[seqfan] Re: Digital question

[Vladimir Shevelev]

>From b^(t-1)*n-n_1==0 (mod d) we conclude that d | n => d | n_1. Since {n_1 is inv. n}<=>
{n is inv. n_1}, then also b^(t-1)*n_1-n==0 (mod d) and we conclude that d | n_1 => d | n.

Regards,
Vladimir
----- Original Message -----
From: David Wilson <davidwwilson at comcast.net>
Date: Tuesday, February 22, 2011 9:22
Subject: [seqfan] Re: Digital question
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> d | b^2 - 1 => forall n (d | n => d | n') is easy.
> 
> Perhaps the converse is indeed evident, in which case a proof 
> should be no 
> problem. So where is it?
> 
> ----- Original Message ----- 
> From: "Vladimir Shevelev" <shevelev at bgu.ac.il>
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Sent: Sunday, February 20, 2011 4:49 PM
> Subject: [seqfan] Re: Digital question
> 
> 
> > Let us prove that, if d | b^2-1, then d | n iff d | base-b 
> reverse of n.
> >
> > Indeed, let n=sum{i=1,...,t}a_i*b^(t-i), 
> n_1=sum{i=1,...,t}a_i*b^{i-1}. 
> > Then b^(t-1)*n-n_1=
> > sum{i=1,...,t-1}a_i*(b^(2*t-i-1)-b^(i-1))=sum{i=1,...,t-
> 1}a_i*((b^2)^(t-i)-1)*b^(i-1)==0 
> > (mod d).
> > Quite analogously we obtain that n-n_1* b^(t-1)==0 (mod d). 
> Now the 
> > statement follows.
> >
> > The inverse statement is evident: there exist many examples 
> when d divides 
> > n and
> > divide neither b^2-1 nor n_1.
> >
> > Regards,
> > Vladimir
> > 
> 
> 
> -----
> No virus found in this message.
> Checked by AVG - www.avg.com
> Version: 10.0.1204 / Virus Database: 1435/3458 - Release Date: 
> 02/21/11
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