Re: problem of arithmetic modulo

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
1322070819480806636890455259752144365965422032752148167664920368226828597346704
89954077831385060806196390977769687258235595095458210061891186534272525795367402
76202251983208038780147742289648412743904001175886180411289478156230944380615661
73054086674490506178125480344405547054397038895817465368254916136220830268563778
58229022841639830788789691855640408489893760937324217184635993869551676501894058
8109060426089671438864102814350385648747165832010614366132173102768902855220001,

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.

--
Helmut Richter
.

Relevant Pages

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