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From: T.H. Ray on 13 Nov 2009 03:03
Dan wrote

Herman wrote

> > Years ago I read a puzzle (I think it was in a
> book
> > by Halmos, but I'm
> > not sure) about five honest pirates who had stolen
> a
> > number of coconuts.
> > Before going to sleep, they made the arrangement
> to
> > meet in the morning
> > and divide the coconuts evenly among them.
> > But during the night, one pirate woke up, took 1/5
> of
> > the coconuts, and
> > left; then another woke up, took 1/5 of whatever
> he
> > found present (not
> > knowing about what the first had done), and so did
> > the other three.
> > Puzzle: assuming that no pirate had to chop pieces
> > off any coconut,
> > what's the least possible number of coconuts left
> > over at the end, and
> > with (at least) how many did they start?
> >
> > The puzzle itself is not very exciting; but it was
> > accompanied by the
> > remark that this puzzle can be elegantly solved
> using
> > /eigenvalues/.
> >
> > My question: what could have been meant with that
> > remark?
> >
> > --
> > Cheers,
> > Herman Jurjus
> >
>
> I have no idea but is the total number of coconuts
> stolen = 5^5 ?
> Leaving the least amount of 1024 coconuts after each
> pirate took his 1/5 of the remainder.
>
> Dan

Right idea, wrong answer. The key is that the starting
number and every subsequent division by 5 must end in
the integer 5. Thus, the original number of 3125 (5^4)
divides 1)625, 2)125 3)25 4)5 5)1. So there are 4
coconuts left over (which if the monkey version that
Frederick Williams mentioned were used, the remainder
would be zero.

So far as the eigenvalue solution goes, it would be
worth investigating. But not for me, at the present
moment.

Tom
From: Herman Jurjus on 13 Nov 2009 14:25
Mensanator wrote:
> On Nov 13, 3:20 am, Herman Jurjus <hjm...(a)hetnet.nl> wrote:
>> Years ago I read a puzzle (I think it was in a book by Halmos, but I'm
>> not sure) about five honest pirates who had stolen a number of coconuts.
>> Before going to sleep, they made the arrangement to meet in the morning
>> and divide the coconuts evenly among them.
>> But during the night, one pirate woke up, took 1/5 of the coconuts, and
>> left; then another woke up, took 1/5 of whatever he found present (not
>> knowing about what the first had done), and so did the other three.
>> Puzzle: assuming that no pirate had to chop pieces off any coconut,
>> what's the least possible number of coconuts left over at the end, and
>> with (at least) how many did they start?
>
> The real question is: why are they called "honest"?

Because they didn't take more than 1/5?

BTW, one of the respondents mentioned -4 as a solution.
Stealing -4 coconuts surely counts as /extremely/ honest?
Unless the coconuts were a form of chemical waste, of course.

--
Cheers,
Herman Jurjus

From: dan73 on 13 Nov 2009 04:39
Tom wrote:

>Right idea, wrong answer. The key is that the starting
>number and every subsequent division by 5 must end in
>the integer 5. Thus, the original number of 3125 (5^4)
>divides 1)625, 2)125 3)25 4)5 5)1. So there are 4
>coconuts left over (which if the monkey version that
>Frederick Williams mentioned were used, the remainder
>would be zero.

>So far as the eigenvalue solution goes, it would be
>worth investigating. But not for me, at the present
>moment.

>Tom

By what you are saying I must be interpreting
the original problem wrong but the way I see it
is ---

First pirate takes 1/5 of 5^5 coconuts = 625
3125 - 625 = remaining 2500
Second pirate takes 1/5 of 2500 = 500
2500 - 500 = remaining 2000
Third pirate takes 1/5 of 2000 = 400
2000 - 400 = remaining 1600
Fourth pirate takes 1/5 of 1600 = 320
1600 - 320 = remaining 1280
Fifth pirate takes 1/5 of 1280 = 256
1280 - 256 = remaining 1024

This I believe is the smallest remainder possible
given the restrictions by the OP.

