The LCM of two integers n1 and n2 is the smallest positive integer that is perfectly divisible by both n1 and n2 (without a remainder). For example, the LCM of 72 and 120 is 360.

LCM using while and if

#include <stdio.h>
int main(){
    int n1, n2, min;
    printf("Enter two positive integers: ");
    scanf("%d %d", &n1, &n2);

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

    while (1) {
        if (min % n1 == 0 && min % n2 == 0) {
            printf("The LCM of %d and %d is %d.", n1, n2, min);
            break;
        }
        ++min;
    }
    return 0;
}

Output

Enter two positive integers: 72
120
The LCM of 72 and 120 is 360.

In this program, the integers entered by the user are stored in variable n1 and n2 respectively.

The largest number among n1 and n2 is stored in min. The LCM of two numbers cannot be less than min.

The test expression of while loop is always true.

In each iteration, whether min is perfectly divisible by n1 and n2 is checked.

if (min % n1 == 0 && min % n2 == 0) { ... }

If this test condition is not true, min is incremented by 1 and the iteration continues until the test expression of the if statement is true.

The LCM of two numbers can also be found using the formula:

LCM = (num1*num2)/GCD

LCM Calculation Using GCD

#include <stdio.h>
int main(){
    int n1, n2, i, gcd, lcm;
    printf("Enter two positive integers: ");
    scanf("%d %d", &n1, &n2);

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

    lcm = (n1 * n2) / gcd;

    printf("The LCM of two numbers %d and %d is %d.", n1, n2, lcm);
    return 0;
}

Output

Enter two positive integers: 72
120
The LCM of two numbers 72 and 120 is 360.