Newsgroups: sci.math, rec.puzzles, alt.math.recreational, comp.dsp, sci.crypt
From: Wally W. <ww8...@aim.com>
Date: Wed, 07 Nov 2012 00:59:59 -0500
Local: Wed, Nov 7 2012 12:59 am
Subject: Re: numerical challenge, part 2
On Tue, 6 Nov 2012 21:42:36 -0800 (PST), RichD wrote:What do you mean, "Finally?"
>On Nov 5, Willem <wil...@turtle.stack.nl> wrote:
>>> 4) Estimate the sum of digits of 3^1000
>3 ^ 100
>> Taking into account that it has to be a multiple of 9,
>Finally, someone is on the right track.
This characteristic was noted a week ago:
From: Christian Gollwitzer <aurio...@gmx.de>
Answer: 3^1000 = 9^500, i.e. the sum of digits must be divisible by 9,
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