The 2008 William Lowell Putnam Mathematical Competition officially took place this weekend! Yup, six hours of grueling math problems. We had a record number of students take it this year: 21! [That’s just 21, not 21 factorial. That would be an impressive number.]
Here’s one of the videos that our students watched while they were gathered around my dining room table Friday night eating dinner (because we totally bribe our students with food: dinner the night before at my house, bagels for breakfast, and lunch at the local pub in between the two three-hour sessions). It’s “I Will Derive” and I know it’s made its way around the internet, but I still think it’s fabulous:
And here’s the problem (A2) that caused the most discussion over lunch:
Alan and Barbara play a game in which they take turns filling entries of an initially empty 2008×2008 array. Alan plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all the entries are filled. Alan wins if the determinant of the resulting matrix is nonzero; Barbara wins if it is zero. Which player has a winning strategy?
Here’s my favorite problem (A1) because I was able to solve it right away and do you know how often that happens? Not very.
Let f : R2 -> R be a function such that f(x,y)+f(y,z)+f(z,x)=0 for all real numbers x,y, and z. Prove that there exists a function g : R->R such that f(x,y) = g(x)-g(y) for all real numbers x and y.
You can see the rest of the problems here, and pretty soon you should be able to find the answers here. Edited Monday 12/8 to add: yes, the answers are posted!
No, of course Godzilla didn’t really use a calculator on an exam. He’s a stickler for following the rules.
Tags: 2008 Putnam Answers, Putnam
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December 7, 2008 at 9:33 pm |
It’s going to take over 4 million moves to complete that game — I’d say that the winning strategy is not to play. 😉 Just feeling dumb because I could have probably either solved or made a decent effort at solving those problems 20 years ago, but not so anymore.
December 8, 2008 at 5:25 am |
Mathmom — LOL, I’m with you on the winning strategy! I thought about that problem, came up with an answer, and then when we talked about it after lunch [at a place with paper on the tables] I realized that my idea was completely and utterly wrong, since it worked for neither the 1×1 nor the 2×2 case. *sigh*
December 8, 2008 at 1:07 pm |
Ok, I’ve seen the answer for (A1), and given a multi-hour test I can see how I might find it, but how did you get it right away?
December 8, 2008 at 1:50 pm |
Jason: I didn’t do it the way that is written up on the solutions. Here’s what I did:
** I tried to find f(x,x) because that looked simplest. Since f(x,x)+f(x,x)+f(x,x) has to be 0, it turns out that f(x,x)=0 for all real x.
** Then I looked at f(x,y) because that seemed next simplest, to only have two variables. Since f(x,y)+f(y,x)+f(x,x)=0 and f(x,x)=0 from above, it turns out that f(y,x)=-f(x,y) for all real x,y.
** Then I played around with 0 and got f(x,0)+f(0,y)+f(y,x)=0, but f(y,x)=-f(x,y) from above so f(x,0)+f(0,y)-f(x,y)=0. This I rewrote as f(x,y)=f(x,0)+f(0,y)) but f(0,y)=-f(y,0) so f(x,y)=f(x,0)-f(y,0) for all real x,y. And by letting g(x)=f(x,0) it’s done!
December 8, 2008 at 8:16 pm |
Yep, that’s what I did, too. (I haven’t seen the “official” solution yet.) I really wanted g(x)=f(x,0), and that turned out to be the right way to go.
December 10, 2008 at 9:37 pm |
Wild! I did exactly the same for A1. I still found A2 much easier (seconds as opposed to minutes), and A3 took longer only because it’s hard to write. I didn’t take the test officially, and the rest look too hard for a few minutes’ thought 😦