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

**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:

- C++ if, if…else and Nested if…else
- C++ while and do…while Loop

LCM of two integers `a` and `b` is the smallest positive integer that is divisible by both `a` and `b`.

### Example 1: Find LCM

#include <iostream> using namespace std; int main() { int n1, n2, max; cout << "Enter two numbers: "; cin >> n1 >> n2; // maximum value between n1 and n2 is stored in max max = (n1 > n2) ? n1 : n2; do { if (max % n1 == 0 && max % n2 == 0) { cout << "LCM = " << max; break; } else ++max; } while (true); return 0; }

**Output**

Enter two numbers: 12 18 LCM = 36

In above program, user is asked to integer two integers `n1` and `n2` and largest of those two numbers is stored in `max`.

It is checked whether `max` is divisible by `n1` and `n2`, if it’s divisible by both numbers, `max `(which contains LCM) is printed and loop is terminated.

If not, value of `max` is incremented by 1 and same process goes on until `max `is divisible by both `n1` and `n2`.

### Example 2: Find LCM using HCF

The LCM of two numbers is given by:

LCM = (n1 * n2) / HCF

Visit this page to learn: How to compute HCF in C++? {Updating Soon}

#include <iostream> using namespace std; int main() { int n1, n2, hcf, temp, lcm; cout << "Enter two numbers: "; cin >> n1 >> n2; hcf = n1; temp = n2; while(hcf != temp) { if(hcf > temp) hcf -= temp; else temp -= hcf; } lcm = (n1 * n2) / hcf; cout << "LCM = " << lcm; return 0; }