LCM (Least Common Multiple) of two numbers is the smallest number which can be divided by both numbers.
For example LCM of 15 and 20 is 60 and LCM of 5 and 7 is 35.
A simple solution is to find all prime factors of both numbers, then find union of all factors present in both numbers. Finally return product of elements in union.
An efficient solution is based on below formula for LCM of two numbers ‘a’ and ‘b’.
a x b = LCM(a, b) * GCD (a, b) LCM(a, b) = (a x b) / GCD(a, b)
We have discussed function to find GCD of two numbers. Using GCD, we can find LCM.
Below is the implementation of above idea :
C++
// C++ program to find LCM of two numbers #include <iostream> using namespace std; // Recursive function to return gcd of a and b class gfg { public : int gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // Base case if (a == b) return a; // a is greater if (a > b) return gcd(a-b, b); return gcd(a, b-a); } // Function to return LCM of two numbers int lcm(int a, int b) { return (a*b)/gcd(a, b); } } ; // Driver program to test above function int main() { gfg g; int a = 15, b = 20; cout<<"LCM of "<<a<<" and "<<b<<" is "<<g.lcm(a, b); return 0; } |
C
// C program to find LCM of two numbers #include <stdio.h> // Recursive function to return gcd of a and b int gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // Base case if (a == b) return a; // a is greater if (a > b) return gcd(a-b, b); return gcd(a, b-a); } // Function to return LCM of two numbers int lcm(int a, int b) { return (a*b)/gcd(a, b); } // Driver program to test above function int main() { int a = 15, b = 20; printf("LCM of %d and %d is %d ", a, b, lcm(a, b)); return 0; } |
Java
// Java program to find LCM of two numbers. class Test { // Recursive method to return gcd of a and b static int gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return gcd(a-b, b); return gcd(a, b-a); } // method to return LCM of two numbers static int lcm(int a, int b) { return (a*b)/gcd(a, b); } // Driver method public static void main(String[] args) { int a = 15, b = 20; System.out.println("LCM of " + a +" and " + b + " is " + lcm(a, b)); } } |
Python3
# Python program to find LCM of two numbers # Recursive function to return gcd of a and b def gcd(a,b): # base case if a and b are equal if (a == b): return a # if a is greater if (a > b): return gcd(a-b, b) return gcd(a, b-a) # Function to return LCM of two numbers def lcm(a,b): return (a*b) / gcd(a,b) # Driver program to test above function a = 15 b = 20print('LCM of', a, 'and', b, 'is', lcm(a, b)) # This code is contributed by Danish Raza |
C#
// C# program to find LCM // of two numbers. using System; class GFG { // Recursive method to // return gcd of a and b static int gcd(int a, int b) { // Everything divides 0 if (a == 0 || b == 0) return 0; // base case if (a == b) return a; // a is greater if (a > b) return gcd(a-b, b); return gcd(a, b-a); } // method to return // LCM of two numbers static int lcm(int a, int b) { return (a * b) / gcd(a, b); } // Driver method public static void Main() { int a = 15, b = 20; Console.WriteLine("LCM of " + a + " and " + b + " is " + lcm(a, b)); } } // This code is contributed by anuj_67. |
PHP
<?php // PHP program to find LCM of two numbers // Recursive function to // return gcd of a and b function gcd( $a, $b) { // base case if ($a == $b) return $a; // a is greater if ($a > $b) return gcd($a - $b, $b); return gcd($a, $b - $a); } // Function to return LCM // of two numbers function lcm( $a, $b) { return ($a * $b) / gcd($a, $b); } // Driver Code $a = 15; $b = 20; echo "LCM of ",$a, " and " ,$b, " is ", lcm($a, $b); // This code is contributed by anuj_67. ?> |
Output:
LCM of 15 and 20 is 60
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