4 ms·

Not a fan of the questions. 1) is trivia/trick question (obviously designed to catch those who learned the squeeze theorem as the sandwich theorem), 2) is hila

Not a fan of the questions. 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_article.php?id=1461361 http://digitaleditions.walsworthprintgroup.com/display_artic...

parennoob 11y ago
> 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.
danbruc 11y ago
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.
- OscarCunningham 11y ago
  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.
- tromp 11y ago
  The question at the actual competition was slightly more subtle: it asked for m modulo 10. So it would have taken rather more tries to find 18...
laxatives 11y ago
Martin Gardner has a bunch of books with short brain teasers of various levels of complexity. Most of them are like $2 on Amazon.
dsjoerg 11y ago
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.
- Arnavion 11y ago
  (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.
  - eru 11y ago
    > (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.
    - Arnavion 11y ago
      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.
nordsieck 11y ago
You might have a fun time with the "Jewish Problems". http://arxiv.org/pdf/1110.1556.pdf http://arxiv.org/pdf/1110.1556.pdf
titanomachy 11y ago
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.
mathattack 11y ago
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.
recursive 11y ago
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.
- almost 11y ago
  I'm glad I'm not the only one who misread it like that!
IIAOPSW 11y ago
Maybe its like the GRE where all the questions are easy but there's a nasty time limit?
marodox 11y ago
5 is the simplest ... CRT plug and chug

parennoob 11y ago

> 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.

danbruc 11y ago

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.

OscarCunningham 11y ago
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.
tromp 11y ago
The question at the actual competition was slightly more subtle: it asked for m modulo 10. So it would have taken rather more tries to find 18...

laxatives 11y ago

Martin Gardner has a bunch of books with short brain teasers of various levels of complexity. Most of them are like $2 on Amazon.

dsjoerg 11y ago

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.

Arnavion 11y ago
(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.
- eru 11y ago
  > (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.
  - Arnavion 11y ago
    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.

nordsieck 11y ago

You might have a fun time with the "Jewish Problems". http://arxiv.org/pdf/1110.1556.pdf http://arxiv.org/pdf/1110.1556.pdf

titanomachy 11y ago

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.

mathattack 11y ago

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.

recursive 11y ago

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.

almost 11y ago
I'm glad I'm not the only one who misread it like that!

IIAOPSW 11y ago

Maybe its like the GRE where all the questions are easy but there's a nasty time limit?

marodox 11y ago

5 is the simplest ... CRT plug and chug