Spivak Chapter 2, problems 27 (and 28) To be honest, I have no idea how to even start this problem. I'm sorry I don't have any work to show, but I'm just at a blank. Help?

Chapter 2: Problem 27:

University B, once boasted $17$ tenured professors of mathematics. Tradition prescribed that at their weekly luncheon meeting, faithfully attended by all $17$, any members who had discovered an error in their published work should make an announcement of this fact, and promptly resign. Such an announcement had never actually been made, because no professor was aware of any errors in her or his work. This is not to say that no errors existed, however. In fact, over the years, in the work of every member of the department at least one error had been found, by some other member of the department. This error had been mentioned to all other members of the department, but the actual author of the error had been kept ignorant of the fact, to forestall any resignations.
One fateful year, the department was augmented by a visitor from another university, one Prof. X, who had come with hopes of being offered a permanent position at the end of the academic year. Naturally, he was apprised, by various members of the department, of the published errors which had been discovered. When the hoped-for appointment failed to materialize, Prof. X obtained his revenge at the last luncheon of the year. "I have enjoyed my visit here very much", he said, "but I feel that there is one thing that I have to tell you. At least one of you has published an incorrect result, which has been discovered by others in the department." What happened the next year?"


Chapter 2: Problem 28:

After figuring out, or looking up, the answer to Problem 27, consider the following: Each member of the department already knew that Prof.X asserted, so how could his saying it change anything?

 A: First simplify the problem to only 2 professors, call them Prof. A and Prof. B (instead of 17).
On the next meeting after Professor X's statement. Prof. A will expect Prof. B to resign since Prof. A knows about Prof. B's error. When Prof. B does not resign, Prof. A will know it is because Prof. B is aware of an error of Prof. A's. Therefore, Prof. A knows about his error and must resign on the next meeting. Similarly Prof. B will have found out his error and will also resign.
Now think about how this works for 3 professors. Then you can generalize it to n professors and use it for your 17 professor problem.

If a professor knows about 0 errors, he must resign as soon as Professor X makes his statement because he would know the error was his.

If a professor knows exactly $k$ errors, and no one has resigned before the $k$th meeting after Professor X's statement, then he knows all members of the department know about at least $k$ errors which are not their own. So the professor now knows there must be at least $k + 1$ errors that everyone, except the error maker, know about. Since he only knows about $k$ errors, one of the errors must be his own and he must resign.

In the case where there are 17 professors where all professors know about exactly 16 errors. All the professors will resign on the 16th meeting after Professor X made his statement.

A: I'm adding some notation and further examples to Danikar's answer.


*

*Denote "A thinks (B thinks C knows no errors) and (D knows no errors)" as
A:(B:(C:0), D:0)
Each & every prof is proud and assumes that she hasn't made a mistake until it is necessarily true.
one prof


*

*A:0 A thinks there are no mistakes.

*Prof X declares that someone made a mistake...

*Of course, A is the only option. -> A:A

*A resigns


two profs


*

*A:(B:0) A thinks B knows of no mistakes.

*Prof X makes announcement...

*A:(B:B) Because A is proud and doesn't think its her mistake.

*But B doesn't resign, so A realizes B must have known of a mistake by A. -> A:A

*A resigns


But the same argument can be made for B. So the last bullet should read:


*

*Everyone resigns


three profs


*

*A:(B:(C:0), C:(B:0))
The problem is completely symmetric, so we can ignore every chain but the first.
A:(B:(C:0))

*Prof X takes revenge...

*A:(B:(C:C)) Of course A thinks that B thinks that C thinks it's C. Otherwise A wouldn't be treating B as a proud prof.

*C doesn't resign, so C must have known of another mistake. A assumes it was B's. A:(B:B)

*B doesn't resign. A:A

*A resigns.


five profs


*

*Simplified: A:(B:(C:(D:(E:0))))

*Prof X ruins a good thing...

*A:(B:(C:(D:(E:E)))) Of course D wouldn't think that E thinks it's D or D wouldn't be being proud. Likewise for C, B, A.

*E doesn't resign A:(B:(C:(D:D)))

*D doesn't resign A:(B:(C:C))

*C doesn't resign. A:(B:B)

*B doesn't resign. A:A

*A resigns.

