Tuesday, October 25, 2005

Like a Rat in a Maze

The whole family has been sucked into a math POW (problem of the week) that has been vexing Miriam. We have tried solving the problem in 27 different ways, using color-coded charts, excel spreadsheets, algebraic formulae, playing with fibonacci numbers -- you name it, we've tried it. We always arrive at the same answer. You'd think that would suggest convincingly that we've done it right. But no. Miriam's teacher insists that answer is wrong. Grrr.

Here's the problem:

Two rats, one male and one female, scampered on board a ship that was anchored at a local dock. The ship set sail across the ocean. When it anchored at a deserted island in late December, the two rats abandoned the ship to make their home on the island. Under these ideal conditions, it might be interesting to estimate the number of offspring produced from this pair in one year. You should make these four assumptions:
  • The number of young produced in every litter is six, and three of those six are females.
  • The original female gives birth to six young on January 1 and produces another litter of six 40 days later and every 40 days thereafter as long as she lives.
  • Each female born on the island will produce her first litter 120 days after her birth and then produce a new litter every 40 days thereafter.
  • The rats are on an island with no natural enemies and plenty of food so no rats will die in this first year.
What will be the total number of rats by the following January 1, including the original pair?

We'd love any assistance on this one. Remember to show your work.


Mary said...

Answer: Buy two cats and send them to the island.

Phil Smith said...

Hmmm, this is an interesting problem. Can I ask how far off your answer was from the original? Here are some ideas about how to go about solving it, off the top of my head.

* I would only count females, then at the end multiply by two.
* There will be 10 birth-days on the island in the year of interest, since you can get 9 intervals of 40 days with the last day being around December 26. It might be helpful to list these dates.
* For each birth-day, look at how many children each of these females will have. For example, on Jan 1 there are 3 females born, they will have 7 litters or 21 females each (total of 63 females) by the end of the year. On Feb 9 there will be just 3 females born, they will have 6 litters or 18 females each (total of 54 females). On March 21 there will be just 3 females born (to the original female again, call her Eve-rat?)
* It is very easy to overcount since we have many generations having kids at the same time. It may make sense to look at first births, that is the first time each new generation has a litter. I think that there will be large increases in births on the 4th birth-day, the 7th birth-day, and the 10th birth-day.
* Don't forget that the litters that Eve-rat has on Feb 9, and March 21 will come of age on the 5th birth-day, 6th birth-day, etc. Also don't forget after you count all the females to multiply by two at the end!

Phil Smith said...

One more thing and then I will leave it. If you give the females names based on their age, it might be helpful. For example, call them "newborn", "kid", "teen", and "adult". When they get to be adults, they have newborns (still counting just the females). So on January 1, you have:

January 1: 1 adult, 3 newborns
February 9: 1 adult, 3 kids, 3 newborns
March 21: 1 adult, 3 teens, 3 kids, 3 newborns
April 30: 4 adults, 3 teens, 3 kids, 12 newborns
June 10: 7 adults, 3 teens, 12 kids, 21 newborns
July 20: 10 adults, 12 teens, 21 kids, 30 newborns

My dates aren't quite right but I think you can age them this way. I don't know if there is a series associated with this, perhaps we can call it the Fiboratti series?

BTW, the genus and species of the common rat? Ratus ratus. Just so get the point.

Ron Rienstra said...

Yeah -- thanks for the help, Phil!

Working through your method results in 1808 rats. This is the answer Mia's teacher tells us is incorrect. I am becoming firmer in my conviction that we are correct and Mia's teacher, like Harriet Miers, will need to take a mulligan.

Mary said...

Would you solve this dang problem, post something else, and move the rat picture down the blog?! It freaks me out.