Another logic riddle - Page 2

slickpoppa · #11 07-25-2005, 09:03 AM

The answer is:

 7. At the first breakfast, if there is only one BEM, then he will leave because he will look around and not see any BEMs and will know that he must be the one.

If there were two BEMs they will both leave at the first lunch. Each BEM will only see one BEM, but they will know that there must be two because if there were only one he would have left at breakfast.

And so on until the 7th meal when 7 BEMs leave

SheetWise · #12 07-25-2005, 09:14 AM

[ QUOTE ]
7. At the first breakfast, if there is only one BEM, then he will leave because he will look around and not see any BEMs and will know that he must be the one.

[/ QUOTE ]

Can't be. The stranger said "some..." - if there is only one BEM, every other monk will assume they are the other and leave. The one BEM will assume the stranger was lying when he said "some of you are blue eyed". The next meal will have only one person -- the BEM.

gumpzilla · #13 07-25-2005, 09:33 AM

This is a very standard kind of riddle, which is probably most of the reason that jason_t answered quickly and with no explanation.

Here's something you might enjoy thinking about: suppose in your community of monks there are at least two blue-eyed monks. What information is the stranger putting into the system, then? It's clear that all of the monks already know that there are some blue-eyed monks.

SheetWise · #14 07-25-2005, 10:01 AM

[ QUOTE ]
suppose in your community of monks there are at least two blue-eyed monks. What information is the stranger putting into the system, then? It's clear that all of the monks already know that there are some blue-eyed monks.

[/ QUOTE ]

I don't think it's clear that all of the monks already know that there are some blue-eyed monks. Two of them know that there is at least one, the rest know that there are at least two.

I had to look up "some". Could this be (one of) my fatal flaw(s)? -

[ QUOTE ]
adj 1: quantifier; used with either mass nouns or plural count nouns to indicate an unspecified number or quantity; "have some milk"; "some roses were still blooming"; "having some friends over"; "some apples"; "some paper"

[/ QUOTE ]

adj 1: quantifier; used with either mass nouns or plural count nouns to indicate an unspecified number or quantity; "have some milk"; "some roses were still blooming"; "having some friends over"; "some apples "; "some paper"

gumpzilla · #15 07-25-2005, 10:13 AM

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I don't think it's clear that all of the monks already know that there are some blue-eyed monks. Two of them know that there is at least one, the rest know that there are at least two.

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Nobody here means some in the context you're taking it; take some to mean at least one. Alternatively, assume you have 3+ monks with blue eyes and then even your usage is satisfied.

theTourne · #16 07-25-2005, 10:26 AM

Answer:
 7. If there was only 1 he'd be gone at lunch, if 2 they'd be gone at dinner, etc.

durron597 · #17 07-25-2005, 11:01 AM

The stranger is clearly lying, because the monks would already know if they had blue eyes by looking in a mirror, a window, or a pool of water. Also even if there were no reflective surfaces the monks surely would told the other monks that they have blue eyes unless there is a vow of silence, which you also did not specify. If neither of these things are true, then there are (answer in white) 7 monks by the principle of mathematical induction.

SheetWise · #18 07-25-2005, 11:05 AM

Ooops.

On the revised assumption that some can mean one. [img]/images/graemlins/wink.gif[/img]
I still can't assume the stranger was telling the truth.

If there was no blue-eyed-monk (BEM), every other monk, seeing no pair of blue eyes, would assume it was them and leave. Nobody at lunch.

If there was only 1 BEM, he would realize who he was at breakfast, and not show up at lunch. The others, realizing there were some, and some saw no others, stay. None at lunch day 1.

If there were 2 BEMs, each would only see 1 other at breakfast -- and wait for them to leave before lunch, At lunch day 1, seeing the other return, conclude there was another, it was them, and both would leave. None at dinner day 1.

If there were 3 BEMs, each would see 2 others at breakfast, and wait for them to show up to lunch day 1, see the other return, realize they were 1 of 2, and leave before dinner day 1. When they show up at dinner day 1, conclude the two were seeing more than 1 other, Conclude it was them, and all 3 leave. None at breakfast day 2.

Etc.

4 = None at lunch, day 2.
5 = None at dinner day 2.
6 = None at breakfast day 3.
7 = None at lunch day 3.

So there may have been 2, 3, 4, 5, 6, or 7 BEMs, again, since the problem doesn't state lunch day 3 is the first time the pattern occurred.

Where did I screw up this time? [img]/images/graemlins/wink.gif[/img]

-SheetWise

SheetWise · #19 07-25-2005, 11:12 AM

Correction. And .... the revised option of 1.

#20 07-27-2005, 02:31 AM

The only thing that make sense is: 18 of the monks are blind. Thats the only thing which fit with the statemaent: a stranger says some are BEMs. the starnger comes with new info, the alarm goes off and two monks that are not blind will each know that they both are blue-eyed because the stranger said "some"

Alternative: 19 are blind. last one is blue-eyed. The word "some" is not taken into consideration.