**News flash: **correct solutions (**9** edges) were received by the deadline from Donny Huang, Kristen Layden, David Moore, Ralph Morrison, Wei Sun, and Sharon Ron’s brother, all eligible for the prize drawing at colloquium Monday, November 30, 1 pm. Morrison solution.

The Turkey prize gift basket went to Donny Huang, presented by Jo Proctor for the Office of Public Affairs. At math colloquium November 30, 2009, Prof. Botts pulled Huang’s name from Prof. Beeson’s hat from among the Special Thanksgiving Conundrum (see solution) solvers Huang, Kristen Layden, David Moore, Ralph Morrison, Wei Sun, and Sharon Ron’s brother. Prize gift basket from Where’d You Get That. Mentioned in Berkshire Eagle account of Dodd Thanksgiving dinner. Colloquium continued with Prof. Johnson on “Sex Ratios.”

[Adapted from Martin Gardner’s *The Colossal Book of Short Puzzles and Problem**s*]

Suppose you have a four-by-four array of square cages, such that contiguous cages share their edges and each cage contains exactly one fat turkey awaiting its Thanksgiving fate! The problem is to remove the smallest number of cage-edges that will break the perimeter of every square of caging material. That is, remove a minimal set of edges from the array so as to break the perimiter of each of the sixteen one-by-one squares, each of the nine two-by-two squares, the four three-by-three squares and the one large four-by-four square that is the outside border—30 squares in total! Then PROVE your result is minimal. How many turkeys will escape? If you answer this, you will have a wonderful Thanksgiving!

If you think you can solve this special Thanksgiving conundrum send Professor Beeson your solution and, if you are correct, the smug satisfaction of being correct is all yours!

Correct answers received by November 28 will be put into a lottery and a winner drawn, to be announced here and in Daily Messages on Monday, November 30, to be awarded at math colloquium, 1-1:45 pm, Bronfman 106. The Office of Public Affairs is providing a prize gift basket from Where’d You Get That?

Prof. Morgan will answer questions and provide hints at the Thanksgiving Day lunch at Dodd, 11:30 am – 1:30 pm.

If you still haven’t had enough you can try to answer the same question for a five-by-five square of cages or an n-by-n square for any n. How about if you have to break the perimeter of every rectangle? Gardner notes that he knows of no work done on any of these later questions!

The October conundrum was to show that if f is a real valued function satisfying

f(x^{3} + y^{3}) = (x+y) [f(x)^{2} – f(x)f(y) + f(y)^{2}]

then f(2009x)=2009f(x) for all real x.

A solution assuming f(x) is continuous was sent in by Daniel Phelps, the author of last month’s solution and a local high school student! Congratulations, Daniel! Here is his solution, edited slightly:

First I take the given f(x^{3} + y^{3}) = (x+y) [f(x)^{2} – f(x)f(y) + f(y)^{2}] and set x = y = 0 to get f(0)=0. Next I set y equal to 0 to get f(x^{3}) = xf(x)^{2}. So f(x) = x^{1/3} f(x^{1/3})^{2} = x^{1/3} x^{2/9} f(x^{1/9})^{4} = …. We get in total that, if f is continuous, f(x) = x f(1), in other words f(x) is linear. Thus, f(2009x) = 2009f(x).