Examples on different ways to calculate the LCM (Lowest Common Multiple) of two integers using loops and decision making statements.

To understand this example, you should have the knowledge of following C programming topics:

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.

```
#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

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

```
#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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