Mind-boggling property of a prime As you have already probably known, an Emirp is a prime whose reversal give a different prime i.e: 37 is an Emirp because 37 is prime and its reversal 73 is also a prime, 79 is also an Emirp. Now I have this 100 digit Emirp with a mind-boggling property (!!!): 31399 71973 78663 47113 91448 65157 72694 85891 75941 91229 38744 59187 76569 25789 74797 49143 19422 88961 13739 39731, write down that 100 digit Emirp into a 10-by-10 square, and you will find that all rows, columns, and the 2 main diagonals are all Emirps (!!!) and distinct (!!). I call such numbers a Magical Emirp. My question is: Are there only finitely many magical Emirps?
 A: A very heuristic argument says that there should be an infinite number.  If we assume the primes are randomly distributed with $p$ having $\frac 1{\ln p}$ chance of being prime, we can ask what is the chance there is a Magical Emirp with $n^2$ decimal digits.  You need two numbers of $n^2$ digits and $4n+2$ numbers of $n$ digits to be prime.  We will take all the probabilities for $n$ digits to be $\frac 1{\ln 10^{n-1/2}}=\frac 1{(n-1/2)\ln 10}$  For $n$ large, we can ignore the $1/2$ and say the chance is $\frac 1{n \ln 10}$  This says the chance a random number of $n^2$ digits is a Magical Emirp is about $\frac 1{n^{4n+2}(n^2)^210^{2n+2}}$ As there are about $10^{n^2}$ numbers with $n^2$ digits we would expect $\frac {10^{n^2}}{n^{4n+2}(n^2)^210^{2n+2}}$ with $n^2$ digits.  This grows without bound, so we would expect more and more of them each square number of digits.  They still become rarer and rarer, so finding them is rather difficult.  Plugging in $n=10$, there should be $10^{32}$ hundred digit ones.  That is a lot, but a small fraction of the hundred digit numbers.
