- There's a monastery that is inhabited by Monks.
- They are sworn not to communicate with each other in any way.
- They have no mirrors, and no other means by which a monk can see his own face.
- They see each other only once each day when they all gather together for afternoon prayers.
- One day the head monk shows up to inform them that at least one of them is possessed by Satan. (The head monk can talk, obviously.)
- They are informed that if one is possessed it will be indicated with an X on the forehead.
- A possessed monk must leave the monastery.
- Day 1 goes by, no one leaves. Day 2 arrives, no one has left. Day 3, all have left.
Question: How many monks inhabited the monastery?
(Apparently the possession occurred after a visit from the Monk pictured.)
Update: the solution
________
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4 comments:
Ok, I'll have a guess at it.
Three monks, two possessed.
Each of the possessed monks sees one possessed monk and a clean monk and assumes that the possessed one is the only possessed one.
The clean one assumes he is clean because he sees two possessed monks.
On day two the two possessed monks see that the other possessed monk has not left and figure that the only reason he wouldn't leave is if he sees a possessed monk. This leads each of them to (correctly) assume that they are possessed and they each decide to leave.
The clean monk on day two sees both the possessed monks and assumes that both the others are also seeing two possessed monks which is why they haven't left. Since this would have to mean that he is possessed, he decides to leave.
The number of monks (three) is right. But your explanation is slightly off. With your version the possessed monks would leave on day two, and the “clean” monk would therefore, not leave on day three.
Your version:
X = possessed, Y=clean
Three monks, A(X), B(X), and C(Y).
Day one:
A sees B & C. He doesn’t leave after prayers as he reasons that B is definitely possessed, and hopes that B is the only one. (Ie, that A and C are not possessed.)
B sees A and C. He also doesn’t leave for the same reason as A.
C sees A and B so he doesn’t leave as they are both possessed.
Day two:
They all see they are all still there.
A reasons that B mustn’t have seen two Ys [A and C] as if B had seen two Ys, he would know he was an X and leave.
A realises that B has seen an X on either himself [A] or on C.
Given A knows that C doesn’t have and X, A knows the he must have an X.
B is in the exact same position as A.
A and B would leave on day two, straight after prayers.
C would not leave as he would know that both A and B are possessed, and he may or may not be.
If C isn’t possessed, he knows that A and B would be in the situation described above. If C is possessed, he knows that both A and B would be looking at two possessed monks (as he is). So they would not leave on day two.
That’s all I’ll say for now. A pretty big hint to solving (and explaining) it fully. (I find drawing a diagram helps solve these ones.)
Spoken like a true antidisestablishmentarianist, i.e. Iconoclast.
Working on the complete solution ATM. Will post soon. Meanwhile, here's a similar one at maggie's farm. Thanks to it I could actually solve the satanic monk one. Here's the answer to the maggie's farm brainteaser.
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