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/

by sonabinu • 10 years ago

85 48 10 years ago

blogs.wsj.com

ikeboy• 10 years ago

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).

pflats• 10 years ago

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.

zem• 10 years ago

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

valine2• 10 years ago

Problem 5 was fun. My solution was less complex: if (((i * i) + 7 * i + 89) % 77 == 0) System.out.print(i);

Bahamut• 10 years ago

That's cheating though - also it likely is faster to do it mentally than typing all that out for the brightest.

valine2• 10 years ago

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.

harryh2• 10 years ago

I think the real answer is that in the competition only pencil & paper is allowed. No computers.

owenversteeg2• 10 years ago

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_artic...

parennoob• 10 years 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• 10 years 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• 10 years 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.

tromp2• 10 years 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• 10 years 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• 10 years 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.

Arnavion2• 10 years 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.

jacobolus• 10 years ago

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

owenversteeg2• 10 years ago

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.

nordsieck• 10 years ago

You might have a fun time with the "Jewish Problems".

http://arxiv.org/pdf/1110.1556.pdf

titanomachy• 10 years 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• 10 years 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• 10 years 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• 10 years ago

I'm glad I'm not the only one who misread it like that!

IIAOPSW• 10 years ago

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

marodox• 10 years ago

5 is the simplest ... CRT plug and chug

octonion• 10 years ago

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).

baddox• 10 years ago

That's precisely how the article explains it.

imaginenore• 10 years ago

Read the article.

vanattab• 10 years ago

Hermel• 10 years ago

And how does the ability to solve these puzzles make you a good quant?

baconizer• 10 years ago

because it takes years of training/experience/luck/success to be invited sitting to solve them while sipping champagne on the house. :)

officialchicken• 10 years ago

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...

eru2• 10 years ago

Warren Buffet does lots of quantitative reasoning as well: on balance sheets rather than fluctuation of market prices.

random_coder• 10 years ago

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.

theoh• 10 years ago

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.

vinchuco• 10 years ago

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.

eachro• 10 years ago

Somehow I'm sure that some of the other problems must be a lot harder. Why else would heavy hitters like Poh Shen Lo and other past IMO competitors bother with this?

ipince• 10 years ago

Because it's for charity.

newman314• 10 years ago

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.

idontgetit_• 10 years ago

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

maxander• 10 years ago

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!

mrcactu5• 10 years ago

Very surprised to see Manjul at one of the tables

Are You Smarter Than a Quant? Questions from the MoMath Masters Contest

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