The Desk Calendar

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.

Screen Shot 2016-08-28 at 8.09.20 AM

 

2 thoughts on “The Desk Calendar

  1. 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).

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