Today is the day of the 67th Annual William Lowell Putnam Mathematical Competition, offered all over the United States and Canada. According to the official website, Mr. Putnam was a strong believer in the value of team competition and so after his death his widow (Elizabeth Lowell Putnam) developed a trust fund which supported some intercollegial competitions. The first one was not in mathematics at all, but in English! The MAA took responsibility for the Putnam Competition after Mrs. Putam’s death in 1935.

Everyone who takes the Putnam these days takes it in two 3-hour blocks; the starting time, however, varies by Time Zone. Here in New York the times are 10am-1pm and 3pm-6pm. Are you an early riser? You might try taking it in at the Pacific coast, where the times are 8am-11am and 1pm-4pm. Are you a night-owl? Then head over to Budapest, perhaps on the Budapest Semesters in Mathematics, where the “morning” session runs 4pm-7pm and the “afternoon” session 9pm-midnight!

Websites with the problems usually pop up shortly after the competition is completed; websites with the answers appear a few days later. We’ll post links to those sites in the Comments as we discover them. Previous problems and solutions can be found here.

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Tags: math competitions, Putnam

This entry was posted on December 1, 2007 at 6:28 am and is filed under Competitions. You can follow any responses to this entry through the RSS 2.0 feed.
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December 4, 2007 at 1:21 pm |

The Exam and Solutions are up! Check out the official site.

December 4, 2007 at 1:54 pm |

It seems there was an error in the statement of problem B1:

Let

fbe a polynomial with positive integer coefficients. Prove that ifnis a positive integer, thenf(n) dividesf(f(n)+1) if and only ifn=1.It should have read, “Let

fbe anonconstantpolynomial,” since otherwise the statement is false.December 4, 2007 at 2:11 pm |

Indeed! If

f(x) is the constant functionf(x)=k, then bothf(n) andf(f(n)+1) are equal tok, sof(n) dividesf(f(n)+1) regardless of whatnis equal to.