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.

## 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;
}
```

**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 `minMultiple`. The LCM of two numbers cannot be less than `minMultiple`.

The test expression of while loop is always true (1). In each iteration, whether `minMultiple` is perfectly divisible by `n1` and `n2` is checked. If this test condition is not true, `minMultiple` is incremented by 1 and the iteration continues until the test expression of if statement is true.

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

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;
}
```