In this
example, you will learn to calculate the LCM (Lowest common multiple) of two
numbers entered by the user.
To understand this
example, you should have the knowledge of the following C
programming topics:
C Programming Operators
C if...else Statement
C while and do...while Loop
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, max;
printf("Enter two positive integers: ");
scanf("%d %d", &n1, &n2);
// maximum number between n1 and n2 is stored in max
max = (n1 > n2) ? n1 : n2;
while (1) {
if (max % n1 == 0 && max % n2 == 0) {
printf("The LCM of %d and %d is %d.", n1, n2, max);
break;
}
++max;
}
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 max. The LCM of two numbers cannot be less than max.
The test expression of while loop is always true.
In each iteration, whether max is perfectly divisible by n1 and n2 is checked.
if (min % n1 == 0 &&
max% n2 == 0) { ... }
If this test condition is
not true, max 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.