I came across a problem some time ago in Martin Gardner’s* The Colossal Book of Short Puzzles and Problems. *It’s nice because, well, you’ll see. Anyway, give it a go, and I’ll post the solution in the next few days.

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# The Desk Calendar

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2 thoughts on “The Desk Calendar”

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I came across a problem some time ago in Martin Gardner’s* The Colossal Book of Short Puzzles and Problems. *It’s nice because, well, you’ll see. Anyway, give it a go, and I’ll post the solution in the next few days.

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Interesting, it seems that there is a unique way of doing this:

* Since 11 and 22 must be present, we know that the right cube also needs a 1 and 2.

* It cannot be the case that only one cube has a 0, as otherwise we can then only make six numbers from 01,…,09. So both cubes contain a 0.

* So we now know the right cube, it is {0,1,2,3,4,5}. We argued that the left cube is of the form {0,1,2,x,y,z}… but we are missing FOUR numbers: 6,7,8,9…so it seems impossible???… ahh… 6 can be obtained from 9 by flipping (that last step made me lol).

It’s a good one hey!? Nice work!