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What is the probability that a number and its digit reversal are relatively prime in base b?

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3 Answers 3

There is some relevant discussion here:

http://forums.xkcd.com/viewtopic.php?f=17&t=23136

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The natural conjecture (stated by someone in @optima's suggested discussion) is that except for the "bad primes" (divisibility by which is "palindromic") the answer should be the "usual" $6/\pi^2.$ For $b=10$ $11$ and $3$ are bad primes, though it is not obvious that there are not others. I am not sure why @optima's answer gets negative feedback, since the discussion is quite relevant, and people there have done numerical experiments consistent with the natural conjecture.

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It wasn't me who downvoted, but my guess is that it is because @optima has a record of spam. See tea.mathoverflow.net/discussion/835/possible-troll/#Item_0 –  Daniel Moskovich Dec 28 '10 at 3:23

I think there are bases $b$ where the probability becomes arbitrarily low. Let $b$ be the product of the first $n$ primes plus one then the difference of $b$ and its palindrome will be divisible by $b-1$ and for them to be relatively prime neither can be divided by the first $n$ primes. So if $n$ is chosen large enough the probability can be made arbitrarily low.

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