A little history - I read about a blue-spotted monk version of this problem on another site. A "better" solution came to me, which I posted on my blog (theSilentKnight.info). A couple of people responded and made reference to this site. It turns out my first answer was flawed, but the commenters gave me the insight to find a better answer. In my new answer (Plan B), I find a way so that the monks don't have to go all the way back to zero to start their logic count-up by using the concepts of modular arithmetic. Specifically, whatever number of blue spots you see, start counting not at 0, but at the last multiple 6. So if you saw 10 blue-spotted monks, you would start counting at 6 on Day 1. The possible person seeing 9 blue spots would start at 6, and the possible monks seeing 11 monks would start at six. All the other potential sightings aren't possible. Since everybody started counting at 6, the logic count-up may proceed as usual, and at the appropriate time, all spotted monks will leave the island. Other details are in the comments that follow. Since I have been wrong before, I wouldn't mind if somebody could again point out any error in my ways. What is wrong with Plan B? Thank you.
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Instead of trying to find a better solution, why not just attempt to find a proof that there is no better solution?
The information given on Day 0 is "There is a monk with a blue spot". Suppose there is a monk that sees no blue spots. Then he knows he has a blue spot, and he can act. However, if a monk sees $n > 0$ blue spots, he needs additional information. How can he get this additional information? By waiting until something happens (or not) that actually depends on the blue spot he has.
If there are $n$ blue spots in total, nothing will happen before day $n$.
Proof by induction over $n$:
$n = 1$: The single spotted monk immediately leaves on the first day.
$n-1 \Rightarrow n$: There are $n > 1$ monks with spots. A spotted monk sees $n-1$ spots. He knows that if he does not have a spot, nothing will happen until day $n-1$, by induction hypothesis, and he also knows that if he has a spot, nothing at all will happen before he himself leaves, since all other spotted monks are in the exact same situation as him.
He can therefore act on day $n$ at earliest, since before day $n-1$ he receives no new information whatsoever.
An unspotted monk won't act anyway, in particular not before day $n$. $~~~~\square$