Are You Smarter Than a Quant? Questions from the MoMath Masters Contest(blogs.wsj.com)
blogs.wsj.com
Are You Smarter Than a Quant? Questions from the MoMath Masters Contest
http://blogs.wsj.com/moneybeat/2016/03/04/are-you-smarter-than-a-quant-here-are-5-questions-from-the-momath-masters-contest/
14 comments
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.
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.
Maybe its like the GRE where all the questions are easy but there's a nasty time limit?
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.
5 is the simplest ... CRT plug and chug
> 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.
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.
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.
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...
So it would have taken rather more tries to find 18...
Martin Gardner has a bunch of books with short brain teasers of various levels of complexity. Most of them are like $2 on Amazon.
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.
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.
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.
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.
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.
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.
You might have a fun time with the "Jewish Problems".
http://arxiv.org/pdf/1110.1556.pdf
http://arxiv.org/pdf/1110.1556.pdf
I got all except 4, which is trivia. (Well 1 is sort of trivia as well, but I happened to know that Fermat proved that one).
I somehow dug Cindy out of the crypts of my brain. The squeeze theorem is sometimes also called the sandwich theorem, so that one threw me off.
It's also sometimes (in Italian and Russian, according to my professors) called the Theorem of The Two Policemen. Imagine the path of a drunk staggering down the street. Then imagine the path of the two cops who come to grab him by the shoulders.
Et voila! The Theorem of the Two Policemen.
It's also sometimes (in Italian and Russian, according to my professors) called the Theorem of The Two Policemen. Imagine the path of a drunk staggering down the street. Then imagine the path of the two cops who come to grab him by the shoulders.
Et voila! The Theorem of the Two Policemen.
> The squeeze theorem is sometimes also called the sandwich theorem, so that one threw me off.
E is also the "ham sandwich theorem", so it looks like they were going for deliberately confusing
E is also the "ham sandwich theorem", so it looks like they were going for deliberately confusing
Smallest positive m such that 77 divides m^2+7m+89 is greatly simplified by realizing (mod 77) m^2+7m+89=m^2+7m+12=(m+3)(m+4).
<nitpick>
m^2+7m+89 != m^2+7m+12
(m^2+7m+89)%77 =(m^2+7m+12)%77
</nitpick>
That's precisely how the article explains it.
Read the article.
What's annoying me about the second question is the distinction I'm thinking about what constitutes a handshake:
are we determining that the number of hand shakes
- grasping of the hands between two people constitutes a handshake; or
- the number of times one moves their hands up and down is the hand shake
so I'm thinking the former, but still finding it difficult to see how the number of handshakes can ever be odd.
You need at least two people, the sum of their hand shakes is even, you add a third, and everyone shakes hands at least once, so the sum of all their handshakes is even.
I'm not sure the rest of the solution can involve any odd numbers.
i have a mental hurdle -- i'm not thinking about it in the right way
are we determining that the number of hand shakes
- grasping of the hands between two people constitutes a handshake; or
- the number of times one moves their hands up and down is the hand shake
so I'm thinking the former, but still finding it difficult to see how the number of handshakes can ever be odd.
You need at least two people, the sum of their hand shakes is even, you add a third, and everyone shakes hands at least once, so the sum of all their handshakes is even.
I'm not sure the rest of the solution can involve any odd numbers.
i have a mental hurdle -- i'm not thinking about it in the right way
If there's only two teachers at the convention and they shake hands, that's just one handshake- an odd number. The count is an "objective" count of all handshake-events, not the sum of everyone's personal handshake counts.
Writing unambiguous math problems is a surprisingly hard thing to do!
Writing unambiguous math problems is a surprisingly hard thing to do!
But what if you prefer qualitative? Can you out-smart Warren Buffet [1]?
[1] http://www.forbes.com/sites/investor/2011/10/11/warren-buffe...
[1] http://www.forbes.com/sites/investor/2011/10/11/warren-buffe...
Warren Buffet does lots of quantitative reasoning as well: on balance sheets rather than fluctuation of market prices.
I have to admit the factoring threw me for a bit. The last time I had to do something like this was over 20(!) years ago.
Although it was gratifying to have the solution drift into mind after a while. What is sad is that I'm acutely aware that I no longer think as quickly as I did in my 20s.
Although it was gratifying to have the solution drift into mind after a while. What is sad is that I'm acutely aware that I no longer think as quickly as I did in my 20s.
For 5), 89 modulo 77 is 12 so we just solve for the least m such that (m + 3) * (m + 4) (= mm + 7m + 12) is divisible by 77. Both 73 and 74 clearly satisfy the condition, but 18 + 3 = 21 divisible by 7 and 18 + 4 = 22 divisible by 11 is the least among the choices.
Maybe there's a subtle irony to reflect on here - which is that intelligence is something diverse and qualitative, not just a single quantitative scale from dumb to smart.
The way I heard it last time was that intelligence is like running a race to get to a place (the solution to a problem). It doesn't matter how fast you run if you're going the wrong direction.
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Very surprised to see Manjul at one of the tables
Problem 5 was fun. My solution was less complex: if (((i * i) + 7 * i + 89) % 77 == 0) System.out.print(i);
That's cheating though - also it likely is faster to do it mentally than typing all that out for the brightest.
That's probably fair. It likely would have taken me a minute to do the mental math. The burst force approach took 15 seconds of typing and very little mental effort.
I think the real answer is that in the competition only pencil & paper is allowed. No computers.
They are using tablets on the pictures. Surely those have calculators.
The ipads are locked down to run the contest app.
Besides, calculators have rather poor programming support...
Besides, calculators have rather poor programming support...
"Speak for your self" said the HP-48 owner, tapping into a decades-old bank of accumulated smug.
Mmmmmm... delicious, delicious RPN smug. I haven't tasted this in years. Not as fresh as it used to be, but still sweet.
Mmmmmm... delicious, delicious RPN smug. I haven't tasted this in years. Not as fresh as it used to be, but still sweet.
I think those are the source of the questions? But ya, you might be right.
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...