Now suppose Alice is telling the truth, so she has at least 2 trophies. Suppose Bob is telling the truth as well, i.e. he has at least 2 trophies too. So we have B>C>A. Convince yourself that there is no possible combination:
- Suppose Alice had 2 trophies. Then Charlie must have 1 trophy and Bob didn't win any -- incorrect.
- Suppose Alice had 3 trophies. Then no matter what their total could not reach 7 -- again incorrect.
- Suppose Alice had 4 trophies. Who has the other 3? Bob must have 2 trophies since he is telling the truth, so Charlie has only 1 trophy. But that would contradict with Bob's true statement (C>A).
- Suppose Alice had 5 (or more) trophies. Then Bob has at most 1 trophy, but that cannot be since truth-tellers have more than 1 trophy.
So, both Alice and Bob cannot be telling the truth at the same time. Suppose Bob was lying and has only one trophy. But from Alice's (assumed true) statement Bob has more trophies than Charlie and Charlie has none, which is impossible since everyone has at least one trophy.
Therefore, since it is impossible for Bob to be not telling the truth and not lying at the same time, Alice must be lying no matter what!
So, Alice has only 1 trophy. From her false statement, either Charlie won more trophies than Bob or they both won the same number. Recall that they have a total of 7 trophies, so between Charlie and Bob the only possible distributions are 1+5, 2+4 and 3+3. Since everyone has at least 1 trophy, Charlie will always have more trophies than Alice (5, 4, or 3), so Bob must be telling the truth. If Bob is telling the truth, then he has either 2 or 3 trophies. In any case, we know that Charlie must be telling the truth since he has more than 1 trophy.
To summarize, we have conclusively identified that Alice is the liar, and both Bob and Charlie are telling the truth. There are two possible trophy distributions: 1,2,4 and 1,3,3.
Today's takeaway: An educated guess will always be more effective than a pure brute force approach.
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