A customer went into a pet shop and bought two rabbits plus half the
remaining rabbits. A second customer went into the shop of bought
three rabbits plus one-third of the remaining rabbits. A third
customer went into the shop and bought four rabbits plus one-quarter
of the remaining rabbits. And so on, until it was no longer possible
to continue without splitting rabbits. How many customers went away
satisfied?

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Request for Question Clarification by rainbow-ga on 31 Mar 2005 03:48 PST

Hi ph3123,

You say "until it was no longer possible to continue without splitting rabbits."

Without knowing the number of rabbits, there can be any number of
customers buying rabbits until t was no longer possible to continue
without splitting rabbits.

For example, let's say (since we don't know how many rabbits there
were) there were 68 rabbits in the beginning. The first customer came
and bought 2 plus half the remaining.
He would have bought 68-2=66-33 (which is 1/2 of 66) leaving 33 rabbits.
The next customer would have bought 33-3=30-10 (which is 1/3 of 30)
leaving 20 rabbits.
The 3rd customer would have bought 20-4=16-4 (which is 1/4 of 16)
leaving 12 rabbits.
The 4th customer would not be able to continue without splitting rabbits. (12-5=7).

According to this example, only 3 customers would have gone away satisfied. 

Instead of "and so on, until it was no longer possible to continue
without splitting rabbits", do you perhaps mean until there were no
more rabbits left?

Waiting to hear your views.

Best regards,
Rainbow

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Clarification of Question by ph3123-ga on 31 Mar 2005 16:36 PST

I think you need to think laterally - it is not a mathematics question!

Hmmm?  A researcher gave up on this, but for $2 that is perhaps understandable.
I cannot answer it either, but will accept the condition that the
problem ends with some rabbits, and not that it ends when no rabbits
are left.
But both may be possibilities, making the problem have at least two solutions.
If, however, the question has NOT been incorectly stated  -as
rainbow-ga suggests - then I feel certain that there is no single
answer, mathematically, disregarding the number of rabbits a pet store
could have.

Some one like Math-talk-ga can probably outline the problem in
mathematical terms, but it seems to me that the problem allows for
innumerable solutions, all ending with a remainer of rabbits probably 
as the sum of to prime numbers.
Of course, I am suggesting that the total number of rabbits at the
beginning would be in the thousands, millions  .... (bunnies are like
that), and the sum of the primes (if that should be a correct
assumption), could also be any such, regardless of the actual number.

As I have said somewhere else, I should probably stay away from math questions!

This is really easy,  You can't split haires....

1.  You haven't said what it took to satisfy a customer.

2.  You haven't said whether rabbits bred in the shop.

Therefore your answer is "spinach".

It may be possible to find a maximum number of satisfied customers
(and I suspect that that is 4), but without knowing the number of
rabbits, no specific answer is possible.

For example, if there are only 8 rabbits available, only 2 customers
could be satisfied, but if there were 20, 3 could be satisfied.

Are we to assume an infinite supply of rabbits and seek the maximum
number of satisfiable customers?  Please clarify.

I think Clint's comment is right on!!

Thanks Steph!

Not to split hares, but I think this is a maths problem, but
apparently not one that you can get a two dollar solution for. And it
doesn't seem like it is constrained by the number of rabbits a normal
pet store customer would buy (which painfully suggests that purchasers
could be involved who might have a use for half or a third of rabbit).

The answer depends on the number of original rabbits.  If the original
number was odd then no customers will be satisfied.  However if the
original number was even then the first customer requesting 2 rabbits
would be the only customer leaving satisfied.

Assuming that there is a infinite supply of rabbits, and that we are
looking for the maximum number of customers (satisfied customer being
one that is able to purchase their specified number of whole rabbits),
then the answer is four customers.

The first number of rabbits that would allow four customers to be
satisifed is 80.  With 80 rabbits, the first customer gets 41 rabbits
(2 + 78/2)and 39 remain, the second customer gets 15 (3 + 36/3) and 24
remain, the third customer get 9 (4 + 20/4) and 15 remain, and the
fourth gets 7 (5 + 10/5) and eight remain.  A fifth customer could not
be satisfied as 8-6 cannot be divided by 6 without "splitting hares".

After 80 rabbits, the next quantity that allows four customers to be
satisfied is 140, then 200, then 260, 320, then 380, then 440, and so
on.  Basically, every 60 rabbits after 80 is sufficient to allow four
customers to buy rabbits without splitting rabbits.

Why not a fifth customer then?  Let's look at how many rabbits remain
after the fourth customer leaves.
With 140 rabbits, 20 remain at the end.  
With 200, 32 remain
With 260, 44 remain
with 320, 56 remain

Just  as the starting number of rabbits increases by sixty, the ending
number after four customers increases by 12.  the problem is that 20,
32, 44, 56 (and so on) are not divisible by six, which would be needed
for a fifth customer to come in and purchase their satisifying amount.

There is no quantity of rabbits that would allow five customers to be
satisfied, no matter now large the population, without taking a
fraction of a rabbit.

Bravo, vpolhemus-ga!