tag:blogger.com,1999:blog-9124539381685751273.post2382547917481982357..comments2023-06-19T04:35:06.263-07:00Comments on Skeptic's Play: Selected Putnam Problemsmillerhttp://www.blogger.com/profile/05990852054891771988noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-9124539381685751273.post-30216055047727530882008-12-07T21:05:00.000-08:002008-12-07T21:05:00.000-08:00In my experience (and from what I've been told), f...In my experience (and from what I've been told), for a <I>very</I> complete answer, you'll get 9 or 10 points, but for a nearly complete answer, you'll only get 1 or 2 points. That said, last year I missed the trivial case on one question and still got full credit on it (I got 12 points total, and I'm assuming 10 of those came from that one question).<BR/><BR/>A1 is actually surprisingly easy (either that or a semester of grad school has boosted my problem-solving ability considerably!)<BR/><BR/>For any x, f(x, x) + f(x, x) + f(x, x) = 0, so f(x, x) = 0. Therefore for any x, y, f(x, y) + f(y, x) + f(x, x) = 0 implies f(x, y) = -f(y, x). So fix z (any real number will do), and let g(x) = f(x, z). Then for any x, y, f(x, y) = -f(z, x) - f(y, z) = f(x, z) - f(y, z) = g(x) - g(y).<BR/><BR/>I considered for a moment the possibility that f(x, y) = 0 for all x, y, but that's not the case: f(x, y) = x - y is a nontrivial example.<BR/><BR/>Thanks for posting these! I'll have to think about the others for a bit.Anonymousnoreply@blogger.com