## Putnam 2010!

For the discussion of math. Duh.

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z4lis
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### Putnam 2010!

I'm certain there's a rule about how long we have to wait until we discuss the thing... I found the 2010 bins problem rather entertaining. I was sort of shocked about how well I did on the A part, though. Managed to write up five solutions (one of which has an error which would be fixed with about three words... sigh. I was working too quickly.) for the A part and two for the B part. Did anyone else find A surprisingly... easy? Also, my proof of the group question (where the group product was the dot product in some cases) was maybe a paragraph at most, using pretty elementary facts about the cross product. I get rather hesitant about Putnam solutions that are so, so short.

So, I'm hoping for 60 points, at least. I'll see how kind the graders are with the last 10.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

mdyrud
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### Re: Putnam 2010!

The Art of Problem Solving forums said to wait until tomorrow evening to discuss specifics. I only took the A section, due to a choir concert later in the afternoon, and I was surprised by how easy it was. Not to say that I did well. But I was almost 100% positive my answer to the first question was airtight. All in all, I'm pleased with my first experience with it.

duckshirt
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### Re: Putnam 2010!

I have a crappy scan of the problems if anyone is interested, but maybe I should hold off on posting them publicly...

I didn't do as well as I hoped or should have given that the problems didn't seem too hard, but I think I did better than last year. I turned in 3 on each - one is wrong for A, two not quite right for B, so I am hoping for a score in the 30's. So I guess my goal of getting top 100 some year will wait...

I did find A a little easier than B, but not a ton. I'm not sure about the test as a whole; I think the other people at my school felt a lot better about it than last year, so we'll see.
Last edited by duckshirt on Sun Dec 05, 2010 6:49 am UTC, edited 1 time in total.
lol everything matters
-Ed

++\$_
Mo' Money
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### Re: Putnam 2010!

Part A was significantly easier than usual, I thought. Part B was perhaps a touch on the easy side of average.

I ended up getting 7 right and 1 very, very wrong.

Qaanol
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### Re: Putnam 2010!

For those interested, the problems (and solutions) can be found here.
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mike-l
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### Re: Putnam 2010!

Looks relatively easier than most years. Normally I can get 3-5 within an hour or two work, this years I can get 5 right away with good progress on a 6th.

I can get A1-3 at a glance, and they don't seem like very interesting tricks:

1
Spoiler:
is just the addition of an arithmetic series
3
Spoiler:
is just the directional derivative combined with the differential equation for e^x.
Presumably there is a nicer solution to 2 than mine, but the principle is easy enough,
Spoiler:
each interval [n,n+1] has to "look" the same in both a scaling and non-scaling way, so they all have to be flat... I imagine it's a good page of work to make that precise though)

A4 looks not fun... I can't feasibly compute even small examples which annoys me. I'd probably just skip it.
A5 I can see a bunch of properties off hand and could probably work to a solution, I may try this later.
A6 I don't really know, I'd probably have spent maybe an hour on 1-3 and 5 and bang my head against this for 2 hours. Why is A6 almost always an annoying integral?

B1 and B2 are trivial. Edit: Actually writing out 2 it's mildly harder than I thought.

1
Spoiler:
Glancing at the solution to make sure I'm not making an obvious error, the Cauchy Schwartz solution is much nicer than mine, which is the second solution posted
2
Spoiler:
Since everybody knows 3,4,5 is the smallest pythagorean triple, you only have to rule out 2 numbers, neither of which appear in a pythagorean triple so the edge in question could not be a diagonal. Thus we are reduced to showing that (without loss of generality) if [imath]a^2 + b^2 = c^2[/imath] , then [imath](a+1)^2 + b^2[/imath] and [imath](a+2)^2 + b^2[/imath] are not squares, but this is trivial considering parity and the fact that c > a

B3 looks really fun, I'll definitely be working on this later.
I can't decide how hard each of B4-6 are without doing more work, though they seem quite feasible.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

GyRo567
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### Re: Putnam 2010!

You can get B5 by noticing that f is infinitely differentiable (though you'll only make use of the 2nd derivative) because each derivative can be written as a composition/product of f. Then since f' > 0 everywhere, you can write f' = e^h for some continuous function h, which you know nothing of---except that since f' is differentiable, then h' must exist, which allows you to write f in terms of h and the exponential function. From there, you can play around with equalities until you force out the two possibilities for h, neither of which allows for f'(x) = f(f(x)).

I may come back & work this out in all its TeXiness, but that should be enough to help anyone curious about it. There's an entirely different approach over at AoPS.
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duckshirt
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### Re: Putnam 2010!

A4 looks not fun... I can't feasibly compute even small examples which annoys me. I'd probably just skip it.

Yeah, I did figure that out long after the test... I'm impressed with anyone who could come up with the answer quickly.
lol everything matters
-Ed

skeptical scientist
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### Re: Putnam 2010!

mike-l wrote:Presumably there is a nicer solution to 2 than mine, but the principle is easy enough,
Spoiler:
each interval [n,n+1] has to "look" the same in both a scaling and non-scaling way, so they all have to be flat…

I imagine it's a good page of work to make that precise though

Not the way I did it:
Spoiler:
For all x, $$f'(x+1)=f(x+2)-f(x+1)=\left(f(x+2)-f(x)\right)-\left(f(x+1)-f(x)\right)$$ $$=2\frac{f(x+2)-f(x)}{2}-\frac{f(x+1)-f(x)}{1}=2f'(x)-f'(x)=f'(x).$$ Therefore f' is periodic, with period 1. Thus $$\frac{d}{dx} f(x+1)-f(x)=f'(x+1)-f'(x)=0,$$ so f(x+1)-f(x) is equal to some constant c, independent of x. But f'(x)=f(x+1)-f(x), so f' is constant, meaning f is a line.
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"With math, all things are possible." —Rebecca Watson

mike-l
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### Re: Putnam 2010!

skeptical scientist wrote:
mike-l wrote:Presumably there is a nicer solution to 2 than mine, but the principle is easy enough,
Spoiler:
each interval [n,n+1] has to "look" the same in both a scaling and non-scaling way, so they all have to be flat…

I imagine it's a good page of work to make that precise though

Not the way I did it:
Spoiler:
For all x, $$f'(x+1)=f(x+2)-f(x+1)=\left(f(x+2)-f(x)\right)-\left(f(x+1)-f(x)\right)$$ $$=2\frac{f(x+2)-f(x)}{2}-\frac{f(x+1)-f(x)}{1}=2f'(x)-f'(x)=f'(x).$$ Therefore f' is periodic, with period 1. Thus $$\frac{d}{dx} f(x+1)-f(x)=f'(x+1)-f'(x)=0,$$ so f(x+1)-f(x) is equal to some constant c, independent of x. But f'(x)=f(x+1)-f(x), so f' is constant, meaning f is a line.

Nice.
addams wrote:This forum has some very well educated people typing away in loops with Sourmilk. He is a lucky Sourmilk.

mdyrud
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### Re: Putnam 2010!

I got my scores! Ten points, which puts me in the top fifty percent. My schools team placed in the top 150, which is the first time that has ever happened. All in all, a great first experience with it.

skeptical scientist
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### Re: Putnam 2010!

Does 1 point still put you in the top 50%? When I took it, I think most of the years it did, but one year you needed 2 points.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

mdyrud
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Joined: Fri Jun 13, 2008 10:34 pm UTC

### Re: Putnam 2010!

I feel that the cutoff will be a little higher this year. The first problem was pretty easy, definitely easy enough to have a lot of people score a point or two.