# Program to Find LCM of two numbers in c

The LCM ( Least common multiple) of two integers n1 and n2 is the smallest positive integer that is perfectly divisible by both n1 and n2 (without leaving any remainder). For example, the LCM of 2 and 4 is 4.

## Example 1: Finding LCM using while Loop

```#include <stdio.h>
int main(){
Program to Find LCM of two numbers in c
int n1, n2, minMultiple;
printf("Enter two positive integers: ");
scanf("%d %d", &n1, &n2);
// maximum number between n1 and n2 stored in minMultiple
minMultiple = (n1>n2) ? n1 : n2;
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: 24
36
The LCM of 24 and 36 is 72.```

In this program, the integers entered by the user are stored in variables named  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;
}```
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