Remember no fractional coconuts allowed.;-)

Dan
From: Mike Terry on 13 Nov 2009 14:57
"T.H. Ray" <thray123(a)aol.com> wrote in message
news:43611801.12623.1258135454546.JavaMail.root(a)gallium.mathforum.org...
> Dan wrote
>
> Herman wrote
>
> > > Years ago I read a puzzle (I think it was in a
> > book
> > > by Halmos, but I'm
> > > not sure) about five honest pirates who had stolen
> > a
> > > number of coconuts.
> > > Before going to sleep, they made the arrangement
> > to
> > > meet in the morning
> > > and divide the coconuts evenly among them.
> > > But during the night, one pirate woke up, took 1/5
> > of
> > > the coconuts, and
> > > left; then another woke up, took 1/5 of whatever
> > he
> > > found present (not
> > > knowing about what the first had done), and so did
> > > the other three.
> > > Puzzle: assuming that no pirate had to chop pieces
> > > off any coconut,
> > > what's the least possible number of coconuts left
> > > over at the end, and
> > > with (at least) how many did they start?
> > >
> > > The puzzle itself is not very exciting; but it was
> > > accompanied by the
> > > remark that this puzzle can be elegantly solved
> > using
> > > /eigenvalues/.
> > >
> > > My question: what could have been meant with that
> > > remark?
> > >
> > > --
> > > Cheers,
> > > Herman Jurjus
> > >
> >
> > I have no idea but is the total number of coconuts
> > stolen = 5^5 ?
> > Leaving the least amount of 1024 coconuts after each
> > pirate took his 1/5 of the remainder.
> >
> > Dan
>
> Right idea, wrong answer. The key is that the starting
> number and every subsequent division by 5 must end in
> the integer 5. Thus, the original number of 3125 (5^4)
> divides 1)625, 2)125 3)25 4)5 5)1. So there are 4
> coconuts left over (which if the monkey version that
> Frederick Williams mentioned were used, the remainder
> would be zero.

I don't get this - Dan's answer looks right to me. Each pirate during the
night would multiply the pile by 4/5, so together their effect of 5 pirates
would be to multiply by 4^5 / 5^5. So to get whole coconuts we must start
with a multiple of 5^5, and the smallest starting value would be just 5^5,
then at the end we would have 4^5 = 1024. I can't see where your 4 is
coming from (even given your sequence of divisions, whose significance I
don't understand...)

Regards,
Mike.




From: Frederick Williams on 14 Nov 2009 06:38
Mike Terry wrote:
>
> "T.H. Ray" <thray123(a)aol.com> wrote in message
> news:43611801.12623.1258135454546.JavaMail.root(a)gallium.mathforum.org...
> > Dan wrote
> >
> > Herman wrote
> >
> > > > Years ago I read a puzzle (I think it was in a
> > > book
> > > > by Halmos, but I'm
> > > > not sure) about five honest pirates who had stolen
> > > a
> > > > number of coconuts.
> > > > Before going to sleep, they made the arrangement
> > > to
> > > > meet in the morning
> > > > and divide the coconuts evenly among them.
> > > > But during the night, one pirate woke up, took 1/5
> > > of
> > > > the coconuts, and
> > > > left; then another woke up, took 1/5 of whatever
> > > he
> > > > found present (not
> > > > knowing about what the first had done), and so did
> > > > the other three.
> > > > Puzzle: assuming that no pirate had to chop pieces
> > > > off any coconut,
> > > > what's the least possible number of coconuts left
> > > > over at the end, and
> > > > with (at least) how many did they start?
> > > >
> > > > The puzzle itself is not very exciting; but it was
> > > > accompanied by the
> > > > remark that this puzzle can be elegantly solved
> > > using
> > > > /eigenvalues/.
> > > >
> > > > My question: what could have been meant with that
> > > > remark?
> > > >
> > > > --
> > > > Cheers,
> > > > Herman Jurjus
> > > >
> > >
> > > I have no idea but is the total number of coconuts
> > > stolen = 5^5 ?
> > > Leaving the least amount of 1024 coconuts after each
> > > pirate took his 1/5 of the remainder.
> > >
> > > Dan
> >
> > Right idea, wrong answer. The key is that the starting
> > number and every subsequent division by 5 must end in
> > the integer 5. Thus, the original number of 3125 (5^4)
> > divides 1)625, 2)125 3)25 4)5 5)1. So there are 4
> > coconuts left over (which if the monkey version that
> > Frederick Williams mentioned were used, the remainder
> > would be zero.
>
> I don't get this - Dan's answer looks right to me. Each pirate during the
> night would multiply the pile by 4/5, so together their effect of 5 pirates
> would be to multiply by 4^5 / 5^5. So to get whole coconuts we must start
> with a multiple of 5^5, and the smallest starting value would be just 5^5,
> then at the end we would have 4^5 = 1024. I can't see where your 4 is
> coming from (even given your sequence of divisions, whose significance I
> don't understand...)

In the version that I know, one coconut is left over after each division
(it is given to a monkey). That allows the starting number to be -4.
I'm using the word "allows" rather loosely here:-).

--
Which of the seven heavens / Was responsible her smile /
Wouldn't be sure but attested / That, whoever it was, a god /
Worth kneeling-to for a while / Had tabernacled and rested.
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