# C++ program to find LCM of two numbers

# Introduction:

In this C++ tutorial program, we will learn how to find the LCM of two user-given number. The program will take these numbers as input from the user and print out the result.

With this example, you will learn how to read user inputs, how to print values on the console, how to use an *if-else* condition and how to use a *while loop* in C++.

## LCM of two numbers :

*LCM* or *lowest common multiplier* of two numbers is the smallest number that is perfectly divisible by both numbers.

For example, *LCM* of *3* and *5* is *15*.

## Algorithm :

To find out the *LCM*, we will use one loop that will run till the *LCM* is found. Suppose, we want to find out the *LCM* for *5* and *3*. Following are the multipliers for each :

5 : 5, 10, 15, 20, 25.... 3 : 3, 6, 9, 12, 15, 18, 21....

So, if we are checking for an *LCM*, we can only check for the multiplier of the bigger number i.e. *5*. If you check the multiplier or the smaller number or all numbers starting from *3*, it will work. But the process will be faster if we check only the multiplier of the bigger number starting from *5* in this example.

The user will enter both numbers and our program will first verify which one is the bigger number. It will then keep checking for the *LCM* by using a loop.

## C++ program :

#include <iostream> using namespace std; int main() { //1 int firstNo, secondNo, lcm; int count = 1; //2 cout << "Enter the first number : " << endl; cin >> firstNo; //3 cout << "Enter the second number : " << endl; cin >> secondNo; //4 if (firstNo < secondNo) { //5 lcm = secondNo; while (count < firstNo) { if (lcm % firstNo == 0 && lcm % secondNo == 0) { break; } lcm += secondNo; count++; } } else { //6 lcm = firstNo; while (count < secondNo) { if (lcm % firstNo == 0 && lcm % secondNo == 0) { break; } lcm += firstNo; count++; } } //7 cout << "LCM is " << lcm << endl; return 0; }

### Explanation :

*The commented numbers in the above program denote the step numbers below :*

*firstNo*,*secondNo*and*lcm*are three integer variables for holding the first number, second number and the LCM value.- Ask the user to enter the first number. Read and store it in
*firstNo*variable. - Ask the user to enter the second number. Read and store it in
*secondNo*variable. - Check if the
*firstNo*is smaller than the*secondNo*or not. If yes, enter inside the*if*block or else enter in the*else*block. - If
*firstNo*is smaller than*secondNo*, assign*secondNo*to*lcm*. Run one*for loop*until the value of*count*is less than*firstNo*. Check if the current value of*lcm*is divisible by both of these numbers or not. If yes, break from the loop or else increment the value of*count*and add*secondNo*to*lcm*.The

*lcm*variable is actually the multiplier or the bigger number. If the*secondNo*or the bigger number is*5*,*lcm*will be*5, 10, 15…*etc. The*count*will stop the operation once the value of*lcm*is equal to the product of both numbers. If no divisor is found, product of the two numbers will be the lowest common divisor always. - Similar to the step above, run one
*while*loop and find the*lcm*if*firstNo*is greater than the*secondNo*. - Finally, print out the value of the
*LCM*to the user.

### Sample Output :

Enter the first number : 3 Enter the second number : 5 LCM is 15 Enter the first number : 3 Enter the second number : 7 LCM is 21 Enter the first number : 100 Enter the second number : 10 LCM is 100 Enter the first number : 13 Enter the second number : 17 LCM is 221

### Conclusion :

Go through the above program and try to run it on your system. If you have any queries, don’t hesitate to drop a comment below.

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