Given the **length**, **breadth**, **height** of a cuboid. The task is to divide the given cuboid in minimum number of cubes such that size of all cubes is same and sum of volumes of cubes is maximum.

Examples:

Input : l = 2, b = 4, h = 6 Output : 2 6 A cuboid of length 2, breadth 4 and height 6 can be divided into 6 cube of side equal to 2. Volume of cubes = 6*(2*2*2) = 6*8 = 48. Volume of cuboid = 2*4*6 = 48. Input : 1 2 3 Output : 1 6

First of all, we are not allowed to waste volume of cuboid as we meed maximum volume sum. So, each side should be completely divide among all cubes. And since each of three side of cubes are equal, so each side of the cuboid need to be divisible by same number, say x, which will going to be the side of the cube. So, we have to maximize this x, which will divide given length, breadth and height. This x will be maximum only if it is greatest common divisor of given length, breadth and height. So, the length of the cube will be GCD of length, breadth and height.

Now, to compute number of cubes, we know total volume of cuboid and can find volume of one cube (since side is already calculated). So, total number of cubes is equal to (volume of cuboid)/(volume of cube) i.e (l * b * h)/(x * x * x).

Below is implementation of this approach:

## C++

// CPP program to find optimal way to break // cuboid into cubes. #include <bits/stdc++.h> using namespace std; // Print the maximum side and no of cube. void maximizecube(int l, int b, int h) { // GCD to find side. int side = __gcd(l, __gcd(b, h)); // dividing to find number of cubes. int num = l / side; num = (num * b / side); num = (num * h / side); cout << side << " " << num << endl; } // Driver code int main() { int l = 2, b = 4, h = 6; maximizecube(l, b, h); return 0; }

## Java

// JAVA Code for Divide cuboid into cubes // such that sum of volumes is maximum import java.util.*; class GFG { static int gcd(int m, int n) { if(n == 0) return m; else if(n > m) return gcd(n,m); else return gcd(n, m % n); } // Print the maximum side and no // of cube. static void maximizecube(int l, int b, int h) { // GCD to find side. int side = gcd(l, gcd(b, h)); // dividing to find number of cubes. int num = l / side; num = (num * b / side); num = (num * h / side); System.out.println( side + " " + num); } /* Driver program */ public static void main(String[] args) { int l = 2, b = 4, h = 6; maximizecube(l, b, h); } } // This code is contributed by Arnav Kr. Mandal.

## Python3

# Python3 code to find optimal way to break # cuboid into cubes. from fractions import gcd # Print the maximum side and no of cube. def maximizecube( l , b , h ): # GCD to find side. side = gcd(l, gcd(b, h)) # dividing to find number of cubes. num = int(l / side) num = int(num * b / side) num = int(num * h / side) print(side, num) # Driver code l = 2 b = 4 h = 6 maximizecube(l, b, h) # This code is contributed by "Sharad_Bhardwaj".

## C#

// C# Code for Divide cuboid into cubes // such that sum of volumes is maximum using System; class GFG { static int gcd(int m, int n) { if(n == 0) return m; else if(n > m) return gcd(n,m); else return gcd(n, m % n); } // Print the maximum side and no // of cube. static void maximizecube(int l, int b, int h) { // GCD to find side. int side = gcd(l, gcd(b, h)); // dividing to find number of cubes. int num = l / side; num = (num * b / side); num = (num * h / side); Console.WriteLine( side + " " + num); } /* Driver program */ public static void Main() { int l = 2, b = 4, h = 6; maximizecube(l, b, h); } } // This code is contributed by vt_m.

## PHP

<?php // PHP program to find optimal way // to break cuboid into cubes. // 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 - $a) ; } // Print the maximum side and no of cube. function maximizecube($l, $b, $h) { // GCD to find side. $side = __gcd($l, __gcd($b, $h)); // dividing to find number of cubes. $num = $l / $side; $num = ($num * $b / $side); $num = ($num * $h / $side); echo $side , " " , $num ; } // Driver code $l = 2; $b = 4; $h = 6; maximizecube($l, $b, $h); // This code is contributed by anuj_67. ?>

Output:

2 6

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