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# Program to find LCM of 2 numbers without using GCD

• Difficulty Level : Easy
• Last Updated : 17 Jan, 2023

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 ``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

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## Javascript

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Output:

`LCM of 5 and 7 is 35`

Time Complexity: O(max(a, b))

Auxiliary Space: O(1)

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