It's a Puzzlement

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Double Squares
by John P. Robertson

Some squares look as if they should be expressed in two halves. For example, 9805² = 96138025 looks like 31²_195² because 31² = 961 and 195² = 38025. Also, 60526² = 3663396676 looks like 1914²_26², with 1914² = 3663396 and 26² = 676.

In this crossnumber puzzle, x² = y²_z² = 1000y² + z². Numbers going across are denoted by capitals, and numbers going down are denoted by lowercase letters. Each cell in the diagram above at the right contains one digit, and there are no zeroes in the completed diagram.

x	y	z
A/2	e/6	E
4F	2b/e	C/4
5a	7d/5	B/3
c	B/3	D/3

This puzzle was created by John Gowland, and is used with permission.

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Card Trick
The last issue's puzzlement was as follows. Amy has seven cards numbered 1, 2, 3, 4, 5, 6, 7. She randomly deals three each to Bill and Celia, keeping one for herself. All three people then look at their cards. Can Bill and Celia communicate with each other, in the presence of Amy, so that Bill and Celia can each determine what cards the other holds, but Amy will not know who holds any given card, other than the one that she herself holds?

I had in mind a very simple, elegant solution, and several readers submitted this solution. But the CAS never ceases to amaze me. Alex Kozmin submitted a solution that is even simpler. The solution that I had in mind was that Bill and Celia each announce the remainder when the sum of their cards is divided by 7. As the sum of all seven cards has a remainder known to all, namely zero, this gives Bill and Celia enough information to deduce Amy's card, and hence the cards held by the other. But, for each possible pair of remainders that Bill and Celia could announce, there are enough ways each pair could arise that Amy cannot determine any card held by either Bill or Celia. For instance, if Bill announces, "6," and Celia announces, "1," then the possible cards for Bill and Celia are {1, 2, 3} and {4, 5, 6}, or {2, 5, 6} and {1, 3, 4}, or {3, 4, 6} and {1, 2, 5}. Amy cannot determine a single card held by either Bill or Celia.

How could this be made any simpler? Kozmin proposes that Bill announce his remainder, and that Celia then simply announce Amy's card.

Robert Ballmer, Bob Conger, Jon Evans, Chris Noble, Yipei Shen, Ed Shoop, John Stenmark, and David Uhland also submitted solutions. There was one anonymous submission.