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Here is a variant on the blue-eyed puzzle. I hope it brings at least some more difficulty than the original one, even if you know the solution to the original.

The Guru makes the following statement:

Of the following two statements, exactly one is true:

  1. The number of islanders with blue eyes is prime.

  2. The number of islanders with blue eyes greater than or equal to 25.

It is common knowledge that this statement is true.

Everything else is left equal, so on the island, there are 100 blue-eyed people, 100 brown-eyed people, and the Guru. Of course, the question is: Who leaves the island, and on what night?

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I'm going to have to disagree with BianB BB's answer, which was formerly the accepted answer:

If there are 98 blue-eyed people, then each of them sees 97 blue eyes. They know the number is not prime, so they leave immediately on day 1.
99 blue-eyed people wait 1 day, see there are not only 98 blue eyes, and leave on day 2.
100 blue-eyed people need therefore to wait only 2 days, see there must be more than 99 blue-eyed islanders, and leave on day 3.

For other values of n, the number of blue-eyed islanders:

If $n$ is smaller than 25, then it must be prime. This means that for every n different from 3, the islanders will see $n-1$, which is not prime, and conclude that they should leave on day 1. The exception is 3, for which $n-1$ is also a prime, so the islanders will need to wait one extra day to rule the possibility that $n=2$ out, and leave on day 2.

If $n$ is 25, then the islanders realize it can't be 24 because it is not a prime, and leave immediately. If $n$ is greater than 25 and one more than a prime, the islanders know it can't be a prime and also leave immediately. For all other $n$ greater than 25, the blue-eyed islanders leave on day $k+1$, where $k$ is the difference between $n$ and the greatest number $d$ such that the islanders leave immediately (either 25 or one more than a prime) and $d<n$.

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  • $\begingroup$ This is indeed the actual correct solution, I made the same mistake as the formerly accepted answer. $\endgroup$ – wythagoras Jul 28 '16 at 12:52
  • $\begingroup$ For completeness you should include the rest of the solution in this answer so it stands alone (I would think the accepted answer should be a full answer, even if that means copying bits from somebody else and crediting them). Its especially complicated by the fact that it isn't even clear which answer you are referring to that you disagree with (no way to tell which was formerly accepted)! $\endgroup$ – Chris Jul 28 '16 at 13:39
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    $\begingroup$ @Chris The question only asked for n=100, so in my eyes solving that case is a complete solution to the question asked. But I added the general case anyway, because it can't hurt... $\endgroup$ – ffao Jul 28 '16 at 13:59
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If one islander sees 1, 4, 6, 10, 12, 16, 18 or 22, he should have blue eyes as well to keep the statement true (less than 25 and is a prime), so in these cases, just one day.

For the same reason, if one islander sees a prime number greater than 25, he should have blue eyes (greater than 25 and non-prime).

If one islander sees 2 blue eyes, you have to wait 1 day (if he hasn't blue eyes, the two of them can see only one, thus going on the first day).

If you see a prime number less than 25, you don't have blue eyes, to keep the thruthful of the statement. One cannot see any other number less than 25, because it would cause the statement to be false. For any other number $n \geq 25$, you should wait $n-25$ minus the number of primes between 25 and $n$.

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  • $\begingroup$ 1 is not prime so each one cannot see 0 people with blue eyes. $\endgroup$ – Marius Jul 28 '16 at 9:42
  • $\begingroup$ This is the correct solution. There is one small mistake in the third paragraph, which again relates to the fact that one is not prime. (Namely if someone sees one blue-eyed person, then he has blue eyes) $\endgroup$ – wythagoras Jul 28 '16 at 10:01
  • $\begingroup$ hehe, again, forgot that :) $\endgroup$ – BianB BB Jul 28 '16 at 10:04
  • $\begingroup$ @bianBBB Actually, it is not correct, as has been pointed out by ffao. $\endgroup$ – wythagoras Jul 28 '16 at 12:52

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