Example #1: LCM using while Loop and if Statement

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

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

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

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

LCM = (num1*num2)/GCD

Example #2: LCM Calculation by Finding 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)
    {
        // Checks if i is 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;

}

GCD EASY CALCULATION

First divide all the numbers by the smallest prime that can divide all of them. The smallest prime that divides 24 or 60 is 2.

2
24 60
12 30


Continue the steps until we can't find any prime number that can divide all the number on the right side.

2
24 60
2
12 30
3
6  15
2  5


The GCD is 2 × 2 × 3 = 12.

LCM EASY CALCULATION

First divide all the numbers by the smallest prime that can divide any of them. The smallest prime that divides 24 or 60 is 2.

2
24 60
12 30


Continue the steps until we have all prime numbers on the left side and at the bottom.

2
24 60
2
12 30
3
6  15
2  5


The LCM is 2 × 2 × 3 × 2 × 5 = 120.