A puzzle by Edward J. Barbeau, from the February 2007 issue of Crux Mathematicorum:
A certain familiar island is inhabited by knights, who can only speak the truth, and knaves, who can only lie. One day a visitor meets three inhabitants, A, B, and C. The visitor asked, “How many knights are there among you three?”
A gave an answer, which the visitor didn’t hear. When the visitor asked B what A had said, B replied, “A said that there is one knight among us.” At this C said, “Don’t believe B. He is lying.”
This solution is by Mandy Rodgers and Josh Trejo. C accuses B of lying, so they can’t both be knights or both knaves. Assume first that B is a knight and C is a knave. That would mean that A really did tell the visitor that there was one knight among the three islanders. But if A’s statement were true then both A and B would be knights, which is a contradiction; and if A’s statement were false then B really would be the only knight, which again is a contradiction. So B can’t be a knight. That means that B is a knave, which makes C a knight.
Note that we don’t know whether A is a knight or a knave. It’s possible that he’s a knight and said that the trio includes two knights, which would be true (in this case A/B/C are knight/knave/knight). And it’s possible that he’s a knave and said that all three were knights or all knaves, both of which statements we know are false. These possibilities are both self-consistent, so we can’t say anything about A’s identity, only that B lies and C tells the truth.
