## Algorithm

Input: Accept two integers (`n1` and `n2`) from the user.

Find Maximum: Determine the maximum of the two input numbers (`n1` and `n2`) and store it in the variable `max`.

Find LCM (Least Common Multiple):

Start a do-while loop that continues indefinitely (`while(true)`).

Check if `max` is divisible by both `n1` and `n2` using the modulo operator (`%`).

If the condition is true, print the value of `max` as the Least Common Multiple (LCM) and break out of the loop.

If the condition is false, increment `max` by 1 and continue the loop.

Output: Display the LCM of the two input numbers.

End: Return 0 to indicate successful completion of the program.

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#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;
}``````
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Output

cmd
Enter two numbers: 12
18
LCM = 36

### #2 Example - C++ Programing Find LCM using HCF . The LCM of two numbers is given by: LCM = (n1 * n2) / HCF

```Code - C++ Programming```

``````#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;
}``````
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## Demonstration

C++ Programing Example to Find LCM-DevsEnv