# 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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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)]```

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 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     ` `} `

## Python

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

## PHP

 ` `

Output :

```232792560
```

The above solution works fine for 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

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