Program to find LCM of 2 numbers without using GCD

Finding LCM using GCD is explained here but here the task is to find LCM without first calculating GCD.

Examples:

Input: 7, 5
Output: 35

Input: 2, 6
Output: 6

The approach is to start with the largest of the 2 numbers and keep incrementing the larger number by itself till smaller number perfectly divides the resultant.

C++

// C++ program to find LCM of 2 numbers 
// without using GCD 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return LCM of two numbers 
int findLCM(int a, int b) 
{ 
    int lar = max(a, b); 
    int small = min(a, b); 
    for (int i = lar; ; i += lar) { 
        if (i % small == 0) 
            return i; 
    } 
} 
  
// Driver program to test above function 
int main() 
{ 
    int a = 5, b = 7; 
    cout << "LCM of " << a << " and " 
         << b << " is " << findLCM(a, b); 
    return 0; 
} 

Java

// Java program to find LCM of 2 numbers 
// without using GCD 
import java.io.*; 
import java.lang.*; 
  
class GfG { 
      
    // Function to return LCM of two numbers 
    public static int findLCM(int a, int b) 
    { 
        int lar = Math.max(a, b); 
        int small = Math.min(a, b); 
        for (int i = lar; ; i += lar) { 
            if (i % small == 0) 
                return i; 
        } 
    } 
      
    // Driver program to test above function 
    public static void main(String [] argc) 
    { 
        int a = 5, b = 7; 
        System.out.println( "LCM of " + a + " and "
            + b + " is " + findLCM(a, b)); 
          
    } 
} 
  
// This dose is contributed by Sagar Shukla. 

Python 3

# Python 3 program to find  
# LCM of 2 numbers without  
# using GCD 
import sys 
  
# Function to return 
# LCM of two numbers 
def findLCM(a, b): 
  
    lar = max(a, b) 
    small = min(a, b) 
    i = lar 
    while(1) : 
        if (i % small == 0): 
            return i 
        i += lar 
      
# Driver Code 
a = 5
b = 7
print("LCM of " , a , " and ",  
                  b , " is " ,  
      findLCM(a, b), sep = "") 
  
# This code is contributed  
# by Smitha 

C#

// C# program to find 
// LCM of 2 numbers 
// without using GCD 
using System; 
  
class GfG  
{ 
      
    // Function to return 
    // LCM of two numbers 
    public static int findLCM(int a,  
                              int b) 
    { 
        int lar = Math.Max(a, b); 
        int small = Math.Min(a, b); 
        for (int i = lar; ; i += lar)  
        { 
            if (i % small == 0) 
                return i; 
        } 
    } 
      
    // Driver Code 
    public static void Main() 
    { 
        int a = 5, b = 7; 
        Console.WriteLine("LCM of " + a +  
                            " and " + b +  
                                 " is " +  
                          findLCM(a, b)); 
          
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to find 
// LCM of 2 numbers 
// without using GCD 
  
// Function to return 
// LCM of two numbers 
function findLCM($a, $b) 
{ 
    $lar = max($a, $b); 
    $small = min($a, $b); 
    for ($i = $lar; ; $i += $lar) 
    { 
        if ($i % $small == 0) 
            return $i; 
    } 
} 
  
// Driver Code 
$a = 5; 
$b = 7; 
echo "LCM of " , $a , " and ",  
                 $b , " is " ,  
                 findLCM($a, $b); 
  
// This code is contributed 
// by Smitha 
?> 

Output:

LCM of 5 and 7 is 35

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My Personal Notes

C++

Java

Python 3

C#

PHP

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