Problem:
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "LeRoy is a frog."
LeRoy: "Chris is a frog."
Mike: "Of the four of us, at least two are toads."
How many of these amphibians are frogs?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
LeRoy and Chris cannot both be frogs, because their statements would be true and frogs lie. Also LeRoy and Chris cannot both be toads, because then their statements would be false, and toads tell the truth. Hence between LeRoy and Chris, exactly one must be a toad.
If Brian is a toad, then Mike must be a frog, but this is a contradiction as Mike's statement would then be true. Hence Brian is a frog, so Brian's statement must be false, and Mike must be a frog. Altogether there are frogs: Brian, Mike, and either LeRoy or Chris.
The problems on this page are the property of the MAA's American Mathematics Competitions