If he wants the multiplication table, he will need each element but two in a query or an answer, so at least n/3 - 1 queries are needed in this case.

Shouldn't the very last term in the very last line have a $- \mid A \mid$ after it?

Even cooler (to the point of freaky) would be to push 18 of the tetrahedra in and pull the other two out just enough to cover. Probably not enough room though.

Can you push the cube to one side and get 21? You can cover a large bit of the corner with one tetrahedron.

Something to try: start with sum p^i for i from log n on up, and see if you can add log n many small powers to get a large multiple of n. Then subtract from M.

This sounds like a subgraph of a graph based on DeBruijn sequences. You might check out the larger graph and its properties.

By looking at (x -a)^3 + (y+a)^3, one is well on the road to showing N less than 5 has no nontrivial solutions.

On rereading my comment, "hint at an answer" seems overly strong, and "make significant progress" seems more appropriate. I'm sorry if I misled you.

No, but between Theorem 1 and the section 1.1 they cite a sequel which extends the Maier matrix method. I may be misreading things, so I ask you, since you seem more able with the analytic number theory literature (compared to me, which isn't saying much).

In the document behind your link, they hint at an answer to Joseph's question. Do you know if that writeup has appeared, even in prearxiv form?

If you compute the adjoint of the matrix, and count the number of nonzero terms of its permanent, that number times n! is what I think is the number of sequences will be, however you may have to compute all the minors to prove this guess.

Actually, some permutation have all zero entries in the diagonal. In fact, J-I has such a regular sequence where all but the 1 by 1 matrix has a zero diagonal. For any regular matrix, determinant computation by Laplace expansion will give you a sequence. In fact, it will give you at least n! many of them, if you count by original index sets.

Welcome to MathOverflow! I am having trouble directly relating your answer to the question. In particular do you know of an X such that a) br(X) is finite and bigger than 1 for X a PCU space, and b) all members bijectively related to X have to be PCU spaces?

I now saw your replying comment; thanks for bearing with my obtuseness.

Ok. The last thing I missed (I hope) is that the tuple of questions asked is an initial fragment of a permutation of the questions handed out. This makes it more of a challenge. I would suggest clarifying the point that no two students will be asked the same question, and that all permutations of (the order of) the n questions are equally likely to occur.

I see one thing I missed: a student can work on more than one question. I see also that there are more questions than students. However, if no students share a question, then we get a product of k terms of the form (n-a)/n, where a is allowed to vary over sizes of a certain partition of n into k parts. Or did I miss more?

Either you are over thinking this, or the actual problem is not well described by the group assignment scenario (or I am missing something). Why is the probability different from (1 -1/n)^k?

Perhaps this argument extends to ordered groups. It is hard for me to see how one would avoid consecutive runs of a subgroup like (Z_2)^2 with methods like this. Maybe estimating the number of partial sequences could benefit from something like this?