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Message from discussion numerical challenge, part 2
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Michael Soyka  
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 More options Oct 31 2012, 10:44 pm
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
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On 10/30/2012 10:22 PM, RichD wrote:
> I saw this posed as an interview question:

> What is the sum of the digits of 3 ^ 1000?

> --
> Rich

I'm pretty sure the answer is 9.

To see this, let's rephrase the problem:
What is the sum of the digits of 9 ^ 500?

9^2 = 81 = 8x10+1 => sum is 9
9^3 = 9*(8*10+1) = 72*10+9 = 7*100+2*10+9 => sum is 9

Now, it is easy to verify that 9 times any single digit 1-9 yields a
number whose digits will sum to 9.

Continuing:

9^4 = 9*(7*100+2*10+9) = 63*100+18*10+81.

Each term is a number whose digits sum to 9:
   63: 6+3=9
   18: 1+8=9
   81: 8+1=9
and so the sum of the digits of 9^4 is 9 as well (recall that 9 times
any single digit yields a number whose digits sum to 9).

I believe I can safely wave my hands at this point and say the pattern
of reasoning is clear and the conclusion is unarguable.  :)
Nonetheless, a mathematically rigorous proof escapes me at the moment
but perhaps in time ...


 
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