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Difference between revisions of "2010 AMC 12A Problems/Problem 12"

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{{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #12]] and [[2010 AMC 10A Problems|2010 AMC 10A #15]]}}
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== Problem ==
 
== 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.
 
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.
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<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
  
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== Solution ==
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== 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.
 
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.
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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.
 
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.
  
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Hence we must have one toad and '''\boxed{C) <math>3</math>}''' frogs.
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Hence we must have one toad and <math>\boxed{\textbf{(D)}\ 3}</math> frogs.
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== Solution 2 (logical reasoning like solution 1, but a different train of thought)==
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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.
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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 <math>n\geq2</math> is <math>n<2</math>, or <math>n=0, 1</math>). 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.
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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 <math>3</math> frogs in total <math>\Longrightarrow \boxed{\textbf{(D) } 3}</math>.
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~JH. L
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Solution 3: Casework/Proof by contradiction
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First, we assume that Brian is a toad. Since his statements are truthful, Mike could not be a toad and must be a frog. Since frogs are liars, Brian must be the only toad in the pond, so Chris and Leroy are both frogs. However, they would then be telling the truth about each other, which only toads do. Hence, this case results in a contraction.
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Hence, Brian is a frog. He is lying, so Mike must also be a frog. Mike's statement is also false, so Chris and Leroy cannot both be toads. As shown in the previous case, them both being frogs is impossible, so one must be a toad and the other a frog. Therefore, there must be a total of 3 frogs.
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By-Monkey_King
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==Video Solution by the Beauty of Math==
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https://youtu.be/kU70k1-ONgM?t=1207
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=11|num-a=13|ab=A}}
 
{{AMC12 box|year=2010|num-b=11|num-a=13|ab=A}}
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{{AMC10 box|year=2010|num-b=14|num-a=16|ab=A}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:40, 28 August 2024

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

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?

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

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

Solution 3: Casework/Proof by contradiction First, we assume that Brian is a toad. Since his statements are truthful, Mike could not be a toad and must be a frog. Since frogs are liars, Brian must be the only toad in the pond, so Chris and Leroy are both frogs. However, they would then be telling the truth about each other, which only toads do. Hence, this case results in a contraction. Hence, Brian is a frog. He is lying, so Mike must also be a frog. Mike's statement is also false, so Chris and Leroy cannot both be toads. As shown in the previous case, them both being frogs is impossible, so one must be a toad and the other a frog. Therefore, there must be a total of 3 frogs. By-Monkey_King

Video Solution by the Beauty of Math

https://youtu.be/kU70k1-ONgM?t=1207

See also

2010 AMC 12A (Problems • Answer Key • Resources)
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 (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

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