LCM of digits of a given number

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


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

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right



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.



My Personal Notes arrow_drop_up

Improved By : Mithun Kumar



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.