Copyright © University of Cambridge. All rights reserved.

'Brush Loads' printed from http://nrich.maths.org/

Show menu


Well, you have a whole collection of cubes and we're going to paint the outside that's visible once we've put them together.
A few simple rules:
  • the cubes must be together;
  • they must not be toppling over.
We'll need to look at how much paint is needed. One Brush Load (a kind of unit that we'll use) will paint one square face.
Well here are $5$ cubes:
n5cubesFlat
Counting the faces to be painted comes to $15$, so $15$ Brush Loads (remember we're only counting visible faces, so not those that are touching the surface where the cubes are placed).

But of course we could have placed the $5$ cubes differently:

n5CubesOrd
Counting the faces to be painted we now have $17$, so $17$ Brush Loads.

And, how about:

n5CubesUp
Now we'll need $21$ Brush Loads.

O.K. Now let's explore different numbers of cubes, say $6, 7, 8, 9,$ etc. Each time we count the number of square faces showing and so find the number of Brush Loads (BLs).
The cubes - as in the case of $5$ - can be arranged in different ways, sometimes giving different BLs.

So I'm setting the challenge of finding ways of arranging cubes so that first of all there are the least number of BLs needed and then the biggest number of BLs needed.

Try first with $5$ cubes.
How about $6, 7, 8$ and $9$ cubes?

If you wish to go further, then increase the number of cubes. See if you can get all the numbers between the largest and the smallest e.g. if $15$ is the smallest number for $5$ cubes and $21$ is the largest, can you get arrangements that'll give $16, 17, 18, 19,$ and $20$ BLs?