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:(adsbygoogle = window.adsbygoogle || []).push({});

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

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 probaly trivial and I'm just not putnam material!

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Putnam questions for fun

Loading...

Similar Threads - Putnam questions | Date |
---|---|

I Question about second order linear differential equations | Aug 21, 2017 |

B Simple double integration of square wave question | Aug 16, 2017 |

I Question regarding integration of an equation | Jul 4, 2017 |

A Some questions regarding the ADI Method | Jun 23, 2017 |

**Physics Forums - The Fusion of Science and Community**