Calendar cubes

A man has two cubes on his desk. every day he arranges both cubes so that the front faces show the current day of the month. what numbers are on the faces of the cubes to allow this?

Solution:

First, to show all possible days, we'd need one of each of the ten digits. We'd also need two 1s and two 2s to show 11 and 22. That's twelve numbers right there. Two cubes, twelve faces, so every face is used. Quite elegant.

We know each cube will need a 1 and a 2. Let's put the 3 on one of them. The 0 has to go on the other. We put 4, 5, and 6 on the 3-cube since we need to show 04 05 06.

But now where do 7, 8, and 9 go? The 0-cube needs them to be on the 3-cube, but it's full. I'm beginning to see why this gets four aha's.

Let's try this: clear both cubes, put the 0 on one of them. Now 1-9 have to go on the other cube to show 01-09. We can't put a 0 on the other cube, too, because that puts us over the 12-digit limit. It seems I have mathematical proof that this cannot be done! What am I missing? It's not like you can turn one of the other numbers sidewise to make another 0...

Heh.

You CAN make a 9 out of a 6, though. That frees up a digit. So you put 0, 1, and 2 on both cubes. Put 3 on one of them. 4 and 5 can go on it too. Put 6, 7, and 8 on the other. Now you can show 01-31 with no problem, and even 00 and 32 if you're feeling weird.

I'd've given it two, maybe three aha's. :)