LCM of Two Numbers in C

In this article, we will learn how to write a C program to find the LCM of two numbers. LCM (Least Common Multiple) of two numbers is the smallest positive number that can be divided by both numbers without leaving a remainder. For example, the LCM of 15 and 25 is 75.

Algorithm to Find LCM in C

• Find the maximum of the two numbers and store them in a variable max.
• Run a loop and check if max is divisible by both numbers.
  • if (max % x == 0 && max % y == 0)
• If the condition is true, it means that max is the LCM of the two numbers.
• If the condition is false, increment max by 1 and continue the loop to check the next number.

C Program To Find LCM of Two Numbers

C

// C program to find LCM of
// two numbers
#include <stdio.h>

// Driver code
int main()
{
    int x = 15, y = 25, max;
    max = (x > y) ? x : y;

    // While loop to check if max variable
    // is divisible by x and y
    while (1) {
        if (max % x == 0 && max % y == 0) {
            printf("The LCM of %d and %d is %d.", x, y,
                   max);
            break;
        }

        ++max;
    }

    return 0;
}

Output

The LCM of 15 and 25 is 75.

Complexity Analysis

• Time complexity: O(x*y)
• Auxiliary space: O(1)

LCM of Two Numbers using GCD

In mathematics, the LCM of two numbers is the product of two numbers divided by their GCD. So,

LCM(a, b) = (a x b)  /  GCD(a, b)

Refer to the article Program to Find GCD or HCF of Two Numbers to learn how to find the GCD of two numbers.

C Program To Find LCM of Two Numbers using GCD

C++

// C++ program to find LCM of two numbers
#include <iostream>
using namespace std;

// Recursive function to return gcd of a and b
long long gcd(long long int a, long long int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

// Function to return LCM of two numbers
long long lcm(int a, int b) { return (a / gcd(a, b)) * b; }

// Driver program to test above function
int main()
{
    int a = 15, b = 20;
    cout << "LCM of " << a << " and " << b << " is "
         << lcm(a, b);
    return 0;
}

Output

LCM of 15 and 20 is 60

Complexity Analysis

• Time complexity: O(log(min(a, b)))
• Auxiliary space: O(1)

Please refer to the complete article Program to find LCM of two numbers for more methods to find LCM of two numbers.