Program to find LCM of two numbers

LCM (Least Common Multiple) of two numbers is the smallest number which can be divided by both numbers. 

For example, LCM of 15 and 20 is 60, and LCM of 5 and 7 is 35.

A simple solution is to find all prime factors of both numbers, then find union of all factors present in both numbers. Finally, return the product of elements in union.

An efficient solution is based on the below formula for LCM of two numbers ‘a’ and ‘b’. 

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

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

We have discussed function to find GCD of two numbers. Using GCD, we can find LCM. 

Below is the implementation of the above idea:

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;
}

C

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

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

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

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

Java

// Java program to find LCM of two numbers.
import java.io.*;
public class Test
{
    // Recursive method to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

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

    // Driver method
    public static void main(String[] args)
    {
        int a = 15, b = 20;
        System.out.println("LCM of " + a +
                           " and " + b +
                      " is " + lcm(a, b));
    }
}

Python3

# Python program to find LCM of two numbers

# Recursive function to return gcd of a and b
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b % a, a)

# Function to return LCM of two numbers
def lcm(a,b):
    return (a // gcd(a,b))* b

# Driver program to test above function
a = 15
b = 20
print('LCM of', a, 'and', b, 'is', lcm(a, b))

# This code is contributed by Danish Raza

C#

// C# program to find LCM
// of two numbers.
using System;
class GFG {

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

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

    // Driver method
    public static void Main()
    {
        int a = 15, b = 20;
        Console.WriteLine("LCM of " + a +
         " and " + b + " is " + lcm(a, b));
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to find LCM of two numbers

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

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

    // Driver Code
    $a = 15;
    $b = 20;
    echo "LCM of ",$a, " and "
         ,$b, " is ", lcm($a, $b);

// This code is contributed by anuj_67.
?>

Javascript

<script>

// Javascript program to find LCM of two numbers

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

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

// Driver program to test above function

    let a = 15, b = 20;
    document.write("LCM of " + a + " and "
    + b + " is " + lcm(a, b));


// This code is contributed by Mayank Tyagi

</script>

Output

LCM of 15 and 20 is 60

Time Complexity: O(log(min(a,b))
Auxiliary Space: O(log(min(a,b))

Another Efficient Approach: Using conditional for loop

C++

// C++ program to find LCM of 2 numbers
#include <bits/stdc++.h>
using namespace std;

// Function to return LCM of two numbers
int LCM(int a, int b)
{
    int greater = max(a, b);
    int smallest = min(a, b);
    for (int i = greater; ; i += greater) {
        if (i % smallest  == 0)
            return i;
    }
}

// Driver program to test above function
int main()
{
    int a = 10, b = 5;
    cout << "LCM of " << a << " and "
         << b << " is " << LCM(a, b);
    return 0;
}

Java

// Java program to find LCM of 2 numbers

import java.util.Scanner;

public class LCM {
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int a = 10;
        int b = 5;
        int lcm = findLCM(a, b);
        System.out.println("LCM of " + a + " and " + b
                           + " is " + lcm);
        sc.close();
    }
    // Function to return LCM of two numbers
    public static int findLCM(int a, int b)
    {
        int greater = Math.max(a, b);
        int smallest = Math.min(a, b);
        for (int i = greater;; i += greater) {
            if (i % smallest == 0)
                return i;
        }
    }
}

Python3

# Python program to find LCM of two numbers

# Function to return LCM of two numbers
def LCM(a, b):
    greater = max(a, b)
    smallest = min(a, b)
    for i in range(greater, a*b+1, greater):
        if i % smallest == 0:
            return i

# Driver program to test above function
if __name__ == '__main__':
    a = 10
    b = 5
    print("LCM of", a, "and", b, "is", LCM(a, b))

C#

// C# program to find LCM of 2 numbers
using System;

class LCMProgram
{

  // Function to return LCM of two numbers
  static int LCM(int a, int b)
  {
    int greater = Math.Max(a, b);
    int smallest = Math.Min(a, b);
    for (int i = greater;; i += greater) {
      if (i % smallest == 0)
        return i;
    }
  }

  // Driver program to test above function
  static void Main()
  {
    int a = 10, b = 5;
    Console.WriteLine("LCM of " + a + " and " + b
                      + " is " + LCM(a, b));
  }
}

Javascript

// Javascript program to find LCM of two numbers

// Function to return LCM of two numbers
function LCM(a, b){
    let greater = Math.max(a, b);
    let smallest = Math.min(a, b);
    for(let i = greater; i <= a*b; i+=greater){
        if(i % smallest == 0){
            return i;
        }
    }
}

// Driver program to test above function
let a = 10;
let b = 5;
console.log("LCM of", a, "and", b, "is", LCM(a, b));

Output

LCM of 10 and 5 is 10

Time Complexity: O(min(a,b))
Auxiliary Space: O(1)

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