MHB What are the palindromic primes with an even number of digits?

[Ackbach]

Here is this week's POTW:

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A palindromic number is one that is the same when written backwards (base 10) like 197791. Find all the palindromic primes that have an even number of digits.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to johng, greg1313, Deveno, kaliprasad, and Kiwi for their correct solutions! It's always nice to see a great level of participation. There were a number of good solutions, but I'll just post greg1313's:

Spoiler

Claim: all palindromic numbers with an even number of digits are divisible by $11$, hence the only prime palindromic number with an even number of digits is $11$.

Proof:

$10^n\equiv10\pmod{11}$ if $n$ is odd and $10^n\equiv1\pmod{11}$ if $n$ is even.

$10^{n-1}\cdot a+10^{n-2}\cdot b+10^{n-3}\cdot c+\dots+10^2\cdot c+10\cdot b+ a\equiv x\pmod{11}$

$10a+b+10c+\dots+c+10b+a\equiv x\pmod{11}$

$10(a+b+c\dots)+(a+b+c\dots)\equiv x\pmod{11}$

$(10+1)(a+b+c\dots)\equiv x\pmod{11}$

$11(a+b+c\dots)\equiv0\pmod{11}$

$\text{Q.E.D.}$