(1) See my signature. The goal is to tell me ALL the errors. Look closely! Also, paradoxical answers are not accepted.

(2)

A: There is a list of 2015 statements:

etc.

B: What happens if you change "Exactly" to "At most"?

C: What if you change "Exactly" to "At least"?

D,E,F: What if you change "True" to "False" in A, B, and C?

NOTE: I don't know the answer to problem 2, as I heard them as 2000 and the answer may be different for odd vs. even.

(For example, if we have statements 0..n-1 where statement i says "Statement (i+1) mod n is false", it's valid for n even but a paradox if n is odd.)]]>

Then, I discuss 5 variations where you change "Exactly" to "At most", then "At least", and if "true" is replaced with "false".]]>

For each of these, consider a generalized list of n statements, then state the n = 2015 case.

A. No two statements can be simultaneously true. Therefore at most one statement is true. So, either no statements are true or only the first is.

n = 2015: Only the first statement is true, or all of them are false.

B. The truth of the kth statement implies the n-k following. For k < n-k+1 this is a contradiction, so for k ≥ (n+1)/2 the kth statements are true and the rest false.

- For n = 2m+1 odd this suggests the first m are false and the remaining m+1 are true.

- For n = 2m even this has the first m are false with the last m true. But the mth statement is false, so there is a contradiction.

n = 2015: Statements 1-1007 are false and 1008-2015 are true.

C. Trivially all statements can be false. The truth of the kth statement implies the k-1 preceding. Hence, each statement implies itself.

Therefore for any k between 0 and n inclusive, the first k statements are true and the rest are false.

n = 2015: There are 2016 solutions as described above.

D. No two statements can be simultaneously true. Then at most one is true; if none are true the last statement is a contradiction, while if one is true it must be the second-last.

n = 2015: Only statement 2014 is true.

E. The last statement is true. This implies the previous statement, and so on. Therefore all statements are true.

n = 2015: All statements are true.

F. The truth of the kth statement implies the k-1 preceding. For k > n-k this is a contradiction, so for k ≤ n/2 the kth statements are true and the rest false.

- For n = 2m+1 odd this suggests the first m are true and the remaining m+1 are false. But the m+1th statement is false, so there is a contradiction.

- For n = 2m even this suggests that the first m are true with the last m false.

n = 2015: There is a contradiction (no solution).

Do tell me if I've made any logical leaps.]]>