Gordon B. Chamberlain CdeP 1956 is this issue's winner.
The year is 1739. Three married couples, all savvy horsemen and horsewomen, go to market and each buy some horses. Each buys as many horses as he or she pays in shillings for one horse. Each husband spends altogether 3 more guineas than his wife. The women are Elouise, Katyana, and Celeste; the men are Harold, Chauncey, and Philip. Harold buys 23 more horses than Katya, while Chauncey buys 11 more horses than Elouise. What is the name of each man’s wife?
The Winning Response
To the Editor(s): In the Spring 2017 puzzle (page 9), which arrived in the mail today, (1) Philip is married to "Elouise" (my generation not only spelled her name right but knew her location at the Plaza). He buys 8 horses @ 8 s., she one @ 1 s.; (8 x 8) - (1 x 1) = 64 - 1 = 63 shillings = 3 guineas @ 21 shillings. (Like the puzzlemaker, I ignore the fact that in 1739, guineas were gold and shillings silver, so that the ratio varied). (2) Chauncey (probably not an accepted Christian name in 1739) is married to Katya (who apparently has crossed time and space to drop in from War and Peace). He buys 12 horses @ 12 s., she 9 @ 9 s.; (12 x 12) - (9 x 9) = 144 - 81 = 63 shillings again. (3) Harold is married to Celeste (sounds like a French maid in a thriller). He buys 32 horses @ 32 s., she 31 @ 31 s.; (32 x 32) - (31 x 31) = 1,024 - 961 = 63 s. Harold's 32 horses are 23 more than Katya's 9; Chauncey's 12 are 11 more than "Elouise's" one. Rah, rah, Casa Piedra! --Gordon B. Chamberlain CdeP 1956.
Once again, thanks go to Kurt Meyer for supplying and adjudicating these confounding conundrums.