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Message from discussion numerical challenge, part 2

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More options Oct 31 2012, 4:32 am
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:

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

I think this question is ill-posed. Using a modern computer, you can do
this in milliseconds; just use bc on a Unix box or even wolfram alpha:

doing the power with pen&paper.

But slightly different questions *can* actually be answered easily. For
example

===================================
1) what is the *repeated* sum of digits of 3^1000, i.e. sum the digits
to base ten until you have a single digit?

Answer: 3^1000 = 9^500, i.e. the sum of digits must be divisible by 9,
and therefore the repeated sum is 9

2) What is the last digit of 3^1000?

Answer: multiplying can be done mod 10

3^0 = 1 (mod 10)
3^1 = 3 (mod 10)
3^2 = 9 (mod 10)
3^4 = 7 (mod 10)
3^5 = 1 (mod 10)

We have a cycle of length four, 1,3,9,7. 1000=0(mod4), therefore the
last digit is 1.

3) What is the sum of digits of 3^1000 to base 3?

3^1000 = digit one+1000x digit 0 (base 3)
therefore the sum is 1
=======================================

Maybe there exists a shortcut for the original queston, too, but I don't
think so - however, I'm not so strong in number theory to be really sure.

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
near to 4.5 (=(1+2+3+4+5+6+7+8+9+0)/10), this means

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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