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

[MeaCulpa]

Hmm...I agree with emerald's f_9 value:

1 female initially (reproduces 10 times)
3 born on 0 (each reproduce 7 times)
3 born on 40 (each reproduce 6 times)
3 born on 80 (each reproduce 5 times)
12 born on 120 (each reproduce 4 times)
21 born on 160 (each reproduce 3 times)
30 born 200 (each reproduce 2 times)
66 born on 240 (each reproduce once)
___
139 reproductive females by day 360.

But I don't think 974 is high enough of a value, considering the number of times each can reproduce:

(1 x (10 x 6)) + (3 x (7 x 6)) + (3 x (6 x 6)) + (3 x (5 x 6)) + (12 x (4 x 6)) + (21 x (3 x 6)) + (30 x (2 x 6)) + (66 x (1 x 6)) = 1806.

[igotrhythm]

Haha, this is good stuff...all the different solutions we've come up with. XD

You should show your teacher this on Monday and give out credits to anyone that ends up being right.

[aTzUeLo1191]

Bumping w/ answer.

Teacher finally gave it back today, one or two people in our class got it right - 1808. Sadly, no one here did, and neither did I.

I didn't get a method on how he or anyone else found it, but if you all want, I could ask him.

Edit: Damn I just noticed Mea was pretty close...did you forget to add the two parents?

[MeaCulpa]

Oh, yeah I guess I did. Just looking at my calculation in my last post, I calculated the total progeny without Adam and Eve. I guess it wouldn't hurt to find out how exactly he solved it, and if there is a proper shortcut after all.

[Commandersa1]

I would like to know the actual process in solving this. :/

[Patashu]

Originally Posted by 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.

Looks like an iterative function to me. A pair of mice produce three reproducing pairs that kick in three steps later, so let's have f_n counting the number of reproducing mice pairs and g_n counting the total number of mice at any given 40 day interval.
f_0, f_1, f_2 = 1
g_0 = 8

f_n = f_(n-1) + 3*f_(n-3) (the mice pairs from exactly 120 days ago are considered breeders from this term on)
g_n = g_(n-1) + 6*f_n (calculate this after calculating f_n)

f_3 = 4
f_4 = 7
f_5 = 10
f_6 = 22
f_7 = 43
f_8 = 73
f_9 = 139

g_1 = 14
g_2 = 20
g_3 = 44
g_4 = 86
g_5 = 146
g_6 = 278
g_7 = 536
g_8 = 974
g_9 = 1808

So after 360 days there are 1808 mice. After 365.251 days there will still be 1808 mice

[QED Stepfiles]

Ah ****... I forgot that we start at time zero, not at time one...

If you take the matrix that I have in my post and take it to the 7th power rather than the 6th, you get the correct answer.

Whoops...

[emerald000]

And using m_10 with my recursion formula. At least I know my technique works (and the one Patashu posted since it is the exact same one).