2010 AMC 12A Problems/Problem 12

Problem 12

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?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution

We can begin by first looking at Chris and LeRoy.

Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.

Clearly, Chris and LeRoy are different species, and so we have at least $1$ frog out of the two of them.

Now suppose Mike is a toad. Then what he says is true because we already have $2$ toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.

Therefore, Mike must be a frog. His statement must be false, which means that there is at most $1$ toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.

Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have $3$ frogs total. $\boxed{\textbf{(D)}}$

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2010 AMC 12A Problems/Problem 12

Problem 12

Solution