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++

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// 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;
}

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Java

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// 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.

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Python 3

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# 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

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C#

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// 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.

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PHP

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<?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
?>

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Output:

LCM of 5 and 7 is 35


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Improved By : vt_m, Smitha Dinesh Semwal