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LCM of digits of a given number
• Difficulty Level : Easy
• Last Updated : 24 Mar, 2021

Given a number n, find LCM of its digits.
Examples:

```Input : 397
Output : 63
LCM of 3, 9 and 7 is 63.

Input : 244
Output : 4
LCM of 2, 4 and 4 is 4.```

We traverse the digits of number one by one below loop
digit = n mod 10;
n = n / 10;
While traversing digits, we keep track of current LCM and keep updating LCM by finding LCM of current digit with current LCM.

## C++

 `// CPP program to find LCM of digits of a number``#include``#include``using` `namespace` `std;` `int` `digitLCM(``int` `n)``{``    ``int` `lcm = 1;``    ``while` `(n > 0)``    ``{``        ``lcm = boost::math::lcm(n%10, lcm);` `        ``// If at any point LCM become 0.``        ``// return it``        ``if` `(lcm == 0)``            ``return` `0;` `        ``n = n/10;``    ``}``    ``return` `lcm;``}` `// driver code``int` `main()``{``    ``long` `n = 397;``    ``cout << digitLCM(n);``    ``return` `0;``}`

## Java

 `// Java program to find LCM of digits of a number` `class` `GFG``{``// define lcm function``static` `int` `lcm_fun(``int` `a, ``int` `b)``{``    ``if` `(b == ``0``)``        ``return` `a;``    ``return` `lcm_fun(b, a % b);``}` `static` `int` `digitLCM(``int` `n)``{``    ``int` `lcm = ``1``;``    ``while` `(n > ``0``)``    ``{``        ``lcm = (n % ``10` `* lcm) / lcm_fun(n % ``10``, lcm);` `        ``// If at any point LCM become 0.``        ``// return it``        ``if` `(lcm == ``0``)``            ``return` `0``;` `        ``n = n/``10``;``    ``}``    ``return` `lcm;``}` `// driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``397``;``    ``System.out.println(digitLCM(n));``}``}``// This code is contributed by mits`

## Python3

 `# Python3 program to find``# LCM of digits of a number` `# define lcm function``def` `lcm_fun(a, b):` `    ``if` `(b ``=``=` `0``):``        ``return` `a;``    ``return` `lcm_fun(b, a ``%` `b);` `def` `digitLCM(n):` `    ``lcm ``=` `1``;``    ``while` `(n > ``0``):``        ``lcm ``=` `int``((n ``%` `10` `*` `lcm) ``/``              ``lcm_fun(n ``%` `10``, lcm));` `        ``# If at any point LCM``        ``# become 0. return it``        ``if` `(lcm ``=``=` `0``):``            ``return` `0``;` `        ``n ``=` `int``(n ``/` `10``);``    ` `    ``return` `lcm;` `# Driver code``n ``=` `397``;``print``(digitLCM(n));` `# This code is contributed by mits`

## C#

 `// C# program to find LCM of digits``// of a number``class` `GFG``{``    ` `// define lcm function``static` `int` `lcm_fun(``int` `a, ``int` `b)``{``    ``if` `(b == 0)``        ``return` `a;``    ``return` `lcm_fun(b, a % b);``}` `static` `int` `digitLCM(``int` `n)``{``    ``int` `lcm = 1;``    ``while` `(n > 0)``    ``{``        ``lcm = (n % 10 * lcm) / lcm_fun(n % 10, lcm);` `        ``// If at any point LCM become 0.``        ``// return it``        ``if` `(lcm == 0)``            ``return` `0;` `        ``n = n/10;``    ``}``    ``return` `lcm;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n = 397;``    ``System.Console.WriteLine(digitLCM(n));``}``}` `// This code is contributed by mits`

## PHP

 ` 0)``    ``{``        ``\$lcm` `= (int)((``\$n` `% 10 * ``\$lcm``) /``              ``lcm_fun(``\$n` `% 10, ``\$lcm``));` `        ``// If at any point LCM``        ``// become 0. return it``        ``if` `(``\$lcm` `== 0)``            ``return` `0;` `        ``\$n` `= (int)(``\$n` `/ 10);``    ``}``    ``return` `\$lcm``;``}` `// Driver code``\$n` `= 397;``echo` `digitLCM(``\$n``);` `// This code is contributed by mits``?>`

## Javascript

 ``

Output:

`63`

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