
Are You Smarter Than a Quant? Questions from the MoMath Masters Contest - sonabinu
http://blogs.wsj.com/moneybeat/2016/03/04/are-you-smarter-than-a-quant-here-are-5-questions-from-the-momath-masters-contest/
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owenversteeg
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...](http://digitaleditions.walsworthprintgroup.com/display_article.php?id=1461361)

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dsjoerg
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.

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Arnavion
(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.

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eru
> (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.

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Arnavion
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.

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goldenkey
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.

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ikeboy
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).

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pflats
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.

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zem
> 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

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eachro
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?

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ipince
Because it's for charity.

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octonion
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).

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baddox
That's precisely how the article explains it.

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idontgetit_
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

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maxander
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!

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officialchicken
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...](http://www.forbes.com/sites/investor/2011/10/11/warren-buffetts-
investing-formula-revealed/)

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

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newman314
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.

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random_coder
For 5), 89 modulo 77 is 12 so we just solve for the least m such that (m + 3)
* (m + 4) (= m _m + 7_ m + 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.

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

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

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theoh
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.

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vinchuco
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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mrcactu5
Very surprised to see Manjul at one of the tables

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

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

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

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S4M
They are using tablets on the pictures. Surely those have calculators.

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tromp
The ipads are locked down to run the contest app.

Besides, calculators have rather poor programming support...

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jerf
"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._

