# [seqfan] Re: Digital question

Vladimir Shevelev shevelev at bgu.ac.il
Tue Feb 22 13:33:49 CET 2011

```>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,
----- 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,
> >
>
>
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