Newsgroups: sci.math, rec.puzzles, alt.math.recreational, comp.dsp, sci.crypt
From: Christian Gollwitzer <aurio...@gmx.de>
Date: Wed, 31 Oct 2012 09:32:00 +0100
Subject: Re: numerical challenge, part 2
Am 31.10.12 03:22, schrieb RichD:
> I saw this posed as an interview question:
I think this question is ill-posed. Using a modern computer, you can do > What is the sum of the digits of 3 ^ 1000?
this in milliseconds; just use bc on a Unix box or even wolfram alpha: http://www.wolframalpha.com/input/?i=sum+of+digits+of+3^1000
Lacking access to a computer, I'm afraid there is no answer besides But slightly different questions *can* actually be answered easily. For ===================================
Answer: 3^1000 = 9^500, i.e. the sum of digits must be divisible by 9, 2) What is the last digit of 3^1000?
Answer: multiplying can be done mod 10
3^0 = 1 (mod 10)
We have a cycle of length four, 1,3,9,7. 1000=0(mod4), therefore the 3) What is the sum of digits of 3^1000 to base 3?
3^1000 = digit one+1000x digit 0 (base 3)
Maybe there exists a shortcut for the original queston, too, but I don't PS: Here is another one:
4) Estimate the sum of digits of 3^1000
Answer: 3^1000 = 10 ^ (log(3)*1000)
log 3 ~0.477, therefore 3^1000 has ~477 digits. The mean value is nery sum of digits of 3^1000 ~ 4.5*477 = 2146.5
which is astonishingly near to the true answer 2142.
Christian
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