[seqfan] Re: Digital question

[Vladimir Shevelev]

Let n=(b-1)*b+c, where c is a positive digit in b-base. Then n_1=c*b+b-1=b*n-(b-1)*(b^2-1).
If d does not divide b^2-1 and does divide n, then d does not divide n_1.
 
Regards,
Vladimir

----- Original Message -----
From: David Wilson <davidwwilson at comcast.net>
Date: Wednesday, February 23, 2011 4:38
Subject: [seqfan] Re: Digital question
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Very good. Now show
> 
> (d does not divide b^2-1) => there exists n with d | n but not d 
> | n_1.
> 
> 
> ----- Original Message ----- 
> From: "Vladimir Shevelev" <shevelev at bgu.ac.il>
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Sent: Tuesday, February 22, 2011 7:33 AM
> Subject: [seqfan] Re: Digital question
> 
> 
> > 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
> 
> 
> 
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 Shevelev Vladimir‎