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# palindromic number

An integer $n$ that in a given base $b$ is a palindrome. Given $n$ being $k$ digits $d_{x}$ long in base $b$, its value being

$n=\sum_{{i=0}}^{{k-1}}d_{k}b^{i}$ |

then if the equalities $d_{k}=d_{1}$, $d_{{k-1}}=d_{2}$, etc., hold, then $n$ is a palindromic number. There are infinitely many palindromic numbers in any given base.

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11A63*no label found*

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## Comments

## a note about 11| { }

Let k,n denote positive integer numbers. It is easy to see that the sequences a_k=10^{2k-1}+1 and b_n=10^{2n}-1 are divisible by 11. Hence 11|a_k+b_n. From this, it is easy to prove that 11 divides every even length palindromic number.

perucho

## Re: a note about 11| { }

Good point, but I'd prefer to generalize 11 to b + 1.

## Re: a note about 11| { }

That's correct Wilfredo, but my note arose due to a suggestion that drini did to akrowne in his entry ``palindrome''. So now you can construct (if you want it) a general proof.

Cheers,

Pedro

## Re: a note about 11| { }

Thanks.

So how do I create a link to the proof (the one about ELPNs)? It shows up as an attachment, but I'd like to link to it in the body text, too.

Wil

## Re: a note about 11| { }

Thanks to you! It was your idea!

No problem; I already gave you full access to the entry and you can make the changes that you consider convenient.

Also, I have checked $(ELPN)_n$ (Eq.(6)) for a binary base and 6-length. So,

$b=2$, $n=3$. I took: $b_0=b_1=1$,$b_2=0$. By using $3=1.2^0+1.2^1=11$, I did the multiplication $10001x11$ and I got the Even Length Palindromic Number=$(ELPN)_3= 110011$. If you like you may add this example.

perucho

## Re: a note about 11| { }

Thanks. I'll use that access carefully and judiciously. I'm at work right now but I'll be pondering your new example later.