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

Learn how to find the GCD of two numbers in C programming.

## 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.