2010 AMC 10A Problem 15

P11-15~4 minPrint

Logic Puzzles Split into exhaustive cases and handle each

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?

Show solution without answering

Key insight

Chris and LeRoy call each other frogs, so exactly one is a toad. Nobody could say Brian's sentence if Mike were a toad, so Mike is a frog.

Solution

Toads always tell the truth and frogs always lie. Look for statements that pin each other down.

Step 1: Chris and LeRoy. They each say the other is a frog.

If Chris is a toad, he tells the truth, so LeRoy is a frog. LeRoy then lies when he says "Chris is a frog." That fits.
If Chris is a frog, he lies, so LeRoy is actually a toad. LeRoy then truthfully says "Chris is a frog." That fits too.

Either way, exactly one of Chris and LeRoy is a toad, and the other is a frog. We do not need to know which.

Step 2: Mike. Suppose Mike were a toad. Test Brian both ways:

If Brian is a toad, he tells the truth. But he and Mike would both be toads, so "Mike and I are different species" is false. A toad cannot say something false. Contradiction.
If Brian is a frog, he lies. But he and Mike really would be different, so his statement is true. A frog cannot say something true. Contradiction.

Neither works, so Mike is a frog.

Step 3: Brian. Mike is a frog, so his statement "at least two of us are toads" is false. That means at most one of the four is a toad. Chris or LeRoy is already that one toad, so Brian is a frog.

Check Brian. Brian and Mike are both frogs, so "Mike and I are different species" is false. A frog saying something false fits.

Count. Frogs: Brian, Mike, and one of Chris or LeRoy. That is $3$ frogs.

The answer is $\boxed{\textbf{(D)}\ 3}$ .

Why this works

Two speakers who each call the other a liar are always one truth-teller and one liar. That gives you a count even before you know who is who. For everything else, try each possibility and throw out any that leads to a contradiction. Brian's sentence "Mike and I are different" is impossible for anyone to say if Mike tells the truth, which settles Mike at once.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Chris and LeRoy. If Chris is a toad, his statement is true, so LeRoy is a frog; LeRoy's claim "Chris is a frog" is then false, as a frog's must be. If Chris is a frog, his statement is false, so LeRoy is a toad, and LeRoy's true claim matches. Either way, exactly one of Chris and LeRoy is a toad.

Mike. Suppose Mike is a toad. Then consider Brian. If Brian is a toad, he and Mike are the same species, so his statement is false, contradiction. If Brian is a frog, he and Mike really are different species, so his statement is true, again a contradiction. Hence Mike is a frog.

Consequences. Mike's statement is false, so fewer than two are toads. One toad already sits among Chris and LeRoy, so Brian must be a frog. Check Brian: he and Mike are both frogs, so "we are different species" is false, as required.

The frogs are Brian, Mike, and one of Chris or LeRoy: three frogs.

The answer is $\boxed{\textbf{(D)}\ 3}$ .

Review in the AoPS books

Intermediate Counting & Probability, Chapter 2: Sets and Logic (§2.4)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Prealgebra, Chapter 14: Counting (§14.3)