C++ Program To Find LCM of Two Numbers

The Least Common Multiple (LCM) of two numbers is the smallest positive number that is divisible by both numbers. For example, the LCM of 15 and 20 is 60.

• LCM is useful in problems involving common multiples and repeating intervals.
• It can be calculated using a simple search, the std::lcm() function, or the GCD-based formula.

Illustration

Input: a = 15, b = 20
Output: LCM = 60

Explanation: 60 is the smallest positive number that is divisible by both 15 and 20.

Methods to Find LCM of Two Numbers

The LCM can be calculated using the following approaches:

Method 1: Using Simple Iteration

Start from the larger of the two numbers and check each successive number until finding one that is divisible by both numbers.

Approach:

• Initialize two numbers a and b.
• Start checking from max(a, b).
• If the current number is divisible by both a and b, it is the LCM.
• Otherwise, increment the number and continue checking.

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

int main()
{
    int a = 15, b = 20;

    int lcm = max(a, b);

    while (lcm % a != 0 || lcm % b != 0)
        lcm++;

    cout << "LCM of " << a << " and " << b << " is "
         << lcm;

    return 0;
} 

LCM of 15 and 20 is 60

Method 2: Using Built-in std::lcm()

C++17 provides the std::lcm() function in the <numeric> header to directly calculate the LCM of two integers.

Syntax

std::lcm(a, b);

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

int main()
{
    int a = 15, b = 20;

    cout << "LCM of " << a << " and " << b << " is "
         << lcm(a, b);

    return 0;
}

LCM of 15 and 20 is 60

Method 3: Using GCD

The LCM of two numbers can be efficiently calculated using their GCD. The relationship is:

LCM(a, b) × GCD(a, b) = |a × b|

Therefore:

LCM(a, b) = |(a / GCD(a, b)) × b|

Dividing before multiplying helps reduce the chance of intermediate overflow.

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

int main()
{
    long long a = 15, b = 20;

    long long g = gcd(a, b);
    long long lcm = (a / g) * b;

    cout << "LCM of " << a << " and " << b << " is "
         << lcm;

    return 0;
}