 # 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?