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C++ Program to Find LCM

Program to Calculate LCM

In this program, you’ll learn how to Calculate LCM (Lowest Common Multiple) of two integers using loops and decision-making statements in C++ Programming.

To understand this example to Calculate LCM, 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.

Program to 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

C++ Program to find least common multiple

In above program, the 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 the loop is terminated.

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

Find LCM using HCF

The formula to find LCM:

LCM = (n1 * n2) / HCF

Program to Find LCM using HCF

#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;
}

Output

Enter two numbers: 12
23
LCM = 276

C++ Program to Find LCM using HCF

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Ask your questions and clarify your/others doubts on how to calculate LCM by commenting. Documentation

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