` ` ` `long` `n = ` `20` `;` ` ` `System.out.println(lcm(n));` `}` `}` |

## C#

`// C# program to find smallest number` `// evenly divisible by` `// all numbers 1 to n` `using` `System;` `public` `class` `GFG{` ` ` `static` `long` `gcd(` `long` `a, ` `long` `b)` `{` `if` `(a%b != 0)` ` ` `return` `gcd(b,a%b);` `else` ` ` `return` `b;` `}` `// Function returns the lcm of first n numbers` `static` `long` `lcm(` `long` `n)` `{` ` ` `long` `ans = 1; ` ` ` `for` `(` `long` `i = 1; i <= n; i++)` ` ` `ans = (ans * i)/(gcd(ans, i));` ` ` `return` `ans;` `}` `// Driver program to test the above function` ` ` `static` `public` `void` `Main (){` ` ` `long` `n = 20;` ` ` `Console.WriteLine(lcm(n));` ` ` `}` `//This code is contributed by akt_mit ` `}` |

## Python3

`# Python program to find the smallest number evenly ` `# divisible by all number 1 to n` `import` `math` ` ` `# Returns the lcm of first n numbers` `def` `lcm(n):` ` ` `ans ` `=` `1` ` ` `for` `i ` `in` `range` `(` `1` `, n ` `+` `1` `):` ` ` `ans ` `=` `int` `((ans ` `*` `i)` `/` `math.gcd(ans, i)) ` ` ` `return` `ans` ` ` `# main` `n ` `=` `20` `print` `(lcm(n))` |

## PHP

`<?php` `// Note: This code is not working on GFG-IDE` `// because gmp libraries are not supported` `// PHP program to find smallest number` `// evenly divisible by all numbers 1 to n` `// Function returns the lcm` `// of first n numbers` `function` `lcm(` `$n` `)` `{` ` ` `$ans` `= 1;` ` ` `for` `(` `$i` `= 1; ` `$i` `<= ` `$n` `; ` `$i` `++)` ` ` `$ans` `= (` `$ans` `* ` `$i` `) / (gmp_gcd(` `strval` `(ans),` ` ` `strval` `(i)));` ` ` `return` `$ans` `;` `}` `// Driver Code` `$n` `= 20;` `echo` `lcm(` `$n` `);` `// This code is contributed by mits` `?>` |

## Javascript

`<script>` `// Javascript program to find the smallest number evenly divisible by` `// all numbers 1 to n` `function` `gcd(a, b)` `{` ` ` `if` `(a%b != 0)` ` ` `return` `gcd(b,a%b);` ` ` `else` ` ` `return` `b;` `}` ` ` `// Function returns the lcm of first n numbers` `function` `lcm(n)` `{` ` ` `let ans = 1; ` ` ` `for` `(let i = 1; i <= n; i++)` ` ` `ans = (ans * i)/(gcd(ans, i));` ` ` `return` `ans;` `}` ` ` `// function call` ` ` ` ` `let n = 20;` ` ` `document.write(lcm(n));` ` ` `</script>` |

**Output :**

232792560

The above solution works fine for a single input. But if we have multiple inputs, it is a good idea to use Sieve of Eratosthenes to store all prime factors. Please refer below article for Sieve based approach.

LCM of First n Natural Numbers

This article is contributed by **Ayush Khanduri**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the **Essential Maths for CP Course** at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**