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Can we assume all three integers are different?
Then . If not, I don't see how to solve it.
You don't have to assume all three integers are different. There is a version of this problem where it is assumed all three integers are different, in which case @Fiat_Revelation's solution would be correct.
However, I don't think that requirement is necessary. There is another logical path that doesn't require the integers to be different. I hope that I'm correct in my assessment!
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The plot thickens~~ I had the same reasoning as @Fiat_Revelation. I think I got it then. Following the same reasoning as outlined by @Fiat_Revelation (but cutting short on the different number assumption), we now need to argue on the relationships being used for each case, namely:
and
If , then Alfred would have known his answer, because only equation (3) gives a valid combination of numbers for both Robin and Alfred. The other valid combination is obtained by Equation (2) and (1), respectively, for Robin and Alfred. However, he did not know the answer, in which case must and that's what Batman says on his second round.
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I'm still not sure if this logic works out. If it does then ultimately Batman can't be sure because Alfred knows his number if Batman is 10 also! Which would mean the problem is either bad or Alfred is not a trained logician
In either case, 500 sats for elucidating the hint :)
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I'm not sure I followed your answer, but in the bottom right box shouldn't it be A -> 70? Not sure if that changes your logic.
Since it seems like few people will be attempting this further, I will give another hint.
Suppose the configuration is (10, 20, 30).
When it gets to Alfred, he knows he's either 10 or 30.
But if he's 10, what would Robin have seen during his first round?
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Yes, 70, not 80. Doesn't change my logic though. I'll give it some more thought later using your hint, and see if it agrees or disagrees with my logic.
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Lmao, I've been laying in bed trying to work this out in my head, but at this point I'm throwing in the towel.
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It's a tough one!
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