Skip to content
Related Articles

Related Articles

Improve Article
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<iostream>
#include<boost/math/common_factor.hpp>
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




<?php
// PHP program to find
// LCM of digits of a number
 
// define lcm function
function lcm_fun($a, $b)
{
    if ($b == 0)
        return $a;
    return lcm_fun($b, $a % $b);
}
 
function 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
$n = 397;
echo digitLCM($n);
 
// This code is contributed by mits
?>

Javascript




<script>
// Javascript program to find LCM of digits of a number
 
    // define lcm function
    function lcm_fun( a, b)
    {
        if (b == 0)
            return a;
        return lcm_fun(b, a % b);
    }
 
    function digitLCM( n)
    {
        let lcm = 1;
        while (n > 0)
        {
            lcm = (n % 10 * lcm) / lcm_fun(n % 10, lcm);
 
            // If at any polet LCM become 0.
            // return it
            if (lcm == 0)
                return 0;
 
            n = parseInt(n / 10);
        }
        return lcm;
    }
 
    // Driver code    
        let n = 397;
        document.write(digitLCM(n));
 
// This code is contributed by gauravrajput1
</script>

Output: 
 

63

This article is contributed by nikunj_agarwal. 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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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.




My Personal Notes arrow_drop_up
Recommended Articles
Page :