Newsgroups: sci.math, rec.puzzles, alt.math.recreational, comp.dsp, sci.crypt
From: Michael Soyka <mssr...@gmail.com>
Date: Wed, 31 Oct 2012 22:44:13 -0400
Local: Wed, Oct 31 2012 10:44 pm
Subject: Re: numerical challenge, part 2
On 10/30/2012 10:22 PM, RichD wrote:
> I saw this posed as an interview question:I'm pretty sure the answer is 9.
> What is the sum of the digits of 3 ^ 1000?
To see this, let's rephrase the problem:
9^2 = 81 = 8x10+1 => sum is 9
Now, it is easy to verify that 9 times any single digit 1-9 yields a
9^4 = 9*(7*100+2*10+9) = 63*100+18*10+81.
Each term is a number whose digits sum to 9:
I believe I can safely wave my hands at this point and say the pattern
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