Extending Fibonacci with Death

I had a student this year play with Fibonacci, then modify the problem, and give a partial solution to the modified problem. The modified problem is well-known and completely solved. You can try your own hand at it, (see next post). but here’s the student’s story:

Late in the fall of this past year (November 2014) I assigned freshmen the task of taking a problem that we had solved and discussed in class, and proposing a new problem as a modification or extension of the original. Some found it fun, and at least one remembered it later. (I’m sure it was more than this one.)

This February I started a special one-day-a-week class for freshmen (number theory and arithmetic, special topics of their choice, I did this once before).

Nancy (not her real name) worked in a team on Euclid’s algorithm. They did a very nice, very clear presentation, most of the students in the room were able to follow and perform the steps and work out a simple example. And then the team broke up.

Nancy decided to play with Fibonacci on her own. I was a little worried about real-world examples, but she stuck to the traditional “a pair of bunnies is born. In its first month it matures. In each month after that it produces a new pair. And she played it out and let the recursion and the problem statement match up fully. (My Ghost the Bunny is just word play)

And then she got bored, and played what-if. Nancy modified the problem – her bunnies would now have 6 month life spans. She carefully worked out what this would mean: 1, 1, 2, 3, 5, 8 all stay the same, but 13 – 1 = 12, and it gets interesting from there. Nancy identified the quantities that needed to be added (the two previous) and subtracted (six back) but had not written up a recursive formula when the class ended (we only met one lunch period per week, ate before we worked, and homework was not allowed).

But see why I’m excited?  She played with a problem, then posed her own problem?  Because she was curious. Ninth grader. Cool, huh?