Re: problem of arithmetic modulo

On Tue, 22 Jan 2008 15:49:40 +0100, Cat wrote:

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.
Regards
Bill

You have

3^0 == 1 (mod 7)

and also

3^6 == 1 (mod 7).

Can you conclude anything about 3^12?

Do you see a way to generalize?


--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
.

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