[seqfan] Re: Digital question

[Vladimir Shevelev]

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

----- Original Message -----
From: David Wilson <davidwwilson at comcast.net>
Date: Sunday, February 20, 2011 12:21
Subject: [seqfan] Digital question
To: math-fun <math-fun at mailman.xmission.com>, Sequence Fans <seqfan at list.seqfan.eu>

> Let d > 0 divide 99. I can then show that d divides n iff d 
> divides the base-10 reverse of n.
> 
> I believe the converse is true as well, that these are the only 
> d with this property.
> 
> In general, I believe that
> 
> ( d | n <=> d | base-b reverse of n ) <=> d | b^2-1
> 
> Supposing this is indeed true, it can't be too hard to prove, 
> but I'm at a loss.
> 
> 
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 Shevelev Vladimir‎