# C++ program to find LCM of Two Numbers

Write a C++ program to find LCM of Two Numbers with an example. LCM means the least common multiple of two or more integers. As per math, it is the smallest integer (positive) that perfectly divides the number (no reminder). In this C++ example, we find the LCM of two numbers using the while loop.

```#include<iostream>
using namespace std;

int main()
{
int number1, number2;

cout << "\nPlease Enter the First Integer for LCM  =  ";
cin >> number1;

cout << "\nPlease Enter the Second Integer for LCM  =  ";
cin >> number2;

int maxValue = (number1 > number2)? number1 : number2;

while(1)
{
if(maxValue % number1 == 0 && maxValue % number2 == 0)
{
cout << "LCM of " << number1 << " and " << number2 << " = " << maxValue;
break;
}
++maxValue;
}
return 0;
}```

## C++ program to find LCM of Two Numbers using GCD

```#include<iostream>
using namespace std;

int main()
{
int number1, number2, lcm, gcd, temp;

cout << "\nPlease Enter the First Integer =  ";
cin >> number1;

cout << "\nPlease Enter the Second Integer  =  ";
cin >> number2;

int a = number1;
int b = number2;

while(number2 != 0)
{
temp = number2;
number2 = number1 % number2;
number1 = temp;
}
gcd = number1;
cout << "\nGCD of " << a << " and " << b << " = " << gcd;

lcm = (a * b) / gcd;
cout << "\nLCM of " << a << " and " << b << " = " << lcm;
return 0;
}```
``````Please Enter the First Integer  =  15

Please Enter the Second Integer  =  40

GCD of 15 and 40 = 5
LCM of 15 and 40 = 120``````

In this C++ LCM of Two Numbers program, the long gcdOfTwoNumbers(long x, long y) method finds the GCD of two numbers. Next, we use that gcd to get the least common multiple.

```#include<iostream>
using namespace std;

long gcdOfTwoNumbers(long x, long y)
{
if(y == 0)
{
return x;
}
else
{
return gcdOfTwoNumbers(y, x % y);
}
}

int main()
{
int n1, n2, result;

cout << "\nPlease Enter the First Value  =  ";
cin >> n1;

cout << "\nPlease Enter the Second Value  =  ";
cin >> n2;

long gcd = gcdOfTwoNumbers(n1, n2);

cout << "\nGCD of " << n1 << " and " << n2 << " = " << gcd;

lcm = (number1 * number2) / gcd;
cout << "\nLCM of " << n1 << " and " << n2 << " = " << result;
return 0;
}```
``````Please Enter the First Value  =  22

Please Enter the Second Value  =  120

GCD of 22 and 120 = 2
LCM of 22 and 120 = 1320``````