Programa para encontrar MCM de 2 números sin usar MCD

Encontrar LCM usando GCD se explica aquí, pero aquí la tarea es encontrar LCM sin calcular primero GCD.
Ejemplos: 
 

Input: 7, 5
Output: 35

Input: 2, 6
Output: 6

El enfoque es comenzar con el mayor de los 2 números y seguir incrementando el número mayor por sí mismo hasta que el número menor divida perfectamente la resultante.
 

// 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 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 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# 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 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
?>

<script>
// javascript program to find LCM of 2 numbers
// without using GCD
 
    // Function to return LCM of two numbers
    function findLCM(a , b) {
        var lar = Math.max(a, b);
        var small = Math.min(a, b);
        for (i = lar;; i += lar) {
            if (i % small == 0)
                return i;
        }
    }
 
    // Driver program to test above function
    var a = 5, b = 7;
        document.write("LCM of " + a + " and " + b + " is " + findLCM(a, b));
 
 
// This code contributed by umadevi9616
</script>

LCM of 5 and 7 is 35

Complejidad del tiempo: O(min(a, b))

Espacio Auxiliar: O(1)