Solution

Let F(x) be the expected number of bands from fusing the ends from x strips of rubber. Obviously, F( 1)= 1. If x1, after you fix one end and choose the second, there is a 1/(2x- 1) chance that these will be ends of the same strip and (2x- 2)/(2x- 1) chance that they won’t.

In the first case, you’ll have a rubber band and x- 1 strips, which means 1 + F(x- 1) expected bands. In the second case, you’ll have x- 1 strips, and F(x- 1) expected bands. Therefore, F(x) = (1/(2x- 1))( 1 + F(x- 1)) + ((2x- 2)/(2x- 1)) F(x- 1)

= F(x- 1) + 1/(2x- 1)

= 1 + 1/3 + 1/5 + 1/7 + ... + 1/(2x- 1) .

For sufficiently large x, F(x) is approximated by 0.635181423 + 0.5ln(2x- 1). Solving F(x)= 10, then, gives x = 68,100,150. Given that it takes a couple of minutes to make a band, Lou effectively (or ineffectively) can’t produce 10 a week.

Solvers
Christopher Allard, Steve Altschuld,
Robert Bartholomew, Rowen Bell,
Andy Blough, Bob Byrne,
Michael Captain, Samantha Casanova,
Andrew Dean, Rob Easton,
David Ellsworth, David Engelmayer,
Bernie Erickson, Mark Evans,
Yan Fridman, Renaud Guilbert,
Rui Guo, Ben Kester, Eric Kovach,
Alex Kozmin, Douglas Levy,
Mike Logan, David Lovit, Tim Luker,
James McGuire, Lee Michelson,
Geoff Moak, Ryan Morgan,
Paul Navratil, Don Onnen,
Patrick Ouyang, David Promislow,
Bruce Rosner, Edmund Scheuermann,
Craig Schmid, Bill Scott, Noam Segal,
Tomasz Serbinowski, Lenny Shteyman,
Walter Shur, John Snyder,
Don Sondergeld, Al Spooner, Glen Stark,
Donn Takebayashi, Jon Turnes,
David Uhland, Michael Vech
Apologies to Robert Bartholomew,
who was left off of the list of solvers for
“National Mathematics Year,” because of
my clerical error.

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