A problem proposed by Mel Stover for the April 1953 issue of Pi Mu Epsilon Journal:
After a meeting of six professors, each man left with another’s hat. The hat that Aitkins took belonged to the man who took Baily’s hat. The man whose hat was taken by Caldwell took the hat of the man who took Dunlop’s hat. And the man who took Easton’s hat wasn’t the one whose hat was taken by Fort. Who took Aitkins’ hat?
|
SelectClick for Answer |
This solution is by Leon Bankoff. Denote each man by his initial and write x → y to indicate that x’s hat was taken by y. We know that B → x → A, where x is the man who took Baily’s hat, and that D → z → y → C, where z is the man who took Dunlop’s hat and y’s hat was taken by Caldwell. These two sequences comprise seven symbols altogether, so x, y, or z must be duplicating A, B, C, or D in one of them. There are two ways to combine the sequences:
- B → D → A → y → C (where x = D and z = A)
- D → z → B → C → A (where x = C and y = B)
Each of these new sequences has five symbols, so each lacks one. Add s to the first sequence and t to the second to provide this missing link and to make each sequence into an unbroken cycle:
- s → B → D → A → y → C → s
- t → D → z → B → C → A → t
The first cycle fits the problem statement if we let y = F and s = E, and the second matches if we let t = F and z = E. In either case, Fort took Aitkins’ hat.
(“Problem Department,” Pi Mu Epsilon Journal 1:8 [April 1953], 329-330.)
|
