A really hard math question.doubt you can get it.?

Question:Two rats, one boy and one girl are on an island.
-the number of young produced in every littler is 6, 3boys 3girls
-the original girl gives birth to another 6 every 40days
-each girl on the island will produce her first babies 120 days after she is born, and then every 40days from then on.
-the rats on the island with no enemy and plenty of food, so none will die in the first year.

Assuming the original girl starts at day 0 and does not give birth on that day:

Original girl has 9 litters of 6, an additional 54 rats to the two already on the island.

Of those 54, 27 will be girls.

The first three of these girls will birth again at day 160 in the year, to 6 more rats, a total of 18, and then another 5 times throughout the year to 18 rats each. So now you've got 6 * 18 for an additional 108 rats.
Just keep following these additional litters now. The first of these litters will birth at day 280 in the year, to 6 rats each, for 9*6 on that day alone, an additional 54 rats. These same rats will then pop another 54 out two more times in the year, so they will kick out an additional 162.
Now, these rats rat's will not have a chance to birth, so you're done with this tree now, a total of
2 + 54 +108 + 162

However now's where it gets complicated. You have to do the same kind of process for each of the litters that our original female kicked out.
The second 3 girls that OrigF kicked out begin their breeding at day 200 (80th they were born, 120th they birth) and will do the same as above, they birth 6 each, and do so again 4 more times throughout the year, so you've got 18*5, an additional 90.
But you only have to consider the first two of THOSE rats births, as the first females (9) birth 6 on day 320, another 54 rats, and again on day 360, a total of 108 more rats.
Total: 162 + 90 + 108 = 360

Moving on to the third set birthed by OrigF;
These are born on day 160, and will birth on day 280. Same thing again, 3 females, 6 each, another 18 rats. These females will be able to do this only two more times, for a total of 18*3 = 48 additional rats. None of their litters will have time to birth.
360 + 48 = 448.

And now the 4th set;
Born on day 200, birth on day 320 to 18 rats, will only birth once more for 36 additional rats
448 + 36 = 484.

And the 5th set;
Born on day 240, birth on day 360 (just in time) to 18 rats.

484 + 18 = 502.

That should do it!
Too darn many!
You're right, cant get it. I like working on math problems and all, but I hate probem solving. Hate it, hate it, hate it. Sorry I couldnt be of much help.
I'm working on it no my computer, I'll post as soon as I am done for you. This one is GREAT!!

UPDATE: Here it is
A pciture on my flickr account because the format here doesn't like tabs http://www.flickr.com/photos/18955982@n0...
i answered this already
because you asked it again
but i will answer again...without all the really hard math

2+972+8748+26244= 36020 rats

it all makes sense and isnt very hard.
you just have to take the time to do it.

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