Solution to Puzzle #118: Rotating Disk

This was a slightly difficult puzzle, and I got relatively few correct solutions. From Alok’s Mathematical Circles class, I know that several people were able to get the answer, that included Smiti Mittal (had a very innovative and simple answer) and Anisha Goyal. Other people to send the correct answer included Suman Saraf (complete answer), Chandu Karan and Umesh Mayekar. Well done all!

I am reproducing multiple answers, the one from Alok’s class I found to be the most complete is the following:

A,B,C,D,E,F have 15,14,13,10,11,12 respectively

Opposite sitting players total at end of 5 games must be 25 for each pair. So A must have at least 14 (because someone in other pairs must have at least 13). This means D can have at most 11, and he already has 9 in first two rounds. So D must have a zero in some round. Since order of rounds is not important, lets say D has a zero in 3rd round. On last two rounds, D’s possibilities (given he can only score 2 more points) are 00, 01, 11, or 02. We can test each of these in turn.

One from Smiti, which I also liked very much ( though I am not able to conclude from this answer that this is the only possibility):

Smiti had a great solution. Once it is clear that D must have 5 in the second round, she just flipped the scores for 3rd and 4th rounds relative to 1st and 2nd round. This brought everyone to a score of 10. Then simply get A a 5 in the last round

Hope you all enjoyed the puzzle!

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