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.