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; }

## Leave a Reply