[High School - Precalc] Mouse Problem - easy shortcut I am missing?

[aTzUeLo1191]

I'll try and sum up the problem so I don't have to type up a long paragraph:

2 mice are on an island with unlimited food and no predators. They have 6 children on the first day, 3 M, 3 F, and reproduce the same every 40 days after that. Children reproduce after 120 days, every 40 days. How many mice will there be after 1 year?

Is there a really simple shortcut I am missing? A problem like this is kinda tough to write out every stage at a time.

[tha Guardians]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Are mice able to inbreed?
Your shortcut has been found xD

[aTzUeLo1191]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

So yeah, there is a ****load of inbreeding going on, and eventually some mice will be born w/o the proper parts to reproduce, but seriously, I'm looking for a shortcut to check my answer.

[tha Guardians]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

So yeah, there is a ****load of inbreeding going on, and eventually some mice will be born w/o the proper parts to reproduce, but seriously, I'm looking for a shortcut to check my answer.

Well I might not have helped you (unless it was a trick question), but I think I helped bring attention to your thread, which in turn may lead you to the answer.

That's right.
You're welcome.

jk

[igotrhythm]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Instead of thinking in terms of the number of mice, try thinking in terms of the number of breeding pairs of mice. Besides making the numbers smaller and thus making it easier to recognize patterns (you mentioned that it was getting tedious doing it step by step), it also leads into discussion of one of my favorite mathematical sequences, which was derived from a very similar problem.

At time 0, there is 1 breeding pair plus 3 immature pairs. The first mature breeding pair kicks out another 3 immature pairs at t=40 and t=80, giving 1 breeding pair plus 9 immature pairs at 80 days.

However, at t=120 days, we have 2 breeding pairs, and mice being the animals they are, they immediately get to work and pop out another 3 immature pairs. So we have 2 mature pairs and 15 immature. At t=160, there's 3 mature and 24 immature, and so on.

Since 1 year is approximately t=360 in this system, we have as follows (white text alert):
Number of mature breeding pairs: 1, 1, 1, 2, 3, 5, 8, 13, 21, 34
Number of immature breeding pairs: 3, 6, 9, 15, 24, 39, 63, 102, 165, 267

If the sequence of numbers of mature breeding pairs for given values of t looks familiar, the pattern should help you with how I derived the number of immature pairs.

[dooey100]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

I'm assuming that by shortcut you mean not brute forcing all the numbers.

This is like the Fibonacci Sequence, where every number is the sum of the two before it. There is a formula for finding any number in the sequence, but if your teacher hasn't given it too you, I doubt they expect you to use it. Here it is:

Note that it works if the series starts 0, 1, 1, 2, 3, 5 so you would have to modify the function a bit to fit your problem.
(the swirl symbol = phi = (1+sqrt(5))/2 = about 1.618)

[QED Stepfiles]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

So, basically, the easiest "non-brute force" way of solving this is to first define the system recursively (i.e. how many mice are present at t_{n+1} as a function of how many mice there are at t_{n} and t_{n-1} and so on), and then to "decouple" the recursion (i.e. be able to write it as just a function of "n") using eigenvalues. Since this is a high school precalculus course, though, that would not really be an option...

In other words, I'm pretty sure you are meant to brute force this problem. If you want to know how to do this problem without just plugging in the numbers, take a linear algebra course in the future. But for now, I think you may have to just be satisfied with being able to enter in numbers into your calculator =/.

[MeaCulpa]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

You don't really have to do accounting every 40 days. Here's how I would interpret this (sorry if I screwed something up):

1 female gives birth 10 times: 0 40 80 120 160 200 240 280 320 360.

The 3 females born on day 0 give birth 7 times: 120 160 200 240 280 320 360.

The 3 females born on day 40 give birth 6 times: 160 200 240 280 320 360.

The 3 females born on day 80 give birth 5 times: 200 240 280 320 360.

There are 6 females born on day 120, each giving birth 4 times: 240 280 320 360.

9 females were born on day 160, each giving birth 3 times: 280 320 360.

12 born on 200, giving birth twice: 320 360.

15 born on 240, giving birth once: 360.


(1 x (10 x 6)) + (3 x (7 x 6)) + (3 x (6 x 6)) + (3 x (5 x 6)) + (6 x (4 x 6)) + (9 x (3 x 6)) + (12 x (2 x 6)) + (15 x (1 x 6))

[aTzUeLo1191]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Ok, thanks for the help all, but I don't really understand the first 3 methods, and the answer for the fourth is too small.

I basically made a chart that had adults, children born that phase, 40 day old children, and 80 day old children. Then I just moved the numbers down the line, with the adults multiplying by 3 to get the newborn children (2 adults per 6 children). I got 1764, so if anyone else got that number, sweet.

[igotrhythm]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

I don't know what one of us is doing wrong...I ended up with only 602. >_<

[aTzUeLo1191]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

I think you are forgetting that the children have children, then those children have children, but the first set still has more children.

3 different classes got the same problem. I've probably asked about 15-20 people, and no 2 answers were alike. It's much trickier than it seems.

[Gilly G]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Mea's method seems correct when elaborated as such. You end up with 924, seems reasonable.

[MeaCulpa]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Oh right, my method is alright, but I undercounted starting day 240. My answer comes out to a little more than yours, though:

The two mice in the beginning = 2
That first female gives birth 10 times = 60
3 give birth 7 times starting on day 120 = 3 x 7 x 6 = 126
3 give birth 6 times starting on day 160 = 3 x 6 x 6 = 108
3 give birth 5 times starting on day 200 = 3 x 5 x 6 = 90

On day 240 though, the 3 females born from the "Adam and Eve" mice, as well as the 9 females born on day 120 (whose 'grandparents' were born on day 0) give birth 4 times each (240, 280, 320, 360)...so 12 x 4 x 6 = 288.

There were a total of 21 females born on day 160, so there are 21 x 3 x 6 = 378 new mice from females who can start giving birth on day 280.

30 females born on day 200, so 30 x 2 x 6 = 360 new mice from females starting on day 320.

Finally, a total of 66 females were born on day 240 (3+9+9+9+36), so 66 x 1 x 6 = 396 mice from females that could being reproducing on day 360.

Add 'em up: 2 + 60 + 126 + 108 + 90 + 288 + 378 + 360 + 396 = 1806 on the 365th day.

Again, if I didn't explain something well, let me know.

[trumaestro]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Exponential growth. Should be able to sub into this formula , where a is the starting amount, b is the rate of growth, and T is the time it takes to achieve the growth rate.
Your question isn't very clear, but I think you should be able to use the formula with 2 for a, 3 for b (as the population triples), and one of either 40 or 120 for T. Put 365 for x and solve for t using logs or graph intersection.
That's the method that makes most sense to me.

[emerald000]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

We would have to use recursion mathematics. Let's be m_n the number of mices at generation n and f_n the number of reproducting females.

Let be m_0=2 and f_0 = f_1 = f_2 = 1.

So, the number of mices m_n is given by:

m_n = m_(n-1) + 6 f_(n-1) for n=1,2,3,4,...

and the number of reproductive females by:

f_n = f_(n-1) + 3 f_(n-3) for n=3,4,5,...

So, in a year, there would be 9 generations. With a little recursion, you can arrive at m_9=974 with f_9=139.

I guess I'm pretty much on the mark, but feel free to bash my theory to death.