AMC 10 · 2010 · #15
Grade 2 logicIn 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?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Four amphibians are each a toad (every statement true) or a frog (every statement false). Brian says he and Mike are different species; Chris says LeRoy is a frog; LeRoy says Chris is a frog; Mike says at least two of the four are toads. Determine how many of the four are frogs.
Givens: A toad's statement is always true; a frog's statement is always false; 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."; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Unknowns: The number of the four amphibians that are frogs
Understand
Restated: Four amphibians are each a toad (every statement true) or a frog (every statement false). Brian says he and Mike are different species; Chris says LeRoy is a frog; LeRoy says Chris is a frog; Mike says at least two of the four are toads. Determine how many of the four are frogs.
Givens: A toad's statement is always true; a frog's statement is always false; 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."; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
Tool #3 (Eliminate Possibilities): each amphibian has only two possible species, so I test a species, check whether it makes the statements consistent, and throw out any choice that forces a contradiction. Tool #16 (Count the Complement): Mike lies about "at least two are toads," so the useful fact is the opposite quantity — "at most one toad" — which caps the whole count. Tool #7 (Identify Subproblems): I settle Mike first, then the Chris-LeRoy pair, then combine those partial results into the final frog count.
Execute — Answer: D
1.OA.D.7 Step 1 Brian's claim forces Mike to be a frog
- Look at Brian both ways.
- If Brian is a toad, his statement is true, so he and Mike really are different species, making Mike a frog.
- If Brian is a frog, his statement is false, so he and Mike are actually the same species — and since Brian is a frog, Mike is a frog too.
- Both cases end the same way.
💡 When both possible species for Brian lead to the same conclusion, that conclusion is locked in.
1.OA.A.1 Step 2 Mike's lie caps the toads at one
- Mike is a frog, so his statement is false.
- He claimed "at least two are toads," and the opposite of "at least two" is "at most one." So among all four amphibians there is at most one toad.
💡 Flipping a false "at least" statement turns it into a hard ceiling on the count.
1.OA.D.7 Step 3 Chris and LeRoy: exactly one is a toad
- Chris and LeRoy accuse each other of being frogs.
- If Chris is a toad, his claim is true, so LeRoy is a frog; then LeRoy's claim "Chris is a frog" is false, which fits Chris being a toad.
- If Chris is a frog, his claim is false, so LeRoy is a toad; then LeRoy truthfully calls Chris a frog.
- Either way one of them is a toad and the other a frog.
💡 Two opposite accusations can't both be true or both be false, so they split into one truth-teller and one liar.
2.OA.A.1 Step 4 Combine the facts and count the frogs
- Step 3 gives exactly one toad among Chris and LeRoy, and Step 2 allows at most one toad in all.
- So that single toad is the only toad, and everyone else is a frog: Brian is a frog, Mike is a frog (Step 1), and whichever of Chris and LeRoy is not the toad is a frog.
- That is $3$ frogs and $1$ toad, so the answer is (D).
💡 One guaranteed toad plus a ceiling of one toad means every remaining amphibian must be a frog.
1.OA.D.7 Look at Brian both ways. If Brian is a toad, his statement is true, so he and Mi 1.OA.A.1 Mike is a frog, so his statement is false. He claimed "at least two are toads," 1.OA.D.7 Chris and LeRoy accuse each other of being frogs. If Chris is a toad, his claim 2.OA.A.1 Step 3 gives exactly one toad among Chris and LeRoy, and Step 2 allows at most o Review
Reasonableness: Test the full assignment Brian frog, Mike frog, Chris toad, LeRoy frog. Brian (frog) says "Mike and I are different" — false, and both are frogs, so same species: correct lie. Chris (toad) says "LeRoy is a frog" — true. LeRoy (frog) says "Chris is a frog" — false, and Chris is a toad: correct lie. Mike (frog) says "at least two are toads" — false, and there is only one toad. Every statement is consistent, confirming $3$ frogs. The mirror case (Chris frog, LeRoy toad) works identically, and any count other than $3$ would need a different number of toads, which the two constraints forbid.
Alternative: Argue by contradiction on Brian. Suppose Brian is a toad. Then Mike is a frog (Step 1), so at most one toad exists (Step 2). But Brian would be a toad and one of Chris/LeRoy is also a toad (Step 3) — that is two toads, contradicting the ceiling. So Brian must be a frog, matching Mike, and the final tally is again $3$ frogs.
CCSS standards used (min grade 2)
1.OA.D.7Understand the meaning of the equal sign, and determine if statements are true or false (Deciding whether each amphibian's statement is true or false based on whether the speaker is a toad or a frog.)1.OA.A.1Solve addition and subtraction word problems within 20 (Turning Mike's false "at least two toads" into the count limit "at most one toad.")2.OA.A.1Solve one- and two-step word problems using addition and subtraction within 100 (Combining the one guaranteed toad and the at-most-one-toad ceiling to count the frogs as 4 minus 1.)
⭐ When both guesses for one speaker lead to the same result, that result is certain — then a liar's "at least" claim flips into a firm ceiling that pins down the rest.
⭐ When both guesses for one speaker lead to the same result, that result is certain — then a liar's "at least" claim flips into a firm ceiling that pins down the rest.
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