most recent 30 from http://mathoverflow.net 2013-05-21T07:42:27Z http://mathoverflow.net/feeds/question/50528 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50528/probability-that-a-number-and-its-digit-reversal-are-relatively-prime David Shor 2010-12-28T02:07:02Z 2010-12-28T20:01:33Z

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

http://mathoverflow.net/questions/50528/probability-that-a-number-and-its-digit-reversal-are-relatively-prime/50529#50529 optima 2010-12-28T02:12:34Z 2010-12-28T02:12:34Z

<p>There is some relevant discussion here:</p> <p><a href="http://forums.xkcd.com/viewtopic.php?f=17&amp;t=23136" rel="nofollow">http://forums.xkcd.com/viewtopic.php?f=17&amp;t=23136</a></p>

http://mathoverflow.net/questions/50528/probability-that-a-number-and-its-digit-reversal-are-relatively-prime/50534#50534 Igor Rivin 2010-12-28T03:17:06Z 2010-12-28T03:17:06Z

<p>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.</p>

http://mathoverflow.net/questions/50528/probability-that-a-number-and-its-digit-reversal-are-relatively-prime/50592#50592 Kristal Cantwell 2010-12-28T20:01:33Z 2010-12-28T20:01:33Z

<p>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.</p>