Cube Calendar Quandary

The Puzzler

RAY: Here's the answer. The hint I gave was to think outside the box. Draw your two cubes as squares or six faces per cube. So you've got 12 faces all together.

We agree that one cube has to have a zero, a one, and a two. The other cube also has to have a zero, a one, and a two.

TOM: Yes, it does.

RAY: And one of the cubes has to have a three. But the other one does not because there's no 33rd of the month. So we're going to put the three on Cube Number One. Cube Number One now has zero, one, two, three. Cube Number Two has a zero, one, and a two. We've used seven faces of the cube.

We still have to put a four, a five, a six, a seven, an eight and a nine, each on a face. That's six more faces but we only have five faces left, and we need six. Can it be done?

TOM: Yes! Because you don't need both a six and a nine!

RAY: You flip the six and it becomes a nine and vice versa! So one cube has a zero, one, two, three, four and a five. And the other has zero, one, two, six, seven and eight. Who's our winner this week?
TOM: The winner is Derward Thompson from Corpus Christi, Texas. And for having his answer selected at random from both of the answers that we got, Derward is going to get a $25 gift certificate to the Shameless Commerce Division. Congratulations Derward!

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