Write a program that asks the user for two numbers then look

Write a program that asks the user for two numbers, then looks for the least common

multiple.

If the program does not find one by 100, stop looking. For example:

Enter a number: 3

Enter another number: 4

2? no. 3? no. 4? no. 5? no. 6? no. 7? no. 8? no. 9? no. 10? no.

11? no. 12? yes!

Solution

Please follow the code and comments for description :

CODE :

#include <iostream>
using namespace std;
int main() // class to run the code
{
    int a, b, lcm; // required initialisations
    cout << \"Enter a number: \"; // prompt to enter the first number
    cin >> a;
   cout << \"Enter another number: \"; // prompt to enter the second number
   cin >> b;
    for(int i = 2; i <= 100; i++) { // iterating over th enumbers till 100 for the LCM to be found
        if (i % a == 0 && i % b == 0) //Checking for the first number which is divisible by both the numbers
        {
            lcm = i; // if found replacing the result to the number
            cout << i << \"? Yes.!\" << endl; // print the output to console
            break; //exiting from the loop, as we don’t need anymore checking after getting the LCM
        }
        else {
            cout << i << \"? No\" << endl; // else iterating over the numbers printing the result to console
        }
    }  
    return 0;
}

OUTPUT :

Enter a number :
3
Enter another number :
4
2? No
3? No
4? No
5? No
6? No
7? No
8? No
9? No
10? No
11? No
12? Yes.!

Hope this is helpful.