It is mainly a more intricate version of the barber paradox, right? The possible solutions do not form a set under ZFC, therefore we cannot assign a cardinality.

Pretty much, yes. I am not familiar enough with ZF to say it myself, but that sounds plausible.

Deciding whether the answer to question 6 is zero or non-zero ends up being equivalent to answering "Is the answer to this question 'no'?" with a 'yes' or a 'no'.

When I devised the questions, my first aim was to find a set of questions 1-5 such that the correct answer to question 6 would be '0', unparadoxically.

I tried out this set of questions, and found it didn't meet that aim; '0' cannot be correct without leading to a paradox.

Then I found the 'solution' (2,6,10,9,4,1) and thought the process of finding it was interesting enough to set this quiz to my work colleagues. Anyway, life is only so long, and I didn't pursue my first aim further. So long as there weren't any other 'solutions' with a '1' at the end, then everything would be ok. Basically everything that faust got in his first reply.

It was a few days later, when I set out to prove the solution, that I came across 6-tuples similar to the ones ghostofmars found. (5,5,10,X,0,1) was the critical one which disproved my first solution attempt, and then we quickly see that nothing else works either.

So we're in paradox land.

However, in spite of the paradox, I think the stated aim is still valid:

The aim is to find a 6-tuple which maximises the number of correct answers.

The number of correct answers is maximised by any 6-tuple which answers Q1-Q5 correctly. I think.

Some ideas for developing this further if I (or anyone else) gets round to it:

- If Q4 is changed to exclude its own answer from the range, then I think (2,6,10,9,4,1) becomes valid. But that seems a clumsy thing to do in a quiz which is meant to be self-referential.

- With different Q1-Q5, there is probably an interesting quiz which has '0' as the correct answer to Q6, as per my original aim.

- With different Q1-Q5, it may be possible to end up with potential solutions (a): (a1,a2,a3,a4,a5,1); (b): (b1,b2,b3,b4,b5,2) and (c): (c1,c2,c3,c4,c5,2). In that case either (a) is correct, or (b) and (c) are correct. That seems similar to deciding how to answer "Is the answer to this question 'yes'?".

- Just leave off questions like Q6 and have fun with other sets of questions which refer to the answers.