The puzzle of the honest pirates
in [Math]
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From: T.H. Ray on 13 Nov 2009 22:29 Dan wrote > 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 D'oh! You're right. Maybe some day I'll learn arithmetic. :-) Tom
From: Herman Jurjus on 16 Nov 2009 07:01 Bart Goddard wrote: > Herman Jurjus <hjmotz(a)hetnet.nl> wrote in news:hdj8cq$37l$1(a)news.eternal- > september.org: > >> The puzzle itself is not very exciting; but it was accompanied by the >> remark that this puzzle can be elegantly solved using /eigenvalues/. >> >> My question: what could have been meant with that remark? > > Assuming that the problem has one left over coconut at each stage: > > Eigenvalues are fixed points. If there are x coconuts in the pile > before a pirate messes with it, then there are f(x) = 4/5(x-1) > coconuts after he messes with it. The easiest solution is the > _fixed point_ of this function, x=-4. A fixed point is an eigenvector with eigenvalue 1; ok - that could explain it. (Although the feeling keeps nagging that Halmos didn't write 'eigenvalue' for nothing.) > Then argue that all integer > solutions must be congruent modulo 5^<something>. How? The function f(.) is a rational function. How does anything about integer solutions follow from this? Still puzzled... > So I guess his point was "invariance." > > Bart Thanks for your answer! -- Cheers, Herman Jurjus
From: Bart Goddard on 17 Nov 2009 10:35
Herman Jurjus <hjmotz(a)hetnet.nl> wrote in news:hdrev9$ude$1(a)news.eternal-september.org: >> Then argue that all integer >> solutions must be congruent modulo 5^<something>. > > How? The function f(.) is a rational function. How does anything about > integer solutions follow from this? Still puzzled... I think this is the original article. http://www.springerlink.com/content/6417952g535uu625/ B. -- Cheerfully resisting change since 1959. |