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  #1  
Old 07-25-2005, 03:14 AM
lastchance lastchance is offline
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Default Another logic riddle

So, there are 20 monks. They only see each other three times a day, during breakfast, lunch, and dinner, where they sit around in a big circle facing everyone else.

Otherwise, they're in meditation.

These monks believe that being blue-eyed is a sin, and any blue-eyed monk must leave. They are also very logical, and all think at a very high level.

At breakfast, the monks welcome a stranger. At the beginning of the meal, he says "some of you are blue-eyed."

Two days later, not a blue-eyed monk is to be found at lunch, while every other monk stays.

How many blue-eyed monks are there? And how do they know to leave?
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  #2  
Old 07-25-2005, 03:20 AM
jason_t jason_t is offline
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Default Re: Another logic riddle

First, being blue-eyed is not a sin.

Second, answer: <font color="white">6</font>
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  #3  
Old 07-25-2005, 05:33 AM
Spaded Spaded is offline
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Default Re: Another logic riddle

And we are to assume that they all have oaths of silence?
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Old 07-25-2005, 06:06 AM
flatline flatline is offline
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Default Re: Another logic riddle

I'm going to go with <font color="white">none. The guest lied. If a monk had blue eyes and believed that any blue eyed monks should leave.. he would already have left.</font>
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  #5  
Old 07-25-2005, 06:11 AM
jason_t jason_t is offline
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Default Re: Another logic riddle

[ QUOTE ]
I'm going to go with none. The guest lied. If a monk had blue eyes and believed that any blue eyed monks should leave.. he would already have left.

[/ QUOTE ]

[ QUOTE ]
Another logic riddle

[/ QUOTE ]
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  #6  
Old 07-25-2005, 06:23 AM
flatline flatline is offline
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Default Re: Another logic riddle

[ QUOTE ]
Another logic riddle

[/ QUOTE ]

Wow, I didn't read that part. It must be <font color="white"> 6 because there are six meals and that is oh so very logical .</font>
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  #7  
Old 07-25-2005, 06:37 AM
lastchance lastchance is offline
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Default Re: Another logic riddle

[ QUOTE ]
And we are to assume that they all have oaths of silence?

[/ QUOTE ]
All the monks took an oath not to talk about being blue-eyed or anything having to do with that. They only see each other at meal-time.

The guy was telling the truth. The monks know that.

And, why do the blue-eyed monks know that they are blue-eyed?
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  #8  
Old 07-25-2005, 07:07 AM
BeatOnRiver BeatOnRiver is offline
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Default Re: Another logic riddle

read below but i dunno how to write in white so dont look, it may be wrong im not too good at these logic riddles
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  #9  
Old 07-25-2005, 07:10 AM
BeatOnRiver BeatOnRiver is offline
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Default Re: Another logic riddle

there are 2, the 1st monk looks around and sees only 1 blue eyed monk and knows he must be the other one, the 2nd monk sees this guy leave and doesnt see another monk with blue eyes therefore he realizes he must have been the other. there cannot be less than 2 because the stranger said SOME of you are blue eyed. <font color="white"> </font> <font color="white"> </font> <font color="white"> </font>
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  #10  
Old 07-25-2005, 08:45 AM
SheetWise SheetWise is offline
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Default Re: Another logic riddle

I have to assume the monks don't know their own eye color.
I 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 the stranger was lying.

If there was only 1 BEM, every other monk, seeing only 1 pair of blue eyes, would assume they were the other -- and at the next meal there would be only 1 BEM.

If there were 2 BEMs, each would only see 1 other -- and at the first meal (breakfast day 1) conclude they were the other, and both would leave. None at lunch.

If there were 3 BEMs, each would see 2 others at the next meal (lunch day 1), and conclude the others were seeing more than 1 (or they would have left). That must be them. All 3 leave. None at dinner.

If there were 4 BEMs, each would see 3 others (dinner day 1). So, why didn't all three leave? There's only one explanation. They all see more than 2. There must be 4. That must be them. All four leave. None at breakfast day 2.

If there were 5 BEMs, each would see four others (breakfast day 2). Why didn't they leave? They must each also see 5. That must be them. All 5 leave. None at lunch day 2.

Etc.

6 = Lunch day 2, none at dinner, day 2.
7 = Dinner day 2, none at breakfast day 3.
8 = Breakfast day 3, none at lunch.

So the question is, how many times must they meet before they come to this conclusion? There are six meals between the breakfast with the stranger and lunch two days later.

The answer is that there may have been 2, 3, 4, 5, 6, 7, or 8 BEMs.

Since the problem states -
[ QUOTE ]
Two days later, not a blue-eyed monk is to be found at lunch, while every other monk stays.

[/ QUOTE ]

It doesn't state that it is the first time this pattern occurred.

-SheetWise
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