1) is trivia/trick question (obviously designed to catch those who learned the squeeze theorem as the sandwich theorem), 2) is hilariously simple, 3) is trivial, 4) is trivia, 5) is interesting.
Anyone have access to the harder, non-trivia(l) questions please?
Hilariously simple and trivial are insulting terms to use for problems that challenge many others. There is no absolute difficulty scale on which problems can be judged. The problems you personally find difficult can be called trivial and hilariously simple by someone much smarter than yourself. Unless of course you're the smartest person alive, in which case in a few hundred years someone will call those problems trivial and hilariously simple, and you will be shamed posthumously for finding them so difficult.
(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.
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.
> 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.
The last one is trivial, too. Is 8 x 8 + 7 x 8 + 89 = 209 divisible by 77? Nope. Is 18 x 18 + 7 x 18 + 89 = 539 divisible by 77? Yes, gives 7, solved it. This isn't even worth trying to look for a more analytical solution.
But since I didn't know in advance that the solution would be one of the smaller possible answers it was faster in expectation to solve the problem using the given method.
5 may be interesting, but it's solvable by trial and error on the 5 answers. It's only interesting if it's fill in the blank. :-)
I believe the event was a fundraiser for a kids-friendly math museum, so I'll cut these guys a lot slack. No matter how they did, they had to whip out their checkbooks at the end of the evening.
> obviously designed to catch those who learned the squeeze theorem as the sandwich theorem
Shouldn't the "Fermat" part rule out the squeeze theorem? I know Fermat did work that contributed to the development of Calculus, but he was gone long before either Newton or Leibniz published.
I couldn't figure out a solution to 3. because of the wording. I thought it proposed drawing a single line segment between a pair of consecutive side midpoints.
Couldn't find the solutions on those 2013 ones. My answers for the first two (the Gardner problems) are [4^3 + 3] and [13^2 - 100] (ASCII codes). If anyone else wants to do them, let me know if you get the same answer.
1) is trivia/trick question (obviously designed to catch those who learned the squeeze theorem as the sandwich theorem), 2) is hilariously simple, 3) is trivial, 4) is trivia, 5) is interesting.
Anyone have access to the harder, non-trivia(l) questions please?
[edit] found a few more from 2013, which I think are more interesting: http://digitaleditions.walsworthprintgroup.com/display_artic...