I think I made a complete fool out of myself on the Putnam. I guess I’ll go through each problem.

A1: After spending a while misreading the problem and making it much harder than it actually was, I came to my senses and found a simple 1-1 correspondence pretty easily. Sergey had a more interesting way of doing it, but I didn’t think of it. Well, mine still works. Yay.

A2: Problem A2 is hard? I really don’t think that’s fair. Since I wasn’t gettng anywhere on the other A problems, I thought I would throw everything I had at this problem. Unfortunately, nothing worked. Only Jeff claims to have a solution, and his solution was apparently really messy.

A3: Everyone thought it was 2. However, it’s 2sqrt(2)-1. How strange.

A4: Too bad. It’s hard. I’m not smart enough to solve hard problems.

A5: How can I fail to solve a problem with Dyck n-paths? That is most certainly unfair. The only American Math Monthly problem I ever solved was a Dyck n-path problem. I was halfway through writing up my solution when I realized that it was wrong. I hate it when that happens.

A6: It has to be true, I suppose. It would be amazing if it stopped being true at 2003. If that is the case, I shall be eternally in awe of the problem-writers. I doubt that’s the case though.

So I solved one problem in the first half. If that’s not shameful, I don’t know what is.

B1: Can a B1 problem really be hard? No one I know solved it (at least, nicely) though. Sure, I can solve a system of 12 equations, but I don’t like doing such things.I enjoy solving problems elegantly. (I suppose everyone does.)

B2: This was definitely the nicest problem I solved. I sat around for a while trying to bound it before I realized that I could evaluate it in closed form. I know what to do when binomial coefficients come my way. Now why weren’t there more problems with binomial coefficients? If there had been, I might have scored more points. Anyway, everything works out when you use the absorption identity and Pascal’s identity and then sum the binomial coefficients.

B3: I think my solution is rather brute-forcish, but it works. I just looked at the power of each prime. It turned out to be switching rows and colums of a Ferrers Diagram, but I wasn’t sure how to spell Ferrers (I was unsure as to whether it was Ferrers or Ferrars), so I just explained it. According to MathWorld, both spellings are used. I guess that’s why I was unsure. I must have seen both.

B4: I imagine one can use Viète’s Formula to prove it, but it wasn’t obvious to me. Why is it necessary that r_1+r_2!=r_3+r_4 even? This problem is probably easy enough that I should have worked on it more.

B5: This problem was in *Mathematical Olympiad Challenges*. It’s a shame I didn’t know that. There go ten easy points.

B6. This looks like a theorem we would have proved in analysis. Unfortunately, we didn’t prove it in analysis. Actually, it doesn’t look like a theorem we would have proved in analysis. We never got to assume that functions were continuous. That would have made life too easy.

Well, I improved. I went up to two problems in the second half. That’s going to be around 30 points. I’m very disappointed. I solved five last year (when it didn’t count, and they didn’t get graded). Today, when it actually counted, for my own personal bragging rights as well as the college’s, I was only able to get three.

Oooh… Dyck paths.

Hey, where can I get a copy of the problems?

They’ll probably be on the AMC website before too long. If you want them sooner, I’ll TeX them for you.

No, I’m sure somebody’s already done that… I just don’t know who. Last year Gabriel Carroll linked me to the problems like the evening after or something. Then again, maybe not. But don’t bother.

Wow, people responded quicky to that post (Brian IMed me to tell me the Dyck path problem.)

Posts get emailed to me, and I check my email often. That’s why I respond quickly.

Yes, and I don’t — I saw your first message on LJ first, actually.

Last year’s Putnam seemed to be obscenely more easy than many other years. 30’s still not too shabby.

How did you do on it?

*cough* We won’t speak of that.

Won’t we? Ok.

a2 rather nifty am-gm. just pick the right thing(s) to apply it on.

a3: f+g+fg.. better be an easier solution

a4 split into cases and eliminate absolute values. there’s probably a better way, though.

b1 there might be a niftier solution, but the “brute force” takes about 10 easy lines. gotta figure out where to look for the contradiction, though, to get anywhere. and use symmetry a lot.

and yeah, as far as putnam’s go, this one was hard. i’m guessing that 30 this year is more impressive than 50 last year. though score isn’t 10*problems solved. it’s 10*problemssolvedwithoutstupidity, which is often less.

and bragging’s silly anyway.

1. Who are you?

2. I agree that Putnam score is 10*(problems solved without stupidity). I checked my solutions tons of times though, so I’m sure they’re right. Whether the graders will agree is another story.

3. I thought something like that might work for B1, but I really didn’t want to be solving some nasty equations, even though any trained monkey could do it.

my heart bleeds for you, Simon. *patpat* (btw, who are these people?)

I have no idea. I wish anonymous posters would stop being so anonymous. What good does an IP address do me?

um… well, if you have hack0rz skills. it is entirely possible to find out who these people are… and if you don’t it requires having friends who do… (Allan and Anderthan) but they refuse to do illegal stuff…

I can sometimes figure out who anonymous posters are if they have posted something less than anonymous earlier with the same IP address. That’s the extent of my hacking skills though.

go ask. I find it almost impossible that all the people you met at MOP adhere to all the legal rules of this country. I’m sure you’ll find someone…

Not to brag or anything, but this years putnam was easier.