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'Brush Loads' printed from http://nrich.maths.org/
We have 5 cubes and we're going to put them together, following a few simple rules:
- the cubes must be together face-to-face;
- they must not be toppling over.
We're going to paint the faces that can be seen. One Brush Load (a kind of unit that we'll use) will paint one square face.
Can you find ways of arranging 5 cubes so that:
Need more information about counting BLs before you start the challenge?
- you need as few BLs as possible?
- you need as many BLs as possible?
Here are $5$ cubes:
Counting the faces to be painted comes to $15$, so $15$ Brush Loads are needed. (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, for example:
Counting the faces to be painted now, we have $17$, so $17$ Brush Loads.
And, how about:
Now we'll need $21$ Brush Loads (BLs).
Can you find arrangements that need all the numbers between the largest and the smallest numbers of BLs?
Take more cubes.......
What happens if you use more cubes, for example 6, 7, 8 ...?
Can you find out the smallest number of BLs and the largest number of BLs possible in each case?
Can you predict the arrangements which need as few BLs as possible and as many BLs as possible?