[Guest]

A number of wooden cubes are to be painted with 6 given colours and each face of a

given cube is to be painted a different colour. Every two cubes are distinguishable from

one another by their colour schemes, no matter how they are placed. How large can the

number of cubes be?

Thank you for your help!

[CaptainEd]

Thank you, heyozzy, for the puzzle.

assuming colors 1,2,3,4,5,6

Look at the face that has color 1.

There are 5 ways the opposite face can be colored. Call it color 2 for the moment (although it could be 2,3,4,5, or 6)

For each of them, consider the sequence of the colors of the other faces as a ring, read out in clockwise order (that is, starting somewhere in the sequence North, East, South, West).

Read them out beginning with 3 (or 2 if 3 is the opposite color)

The only sequences are

3 4 5 6

3 4 6 5

3 5 4 6

3 5 6 4

3 6 4 5

3 6 5 4

So, there are 5 ways to pick the opposite face, and 6 clockwise sequences, so there are 30 colorings, assuming rotational differences do not matter, but handedness does.

[Guest]

Thank you Captain Ed! But wouldn't your technique also have repeated patterns of cubes every once in a while?

[CaptainEd]

I think you're wondering whether the correct number is actually smaller than 30. I can be educated, please show me a pattern that I counted twice.

[Molly Mae]

Since there's no orientation, two cubes you identified that are the same are the ones below, assuming 1 and 2 are opposite sides.

3 4 5 6

3 6 5 4

because there is a way to orient both cubes to become identical....

Actually...no...you're right. They aren't the same.

I just painted the cubes. =P God bless practical application.

3 4 5 6 would be the same as 4 5 6 3 or 5 6 3 4 or 6 3 4 5, but not the same as 3 6 5 4.

A winner is you, CaptainEd.

[Guest]

No CaptainEd I definitely won't be educating you today. I was just wondering what happens when you have color number 2 facing you and color 1 on the reverse...and wouldn't you then have repetition of what you did when you had color 1 facing you and color 2 on the reverse. As I understand your solution a little better I see that your method does not include any possible repetition. Thanks again Captain!

[witzar]

Let's name colors we use: a, b, c, d, e, f.

We can always rotate painted cube so that face a is at bottom. There are 2 cases:

1) Face b is at top.

Now we can rotate cube to have face c in front. Remaining three faces are painted with d, e, f, and each out of 3! cases is possible

2) Face b is not at top.

Now we can rotate the cube to have face b in front. Remaining four faces are painted with c, d, e, f and each out of 4! cases is possible.

So the answer is 3! + 4! = 6 + 24 = 30.

Edited October 6, 2011 by witzar

[Guest]

Thanks Witzar!

[CaptainEd]

Molly Mae, I know what you mean about the value of concreteness. I don't know how many times I've started typing my answer into this box and realized it was dead wrong...Almost as many as the ones where I discovered it after I hit "Post"...

[Guest]

The answer is ten. I can easily point out the double counting made by CaptainEd.

Let face 1 correspond to a particular color. Now the no of ways of painting the opposite face i.e. face 2 is 5. True. But now the no. of ways of painting the remaining four sides such that now all cubes are distinguishable from any angle are :-

3,4,5,6 & (on exchanging opp sides) 3,6,5,4.

which is 2 not 3!.

[all the other four egs. u've mentioned will have an ordered pair of adjacent sides simillar to these two and hence will be simillar to these two cubes w.r.t. some corner.]

For eg. in the examples You've listed:-

3 4 5 6 & 3 4 6 5, will not be distinguishable when you look at the two cubes from the corner corresponding to the faces 2 3 4 or even 3 4 1. hence the double counting.

If i have interpreted the question correctly the answer is definitely 5 X 2 = 10.

Edited October 6, 2011 by Kev Dan

[Molly Mae]

The answer is ten. I can easily point out the double counting made by CaptainEd.

Let face 1 correspond to a particular color. Now the no of ways of painting the opposite face i.e. face 2 is 5. True. But now the no. of ways of painting the remaining four sides such that now all cubes are distinguishable from any angle are :-

3,4,5,6 & (on exchanging opp sides) 3,6,5,4.

which is 2 not 3!.

[all the other four egs. u've mentioned will have an ordered pair of adjacent sides simillar to these two and hence will be simillar to these two cubes w.r.t. some corner.]

For eg. in the examples You've listed:-

3 4 5 6 & 3 4 6 5, will not be distinguishable when you look at the two cubes from the corner corresponding to the faces 2 3 4 or even 3 4 1. hence the double counting.

If i have interpreted the question correctly the answer is definitely 5 X 2 = 10.

I made a similar mistake. While your assumption [[3 4 5 6 & 3 4 6 5, will not be distinguishable when you look at the two cubes from the corner corresponding to the faces 2 3 4 or even 3 4 1. hence the double counting]] is correct, that doesn't mean that the two cubes are indistinguishable in whole. They are very different if you look at the whole cube, not just half of it from a corner.

[Guest]

If i have interpreted the question correctly the answer is definitely 5 X 2 = 10.

Kev Dan is correct if his reading of the question is correct. The key point here is "Every two cubes are distinguishable from

one another by their colour schemes, no matter how they are placed". If you want the cubes to be completely unique then CaptainEd is correct. If you judge the cubes from only one perspective then you will only ever see 3 sides at one time and for most of CaptainEd's cubes the distinguishing colors would not be visible.

[Molly Mae]

Kev Dan is correct if his reading of the question is correct. The key point here is "Every two cubes are distinguishable from

one another by their colour schemes, no matter how they are placed". If you want the cubes to be completely unique then CaptainEd is correct. If you judge the cubes from only one perspective then you will only ever see 3 sides at one time and for most of CaptainEd's cubes the distinguishing colors would not be visible.

In that case, you can only have 1. If you use any two like colours on two cubes, you could set each cube facing directly toward an observer. Regardless of all other sides, they look the same. I assume the OP means that painting two cubes the same and then rotating one of them any direction still counts as the same cube colour scheme.