The knowledge of each islander consists of:
the color of the eyes of every other islander;
any past pronouncement from the guru;
the history of who left the island on previous days (including their eye color), which provides knowledge about other's knowledge (that either they did or did not know their own eye color on previous days).
At the beginning of ...
Every blue-eyed person sees 99 blue-eyed people. Since they don't know that they have blue eyes, they suspect it might be the case that every other blue-eyed person can only see 98 blue-eyed people, and if those people only see 98 blue-eyed people, they might think that each of them only see 97 blue-eyed people. And so it continues, until someone considers a ...
Let's continue the induction, since the jump to 99 blue eyes does seem weird. After all, everyone knows that someone has blue eyes.
If there are 4 blue eyed-people, A will look at B,C,D, thinking :
Now, the issue here is that I, being A, can see that B has blue eyes. Therefore I know that C sees at least D and B as having blue eyes. But this is the ...
I think I have a faster solution than Rubio. (Or I did, when Rubio's solution took one day longer than mine; he's since incorporated my solution into his answer).
Let's start on the time of release and follow the thought process.
We assume both have at least one bar on their window (or the window couldn't be said to be barred, and they're told their windows are the only barred ones).
We further assume that they must determine which total, 18 or 20, is correct.
If they are being asked to determine if it is either (18 or 20) or (not(18 or 20)), that's a whole ...
The whole process is inductive, so it needs a starting point. If there were only one blue-eyed person, he would never know that there is "at least one person with blue eyes," so he would not go the first night. If there are only two, neither of them can know whether the other doesn't go the first night because he only sees brown eyes, so they don't know if ...
The only explanation I've seen that's sufficiently precise to be satisfying is this answer to the corresponding question on math.SE. The key fact that the "oracle" (guru) gives you, that you didn't have before, is that "(everybody knows)N there is at least one blue-eyed person" for any value of N. In particular, you need it to be true for N=100, but the "...
There are only 6 possible configuration of hats.
This would be a better question if you specify that every player is killed if he guesses wrong (my answer) or they must all answer at the same time ($1$ and $2$ always guesses opposite of $3$).
The numbers are
Alice: You can't know my sum.
Bob: Thanks to you, I know your sum.
Alice: Then I know your product.
EDIT: While I show the consistency of the two numbers with the provided statements, please see Mike Earnest's answer for a proof that this is the only possible pair of numbers consistent with the conversation.
The following table gives the numbers of one person and the corresponding possible numbers of the other person.
0 5, 8, 15
1 4, 7, 14
2 3, 6, 13
3 2, 5, 12
4 1, 4, 11
5 0, 3, 10
6 2, 9
7 1, 8
8 0, 7
If one of them has a number greater than 8 he/she would know ...
First, some observations
Alright. Now, let's take a look at different possibilities:
Generalizing this, we get the rule above:
Do note however that the actual answer is actually slightly more complicated, thanks to the effect of asking them in order:
I think considering it backwards might actually be the easier way to understand it.
A given blue-eyed person does not want to leave, so he hopes he has brown eyes and assumes he has brown eyes. He sees 99 blue-eyed people. Because he has assumed he does not have brown eyes himself, he must assume all of those other blue eyed people see 98 other blue eyed ...
This was trickier than it looked. I have a feeling that there should be a quicker way to find the answer than the list of cases I worked through.
Note: I interpret "two of which are factors of the third" to mean that two of the numbers divide the third, and that either of those two factors might be 1.
I think this works, although it results in two possible solutions that I believe meet all the criteria...
The first student says that he knows that the two other students have different numbers, because:
After hearing that the second says now he knows everyone has different numbers, because:
After hearing the statement of the second student the third says ...
X's statement means that his number must be even (if it was odd, Y and Z would have numbers summing to an even number, so they could be the same).
In the first case, Y's statement means that Y has an even number, and neither X nor Z can have the same even number. Therefore Y must have a number that is at least 1008.
Z knows all three numbers, so X and Y ...
Since the puzzle is symmetric in white and black hats, I assume that the first man has a black hat. Then only three cases remain: (1) BWWB; (2) BWBW; (3) BBWW.
($\ast$) If a man sees two hats of color white, then he knows that his hat is either red or black. If a man sees two hats of color black, then he knows that his hat is either ...
The colour of the guru's eyes is not relevant. The guru is allowed to speak about eyes and nobody else is. If any blue eyed person said "I can see someone with blue eyes" where everyone on the island could hear it, the same thing would happen. Also if any brown eyed person did. The moment a blue-eyed person hears that someone else can see some blue eyes, and ...
In other words, we are looking for two positive integers $x$ and $y$ with $x\ge2$, $y\ge2$ and $4\le x+y\le40$ that fit with the conversation.
Let us denote the sum $s=x+y$ and the product $p=xy$.
(1) Danielle's first statement "I don't know what the dimensions of the farm are" gives the following information on the product:
$p$ is not the product of (...
Let $a$ and $b$ be the integers thought of by A and B, and let $N$ and $n$ be the integers given by the teacher, where $N>n$. Assume for contradiction that both A and B keep on saying "no" forever.
If $a\geq n$, then A knows $n$ cannot be the sum and will say "yes" the first time. Thus after A's first "no", B knows $a<n$.
If $b\leq N-n$, then $a+b<...