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# Smallest number divisible by first n numbers

• Difficulty Level : Medium
• Last Updated : 30 Dec, 2022

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 –

1. Initialize ans = 1.

2. 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)]```
1.

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``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

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## Javascript

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Output

`232792560`

Time Complexity: O(n log2n) as the complexity of _gcd(a,b) in c++ is log2n  and that is running n times in a loop.
Auxiliary Space: O(1)
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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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