Solution to Puzzle #199: Two-Cube Calendar

This puzzle was originally posted in April 1969 by Martin Gardner, and is a classic even today. Thanks for an overwhelming response to the puzzle and my apologies for the delayed post as I was traveling on work outside the country over the last two weeks.

Many people sent correct answers – this includes Suman Saraf, Kumz91 (sorry do not know the full name), Naresh Kumra, Mohit Rana, Delhi Scrabble (again, do not know the real name), Gaurav Sharma, Viv423 (do not know the name again), Rishikant, G, Rowan an Pratik Poddar. And sorry, if I have missed out anyone. Well done all.

Please note that each cube must have 0, 1 and 2. This leaves only six faces for the 7 remaining digits. Trick is to realize that 6 and 9 can be on the same face and the cube can be turned around. On the right one, therefore, the three hidden faces will be 0, 1 and 2. On the left cube, one can see 1 and 2. The others will be 0, 6/9, 7 and 8.

Hope you all enjoyed the puzzle!

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