#
Problem
of the Week

Problem
D and Solution

Caen’s
Cubes

## Problem

Caen has a cube with a volume of \(n \text{
cm}^3\). They cut this cube into \(n\) smaller cubes, each with a side length
of \(1\) cm. The total surface area of
the \(n\) smaller cubes is ten times
the surface area of Caen’s original cube. Determine the side length of
Caen’s original cube.

## Solution

Let the side length of Caen’s original cube be \(x\) cm, where \(x>0\). It follows that \(n=x^3\).

Each of the six sides of Caen’s original cube has area \(x^2\) cm\(^2\), so the total surface area of the
original cube is \(6x^2\text{
cm}^2\).

Consider one of the smaller cubes. The area of one the six
faces is \(1\text{ cm}^2\). So, the
surface area of one of these smaller cubes is \(6\text{ cm}^2\). Thus, the total surface
area of the \(n\) smaller cubes is
\(6n\text{ cm}^2\).

Since the total surface area of the \(n\) cubes is ten times the surface area of Caen’s original cube, we have \[6n=10(6x^2)\] Dividing both sides by 6, we
have \[n=10x^2\] But \(n=x^3\), so this tells us that \[x^3=10x^2\] Since \(x>0\), we have \(x^2>0\). Dividing both sides by \(x^2\), we find that \(x=10\).

Therefore, the side length of Caen’s original cube was \(10\) cm.

**Extension**:

If the combined surface area of the \(n\) cubes with a side length of \(1\) cm was \(Q\) times the surface area of the original
uncut cube, then the side length of the original uncut cube would have
been \(Q\) cm. Can you see why?