Off on a Tangent A Fortnightly Electronic Newsletter from the Hope College Department of Mathematics
 November 1, 2010 Vol. 9, No. 5 http://www.math.hope.edu/newsletter.html

 The next two weeks bring two more colloquium opportunities

 Title:  My life as an actuary at Towers Watson Speaker: Bryan Heiser, Towers Watson Time:  Tuesday, November 2 at 4:00 p.m. Place:  VWF 104

Abstract:  I will talk about my experience as an actuary at Towers Watson in Chicago to help students learn about the day-to-day life of an actuary. I will also talk about student preparation, exams and internships. There will be plenty of time for student questions.

 Title:  Counting with Generating Functions Speaker: Dr. Brian Drake, Grand Valley State University Time:  Wednesday, November 10 at 4:00 p.m. Place:  VWF 104

Abstract:  Many types of counting problems are best solved using series expansions of functions.  We will see some basic examples, including how to count tilings with squares and dominos.  Then we will consider some results which allow us to turn difficult counting problems into easy ones by finding a function's inverse.

 Math in the News

 What do Justin Bieber, Monet's Water Lillies, and Euler's identity (ei(π)+1 =0) have in common?  There are people who think each one of these is a thing of beauty.  NPR's Adam Frank asked the question, “Why are so many mathematically inclined folks sent into paroxysms of delight over this string of symbols which seem like gibberish to others?" in his commentary.  To find the answer to this question click here to see the full post of this column.

 Problem Solvers of the Fortnight

 We had the following problem in our last problem of the fortnight:  For what values of n is n3 - 9n2 + 20n divisible by 6? Solution: The polynomial factors into n(n-4)(n-5).  One of (n-4) or (n-5) is even, so the polynomial is divisible by 2 for any natural number n.  One of (n-4), (n-5), (n-6) is divisible by 3.  Since, if (n-6) is divisible by 3, it follows that n is divisible by three, so n(n-4)(n-5) is divisible by both 2 and 3 and thus by 6 for any natural number. The following students gave correct solutions: Erica Budge, Cornelius Smits, and David Dolfin.

 Problem of the Fortnight

 For which natural numbers, n, is (2n+1)(2n-1) divisible by 3? Write a complete solution (not just an answer) in triplicate and drop it off in the Official Problem of the Fortnight Slot outside VWF 212 by 3:00 pm on Wednesday, November 10.  As always, be sure to include your name, the name(s) of your professor(s), and your math class(es) -- e.g. Drew A. Blank, Dr. Jean Poole, Math 333 -- on your solution.

I know it's a rare privilege, but if one can really tackle something in adult life that means that much to you, then it's more rewarding than anything I can imagine.

Andrew Wiles

 Off on a Tangent