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GCD, LCM and Distributive Property

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2022
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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

 

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.
?>


Javascript




<script>
// JavaScript program to compute value of GCD(LCM(x,y), LCM(x,z))
function __gcd( a,  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))
function findValue(x, y, z)
{
    let g = __gcd(y, z);
 
    // Return LCM(x, GCD(y, z))
    return (x*g)/__gcd(x, g);
}
 
    let x = 30, y = 40, z = 400;
    document.write( findValue(x, y, z));
 
// This code contributed by gauravrajput1
 
</script>


Output: 

120

Time Complexity: O(log (min(n))) 
Auxiliary Space: O(log (min(n))) 

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 Aarti_Rathi and Mazhar Imam Khan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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