Re: problem of arithmetic modulo

In article
<Pine.LNX.4.63.0801221602130.7864@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Helmut Richter <hhr-m@xxxxxx> wrote:

On Tue, 22 Jan 2008, Cat wrote:

Date: Tue, 22 Jan 2008 15:49:40 +0100
From: Cat <Ct@xxxxxxxxx>
Newsgroups: sci.math
Subject: Re: problem of arithmetic modulo


Rainer Rosenthal <r.rosenthal@xxxxxx> a écrit dans le message :
5vmbh9F1mnj3lU1@xxxxxxxxxxxxxxxxxxxxx
Bill wrote:

but this is not the case with modulo 7.
10^0=1[7], 10^1=3 [7], 10^2=2[7], 10^3=6, 10^4=4, 10^5=5, 10^6=1 and so
on
[modulo 7]

Please write some more terms of the "and so on". Won't be that random as
you seem to expect. Draw your conclusions.

Regards,
Rainer

I know it's not random.
But how do you compute c modulo 7 with c =sum(a_i)
where a_i are the digits of the numbre 3^1000?
It does'nt seem as straightforward as with c modulo 3.
I even wonder if it's possible to solve the problem.

The unfair solution is to compute 3^1000 which is
13220708194808066368904552597521443659654220327521481676649203682268285973467
04
899540778313850608061963909777696872582355950954582100618911865342725257953674
02
762022519832080387801477422896484127439040011758861804112894781562309443806156
61
730540866744905061781254803444055470543970388958174653682549161362208302685637
78
582290228416398307887896918556404084898937609373242171846359938695516765018940
58
810906042608967143886410281435038564874716583201061436613217310276890285522000
1,

then add up its digits, giving c=2142, then computing the desired values:

c = 0 (mod 2)
c = 0 (mod 3)
c = 0 (mod 7)
c = 0 (mod 17)

Conjecture: c = 0 (mod p) for all prime p.

I trust the conjecture is a joke.

I would suggest the OP calculate the sum mod 7 of the decimal digits
of 1, 3, 9, 27, 81, 243, ..., and see whether there is a pattern and
report back to us. I'm guessing there won't be much of a pattern.

3^n has roughly (log-base-10-of-3) n digits. The average digit
is 4-and-a-half, it's plausible that holds for the digits in 3^n,
in which case the sum of the digits in 3^n should be about
(4.5) (log_10 3) n. But it must be a multiple of 9, so I'd bet on
the multiple of 9 closest to (4.5) (log_10 3) n. As n increases,
every so often that will go up by 9, so the residue mod 7 will go up
by 2, or down by 5. But it won't be so regular.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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