- #1

SeReNiTy

- 170

- 0

Hey guys, I was doing some Putnam questions for fun and came across something strage to me. In the 2005 Putnam competition, question B3, link provided below:

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/

I solved it by guessing a polynomial solution and then verifying the co-efficients, but in their solutions they have a more analytical approach where they let x = a/x. Now I'm a bit uncomfortable with this substitution so i let y = a/x, and preceeded to compute their solution. The point i get stuck is after they calculate the second derivative, they somehow eliminate all the a's. Could something explain that step? Its probably trivial and I'm just not putnam material!

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/

I solved it by guessing a polynomial solution and then verifying the co-efficients, but in their solutions they have a more analytical approach where they let x = a/x. Now I'm a bit uncomfortable with this substitution so i let y = a/x, and preceeded to compute their solution. The point i get stuck is after they calculate the second derivative, they somehow eliminate all the a's. Could something explain that step? Its probably trivial and I'm just not putnam material!

Last edited by a moderator: