GCD, LCM and Distributive Property

Given three integers x, y, z, the task is to compute the value of GCD(LCM(x,y), LCM(x,z)).
Where, GCD = Greatest Common Divisor, LCM = Least Common Multiple

Examples:

Input: x = 15, y = 20, z = 100
Output: 60

Input: x = 30, y = 40, z = 400
Output: 120

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

One way to solve it is by finding GCD(x, y), and using it we find LCM(x, y). Similarly, we find LCM(x, z) and then we finally find the GCD of the obtained results.

An efficient approach can be done by the fact that the following version of distributivity holds true:

GCD(LCM (x, y), LCM (x, z)) = LCM(x, GCD(y, z))

For example, GCD(LCM(3, 4), LCM(3, 10)) = LCM(3, GCD(4, 10)) = LCM(3, 2) = 6

This reduces our work to compute the given problem statement.

C++

// C++ program to compute value of GCD(LCM(x,y), LCM(x,z)) 
#include<bits/stdc++.h> 
using namespace std; 
  
// Returns value of  GCD(LCM(x,y), LCM(x,z)) 
int findValue(int x, int y, int z) 
{ 
    int g = __gcd(y, z); 
  
    // Return LCM(x, GCD(y, z)) 
    return (x*g)/__gcd(x, g); 
} 
  
int main() 
{ 
    int x = 30, y = 40, z = 400; 
    cout << findValue(x, y, z); 
    return 0; 
} 

Java

// Java program to compute value  
// of GCD(LCM(x,y), LCM(x,z)) 
  
class GFG 
{ 
    // Recursive function 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); 
    }  
      
    // Returns value of GCD(LCM(x,y), LCM(x,z)) 
    static int findValue(int x, int y, int z) 
    { 
        int g = __gcd(y, z); 
      
        // Return LCM(x, GCD(y, z)) 
        return (x*g) / __gcd(x, g); 
    } 
      
    // Driver code 
    public static void main (String[] args) 
    { 
        int x = 30, y = 40, z = 400; 
        System.out.print(findValue(x, y, z)); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python program to compute 
# value of GCD(LCM(x,y), LCM(x,z)) 
  
# Recursive function to 
# return gcd of a and b 
def __gcd(a,b): 
      
    # Everything divides 0  
    if (a == 0 or 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) 
  
# Returns value of 
#  GCD(LCM(x,y), LCM(x,z)) 
def findValue(x, y, z): 
  
    g = __gcd(y, z) 
   
    # Return LCM(x, GCD(y, z)) 
    return (x*g)/__gcd(x, g) 
  
# driver code 
x = 30
y = 40
z = 400
print("%d"%findValue(x, y, z)) 
  
# This code is contributed 
# by Anant Agarwal. 

C#

// C# program to compute value  
// of GCD(LCM(x,y), LCM(x,z)) 
using System; 
  
class GFG { 
      
    // Recursive function 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); 
    }  
      
    // Returns value of GCD(LCM(x,y), 
    // LCM(x,z)) 
    static int findValue(int x, int y, int z) 
    { 
        int g = __gcd(y, z); 
      
        // Return LCM(x, GCD(y, z)) 
        return (x*g) / __gcd(x, g); 
    } 
      
    // Driver code 
    public static void Main () 
    { 
        int x = 30, y = 40, z = 400; 
          
        Console.Write(findValue(x, y, z)); 
    } 
} 
  
// This code is contributed by 
// Smitha Dinesh Semwal. 

PHP

<?php 
// PHP program to compute value 
// of GCD(LCM(x,y), LCM(x,z)) 
  
// Recursive function to 
// return gcd of a and b 
function __gcd( $a, $b) 
{ 
      
    // Everything divides 0  
    if ($a == 0 or $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); 
} 
  
// Returns value of GCD(LCM(x,y),  
// LCM(x,z)) 
function findValue($x, $y, $z) 
{ 
    $g = __gcd($y, $z); 
  
    // Return LCM(x, GCD(y, z)) 
    return ($x * $g)/__gcd($x, $g); 
} 
  
    // Driver Code 
    $x = 30; 
    $y = 40;  
    $z = 400; 
    echo findValue($x, $y, $z); 
  
// This code is contributed by anuj_67. 
?> 

Output:

As a side note, vice versa is also true, i.e., gcd(x, lcm(y, z)) = lcm(gcd(x, y), gcd(x, z)

Reference:
https://en.wikipedia.org/wiki/Distributive_property#Other_examples

This article is contributed by Mazhar Imam Khan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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