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It is logically possible to answer the questions without any workarounds like
"I try with blue on the first day". The guru will never speak again or
interact in any way, nor will he ever leave (lets say he has fiery red eyes
for arguments sake).


If there were two people on the island of each colour, the two blue-eyed guys
would look at eachother and expect the other one to leave. If the other one
does not leave with the first boat they both know there must be two blue-eyed
persons on the island, and that one of them is themselves. On the second day
they both leave.

If there were three blue-eyed people on the island, they would see the first
boat leave, and everyone would again know the same as last time, i.e. that
there are at least two blue-eyed people on the island. Still, all of them
expect them to be the two that remain.
When nobody leaves on the second day, everyone know that there are at least
three blue-eyed people on the island, and thus the blue-eyed ones now know
that they have to be the ones, because they can see all the brown-eyed people.

With four of each group, they will require one more day. And so on.

Since nobody know what other colours there could possibly be, all the
brown-eyes stay on the island.

So, 100 days.

Apparently the problem is commonly known. Found a better-written solution at
xkcd (same answer though).
http://xkcd.com/solution.html