# Program to find LCM of 2 numbers without using GCD

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

 ``

## Javascript

 ``

Output:

`LCM of 5 and 7 is 35`

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

Auxiliary Space: O(1)

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