[SheetWise]

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