Difference between revisions of "2010 AMC 10A Problems/Problem 15"
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Now suppose Mike is a toad. Then what he says is true because we already have <math>2</math> 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. | Now suppose Mike is a toad. Then what he says is true because we already have <math>2</math> 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 <math>1</math> toad. Since either Chris or LeRoy is already a toad, | + | Therefore, Mike must be a frog. His statement must be false, which means that there is at most <math>1</math> toad. Since either Chris or LeRoy is already a toad, Brian 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 <math>3</math> frogs total. <math>\boxed{\textbf{(D)}}</math> | Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have <math>3</math> frogs total. <math>\boxed{\textbf{(D)}}</math> | ||
Revision as of 17:50, 25 November 2015
Contents
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?
Solution 1
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 
frog out of the two of them.
Now suppose Mike is a toad. Then what he says is true because we already have 
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 toad. Since either Chris or LeRoy is already a toad, Brian 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 
frogs total. 
Solution 2
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 three frogs.
See also
| 2010 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
