Competition problem — Combinatorics

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Combinatorics Difficulty 4.9 AIME Prove it Philippines

Problem:

Alice, Bob, Charlie and Eve are having a conversation. Each of them knows who are honest and who are liars. The conversation goes as follows:
Alice: Both Eve and Bob are liars.
Bob: Charlie is a liar.
Charlie: Alice is a liar.
Eve: Bob is a liar.

Who is/are honest?

Solution

Solution:

We consider two cases:

Case 1: Alice is honest.
If Alice is honest, both Eve and Bob must be liars. If Eve is a liar, then Bob must be honest. This cannot be the case.

Case 2: Alice is a liar.
If Alice is liar, then either Eve is honest or Bob is honest. Suppose Eve is honest. Then, Bob is a liar. If Bob is a liar, Charlie must be honest, and Alice is a liar. This is a possible case.
Suppose Eve is a liar. Then, Bob is honest. Since Bob is honest, Charlie is a liar, and Alice is honest. This cannot also be the case.

Thus the only possible case is that Alice is a liar, Bob is a liar, Charlie is honest and Eve is honest.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.