Spivak Chapter 2, problems 27 (and 28)

Solution 1:

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.

Solution 2:

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.