# 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
```

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, 20) = 60

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 ` `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

 ` ``\$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:

```120
```

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

## tags:

Mathematical GCD-LCM Mathematical