(2) and (3) really are easy though. Both can be solved by deriving counterexamples for the options until one is left.

For (2), imagine if there were only two teachers in the convention and they shook hands with each other once - that invalidates option C. If they shake again, that invalidates option B and E. A and D both talk about an even number of teachers, so let's imagine there were three teachers. If each teacher shakes hands with the other two once, that invalidates D. So A is the answer. I guess some form of graph theory is involved with solving it properly.

(3) can be reasoned in the same way by imagining quadrilaterals for each option. A quad with two adjacent tiny edges and two long edges symmetric along one axis (like a square with one vertice pulled away) gives a rectangle, which invalidates A, B and E. Making it asymmetric by moving the far vertex invalidates C. So only D is left.

> (2) and (3) really are easy though. Both can be solved by deriving counterexamples for the options until one is left.

Who says there's that exactly one of the answers is right? For (2) the formulation of the question even suggests that any number could be right or wrong.

Nobody, and I didn't assume it until I'd found counter-examples for four of the choices and couldn't find any for the fifth.

Edit: And yes, I know the inability to find counter-examples to something doesn't necessarily mean it's true.

I dont know about that. I found the answer through logic rather than exact example. If a teacher shook hands with an odd number, x, of other teachers. That means that we know there were at least x handshakes that werent his, by symmetry of handshakes. Now, we know that those x handshakes could be divided up into any such partitions between teachers. But whats really cool is that we know some basic properties of partitions of odd numbers. We know that an odd number always breaks into an odd number of odd partitions (carrying the oddness.. ie. The extra 1 mod 2, the bit) You may need a few examples to convince yourself. 9=3+3+3 7=2+5 and well any number 199=196+1+1+1

Now clearly if the main teacher having odd shakes x, implies that an odd number of other teachers have odd total shakes, then the total odd shakers is even.

The “hilariously simple” question can be solved by a clever third grade student in a few minutes. If this “challenges” many “quants”, uh...

> Hilariously simple and trivial are insulting terms to use for problems that challenge many others.

I'd generally be inclined to agree, and I think that there are too many people on HN that constantly call legitimately hard things trivial, which I dislike. But in this case, the two that I referred to as such are actually solvable by an eleven year old. I refuse to believe that any quants -- a profession known to be composed of some of the smartest people on earth -- have any difficulty whatsoever with these problems.

If they were in a book of brain teasers, I wouldn't go through the book and laugh at how "trivial" all the questions were, that would make me a dick. But these are questions asked to goddamn quants, and in that context these questions are certainly hilariously simple.