Pattern Math problem! plz help!?

Question

Francisco was raising African Spotted mice. They were unusual in that every two months the female had exactly two babies, a male and a female, when these babies turn two months they continue to have them every two months after that. Francisco began with one pair of spotted mice on Jan. 1, 1998 and his parents made him sell all that he had on Jan 1, 2000, just after more babies were born. Since none of the mice had died, he had quite a few. How many?

i came up with the answer 4096 but i don't think its right.

please help

Answer 1

Hiya...

t(0) = 2

t(1) = 2

t(2) = 4

t(3) = 4

t(4) = 8

t(5) = 8

t(6) = 16

t(6) = 16

and the pattern goes on that every 2 months (the odd months ie. month 3,5,7,9...) the population doubles.

So since there are 25 months from the time he acquires the mice to the time he gives them up then (25-1)/2 gives us the number of times the population has doubled...this is because they don't double in the first month and only every second month...so they've doubled 12 times...

If we only take into account the months where the population doubles then we have:

t(3) = 4 = 2²

t(5) = 8 = 2³

t(7) = 16 = 2^4

t(9) = 32 = 2^5

........etc

so eventually

t(25) = 2^12 = 4096

You were right!!!

(^_^)

Answer 2

Start with the original two. In two months, counting their offspring, there are four. Two months later, the original parents and their offspring have two more each resulting in eight total. The number keeps expanding by powers of 2 (doubling) each two months.

2 -> 4 -> 8 -> 16 -> 32 -> 64 (after 6 2-month periods)

Repeating the doubling for the second year brings the total to 4096.

In math terms n = 2^p. The number of mice is the increase in their population raised to the power of the number of reproductive periods. Since there are 12 2-month periods in the two years, n = 2^12 = 4096.