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 ?> |
Javascript
<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> |
Output:
LCM of 5 and 7 is 35
Time Complexity: O(max(a, b))
Auxiliary Space: O(1)
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