[SheetWise]

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