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Message from discussion numerical challenge, part 2

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More options Nov 5 2012, 7:08 pm
Newsgroups: sci.math, rec.puzzles, comp.dsp
From: Randy Yates <ya...@digitalsignallabs.com>
Date: Mon, 05 Nov 2012 19:08:14 -0500
Local: Mon, Nov 5 2012 7:08 pm
Subject: Re: numerical challenge, part 2

glen herrmannsfeldt <g...@ugcs.caltech.edu> writes:
> In comp.dsp Willem <wil...@turtle.stack.nl> wrote:
>> christian.bau wrote:
>> ) On Oct 31, 8:32?am, Christian Gollwitzer <aurio...@gmx.de> wrote:
>> )> 4) Estimate the sum of digits of 3^1000

>> )> ? ? ? ? sum of digits of 3^1000 ~ 4.5*477 = 2146.5

>> )> which is astonishingly near to the true answer 2142.

>> ) The last digit is 1 (Powers of 3 end in 3, 9, 7, 1, 3, 9, 7, 1 etc.);
>> ) taking that into account would have made it 2143.

>> Taking into account that it has to be a multiple of 9,
>> you can round that to 2142. But that's not proof that it is 2142.
>> Printing the first 30 digit sums doesn't show any pattern either.

> Not a mathematical proof, but you can use statistical arguments
> to show how likely it is to be correct.

> If you can assume that the digits are uncorrelated, then the
> standard deviation of the sum is about 4.5/sqrt(477),
> for about 0.2 That makes it extrememly unlikely to be
> off by more than one.

> The lowest and highest digits aren't so uncorrelated, though.

> It would take more math than I know to show the digit correlation
> is low enough, though.

Given the number of intelligent people (glen, I mean people like you)
who have viewed this problem and the dearth of solutions, I'd say this
is a crappy interview question. Something like, "Solve Fermat's Last
Theorem."
--
Randy Yates
Digital Signal Labs
http://www.digitalsignallabs.com