A boy received 2 hamsters for Christmas. After one year he had 4 after two years he had 12 and after three years he had 48. How many did he have after four years?

Answer 1

Let U{n} be the number of hamsters he has at the start of n years.

The initial state U{1} = 2 = 2 × (1)

U(2} = 4 = 2 × 2 = U{1} × 2 = 2 × (1 × 2)

U{3} = 12 = 4 × 3 = U{2} × 3 = 2 × (1 × 2 × 3)

U{4} = 48 = 12 × 4 = U{3} × 4 = 2 × (1 × 2 × 3 × 4)

It looks like:

U{n} = U{n-1} × n = 2 × (1 × 2 × ... × n)

Now, 1 × 2 × ... × n is known as factorial n or n!

→ U{n} = 2n!

After 4 years (at the start of year 5):

U{5} = 2 × 5! = 2 × 120 = 240

→ he has 240 after 4 years.

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However, as Mehtamatics has shown, this is not the only possible solution for U{n} and he has come up with a different answer to U{5} by using U{n} = (11x³ - 57x² + 100x - 48)/3 which also gives U{1..4} = {2, 4, 12, 48} but U{5} = 134.

For the given data, a formula can be given so that whatever number of hamsters you want for the next year can be found!

eg the formula: U{n} = (-67n^4 + 714x³ - 2573x² + 3750x - 1800)/12

also gives U{1..4} as {2, 4, 12, 48} but gives U{5} = 0 - after 4 years he had none: he was so annoyed at all the hamsters, cleaning them, feeding them, etc that he gave them all away.

There is also assumptions made that are not stated; eg do all the hamsters continue to live (for at least the end of year 4); are the new hamsters births or extras given to him; etc.

Without further information on how his hamsters are behaving, it is impossible to give a single solution.

For example of a proper thought experiment: Leonardo de Pisa (aka Fibonacci) was considering a field and pairs of rabbits. His assumptions were:

1. There was initially a pair of rabbits in the field
2. They are immature for 1 month and thereafter breed to produce another pair of immature rabbits
3. No rabbits die and every month every pair of immature rabbit mature, and every mature pairs produce another immature pair of rabbits

He investigated this to give the number of pairs of rabbits as:

1 mth → 1 (immature) = 1

2 mth → 1 (mature) = 1

3 mth → 1 (mature) + 1 (immature) = 1 + 1 = 2

4 mth → 2 (mature) + 1 (immature) = 2 + 1 = 3

5 mth → 3 (mature) + 2 (immature) = 3 + 2 = 5

And so on: every month the number of pairs of immature rabbits is the same as the number of pairs of mature rabbits in the previous month; and the number of pairs of mature rabbits is the total number of pairs in the previous month.

He was thus able to give a rule: U{n} = U{n-1} + U{n-2} for n > 2, U{1} = U{2} = 1 which is based on the assumptions and not the numbers so far.

Answer 2

The simplest form for the sequence is U(n) = (11*n^3 - 57*n^2 +100*n - 48)/n which gives 134 after four years.

Note that in the first year the number doubled and after that it trebled. The sequence is, therefore, not a simple geometric progression.