Archive for February, 2010

A problem, and some responses

February 15, 2010

Here at the NRICH project, we are regularly sent ideas for problems which we then develop into resources complete with Teachers’ Notes to be used in the classroom. Last week, Charlie knocked on my door and shared a problem he had been sent, which we solved together, and then we discussed whether it could be made into something suitable for the website. As we were discussing it, and drawing out similarities with existing NRICH problems, I realised that I wanted to try the problem out on others to see whether they would approach it in the same way that we did.

So I decided to post the problem on my Facebook account, to share with an assortment of friends, family, colleagues, people I knew at school and university, people I speak to daily, people I haven’t heard from in years. I was rather surprised by the response it got:

I made a cube from smaller cubes, and then painted entire faces of the larger cube (but maybe not all six faces.)
When I took my cube apart again, there were 45 small cubes with no paint on them at all.
Can you work out how big my cube was, and which large faces I painted?
What I’m most interested in is how people (particularly non-mathematicians) would tackle a problem like this. Let me know!

Firstly, someone picked me up for not being specific – I’d said “Can you work out how big my cube was” meaning how many small cubes I’d used to make it; without knowing the size of those cubes, it could be any size you like. Once we’d clarified that, people started sending me messages explaining their solutions:

It’s a 5x5x5 cube on which you’ve painted four sides, leaving two opposite sides untouched. I think.

I’m not sure why this university student (not a mathematician) was so hesitant about his solution, which was spot on! He went on to explain how he imagined taking off a whole row of cubes as soon as he’d painted a side, as they’re unimportant once they have paint on, and he described the resulting cuboid each time as “roughly a cube itself”, which I rather liked. He also thanked me for providing an excuse to procrastinate (typical student!)

The next answer utilised algebra:

For an nxnxn cube:
Need n^3 > 45 in order to have enough cubes so n >= 4
Need (n-2)^3 < 45 since all the interior ones must be unpainted so n <= 5
If n = 4 then have 19 painted cubes and only one face is 16, so it must be n = 5.

Trial and error gives you two opposite faces unpainted so have block of 5x3x3 unpainted.

This was very concise, and yet having solved it in a similar way myself I followed it easily. I think most adults who are fluent mathematicians would express their solution in this sort of way.

My next solver described a lovely mathematical journey to a solution:

If the big cube is made of n^3 small cubes, then I worked out for each value of ‘faces painted’ how many cubes were left unpainted, in terms of n.
I realised when I got to 4 faces painted that there was more than one way to do it, and went back and did 3 again!
Then looked at all the expressions to see if any would produce 45. All of them had 3 factors. 45 as a product of 3 factors is either 1x5x9 or 3x3x5, so the right one must be n(n-2)^2
So n=5, n^3 = 125, and 4 faces are painted, circling the cube, leaving 2 unadjacent unpainted faces.

The next comment was from someone else who had found all the ways of painting (or leaving blank) faces of a cube. He’d then used a spreadsheet to look for a suitable solution. His comment:

The answer was 4 faces with opposite faces unpainted which was a nice elegant formula n*(n-2)^2 because it’s a block of unpainted cubes through the middle of a ring of painted cubes. This gave a 3x3x5 of 45 unpainted cubes.

I like the way that even after an algebraic and numerical approach, he has visualised the answer and made sense of the algebra in terms of the physical structure of the solution!

Another nice journey to the solution:

Let n be the number of small cubes on each side. The first realisation was that there are (n-2)^3 cubes which are unpaintable because they have no exposed surface.

Then I thought: 45 is not a perfect cube, so you must have painted at least one face. Likewise, you can’t have painted all six faces.

Then I tried to derive a general expression for the number of unpainted cubes for a cube of side n with only one painted face, figuring that I would see if 45 was a solution. This I realised was a cubic equation, and that I would then have to solve one for all the combinations of painting 2, 3, 4 and 5 faces. I thought this would take too long and I’d never get round to my dinner.

Then, I got crafty 😉

n must be at least 4, otherwise there aren’t 45 small cubes at all. n can’t be 6 or more, otherwise there are more than 45 small cubes which are unpaintable. So n is 4 or 5.

Consider n=4: there are 8 unpaintable cubes, and 56 paintable ones. We need 45 unpainted cubes, so we have to paint 56-(45-8)=19 surface cubes. This is not possible. Anytime you paint a side, you are painting an even number of cubes, even if you’ve already painted some of the cubes when painting another side. No set of even numbers will add to 19. So n must be 5.

For n=5, there are 27 unpaintable cubes, and 98 paintable ones. To satisfy the problem, you need to paint 80 surface cubes, leaving 18 surface cubes unpainted.

(At this point, I saw the solution. But I needed to check it was the only solution.)

We know from above you didn’t paint six faces.
Painting five faces can only leave 9 cubes unpainted, not a solution.
Painting four faces, leaving two unpainted faces neighbouring, leaves 21 cubes unpainted, not a solution.
Painting four faces, leaving two unpainted faces opposite, leaves 18 cubes unpainted, SOLUTION.
Painting three faces or fewer can’t paint enough cubes.

I liked this solution, because after trying an algebraic approach, he realised it would take a long time and instead looked at the number of cases he needed to consider. Then he just went through each case until he found the solution. I was also pleased that I wasn’t the only one to NEED to check there was only one solution!

Finally, a solution without any resort to algebra at all – not an n^3 in sight:

It’s definitely solvable without algebra – here goes:

Interior cubes are cubes that don’t form part of a face on the big cube. So they must all be unpainted. If the big cube had 6 to a side then there would be 64 interior cubes, so that’s too big. If the big cube had 3 to a side then there are only 27 cubes in all, so that’s too small. So your cube must have 4 or 5 to a side.

If there are 4 to a side and 45 unpainted cubes, then there must be 64-45=19 painted ones. Painting one faces gives 16 painted cubes. Painting any other face adds more than 3 to the total, so this is impossible: the big cube must have 5 to a side.

Starting from a fully-painted big cube, there are 27 unpainted cubes. ‘Unpainting’ any face adds 9 to the total, leaving 36. ‘Unpainting’ any adjacent face would add more than 9 to the total, but ‘unpainting’ the opposite face adds exactly 9, leaving 45.

So you originally painted 4 faces of a 5x5x5 cube, leaving a pair of opposite faces unpainted. QED.

So there you have it, some similar and yet different approaches to solving what at first glance seems to be a simple problem. This post is already way too long, so I’ll save my weekend thoughts about extensions to the problem and further explorations for another time, but for now, suffice it to say that I’m pleasantly surprised by the response I got to posting some maths on Facebook!