2010 AMC 12A Problems/Problem 12

The following problem is from both the 2010 AMC 12A #12 and 2010 AMC 10A #15, so both problems redirect to this page.


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 1

Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.

As Mike is a frog, his statement is false, hence there is at most one toad.

As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.

Hence we must have one toad and $\boxed{\textbf{(D)}\ 3}$ frogs.

Solution 2 (logical reasoning like solution 1, but a different train of thought)

Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.

Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means that none or one of the amphibians is a toad (the opposite of the statement $n\geq2$ is $n<2$, or $n=0, 1$). However, this is absurd because we assumed Brian is a toad, and we know one of Chris and LeRoy is a toad. So our assumption leads to a contradiction.

Hence Brian must be a frog, and he and Mike are the same species. Mike is also a frog. One of Chris and LeRoy is a frog. There are $3$ frogs in total $\Longrightarrow \boxed{\textbf{(D) } 3}$.

~JH. L

Video Solution by the Beauty of Math


See also

2010 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2010 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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