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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ?>

Output:

63

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