The LCM of two integers is the smallest positive integer that is perfectly divisible by both the numbers (without a remainder).

Example 1: LCM using while Loop and if Statement

public class Main{
  public static void main(String[] args){

    int n1 = 72, n2 = 120, lcm;

    // maximum number between n1 and n2 is stored in lcm
    lcm = (n1 > n2) ? n1 : n2;

    // Always true
    while(true) {
      if( lcm % n1 == 0 && lcm % n2 == 0 ) {
        System.out.printf("The LCM of %d and %d is %d.", n1, n2, lcm);
        break;
      }
      ++lcm;
    }
  }
}

Output

The LCM of 72 and 120 is 360.

In this program, the two numbers whose LCM is to be found are stored in variables n1 and n2 respectively.

Then, we initially set lcm to the largest of the two numbers. This is because, LCM cannot be less than the largest number.

Inside the infinite while loop (while(true)), we check if lcm perfectly divides both n1 and n2 or not.

If it does, we’ve found the LCM. We print the LCM and break out from the while loop using break statement.

Else, we increment lcm by 1 and re-test the divisibility condition.

We can also use GCD to find the LCM of two numbers using the following formula:

LCM = (n1 * n2) / GCD

If you don’t know how to calculate GCD in Java, check Java Program to find GCD of two numbers.

Example 2: Calculate LCM using GCD

public class Main{
  public static void main(String[] args){

    int n1 = 72, n2 = 120, gcd = 1;

    for(int i = 1; i <= n1 && i <= n2; ++i) {
      // Checks if i is factor of both integers
      if(n1 % i == 0 && n2 % i == 0)
        gcd = i;
    }

    int lcm = (n1 * n2) / gcd;
    System.out.printf("The LCM of %d and %d is %d.", n1, n2, lcm);
  }
}

The output of this program is the same as Example 1.

Here, inside the for loop, we calculate the GCD of the two numbers – n1 and n2. After the calculation, we use the above formula to calculate the LCM.