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It seems obvious that infinite many primes can be formed only using two distinct digits.

$67776767776667777777$ is an example for such a prime. Even if we allow only the digits $0$ and $1$, there are $2^{n-2}$ numbers with $n$ digits, for which the first and last digit is $1$. So, we can expect that some of those numbers are prime for every sufficiently large number $n$.

Can it be proven that infinite many primes contain only two distinct digits in the decimal representation ?

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    $\begingroup$ Similar questioh at math.stackexchange.com/questions/162042/…, which is answered by GerryMyerson: the status is not known. $\endgroup$ – Aravind Nov 24 '15 at 14:49
  • $\begingroup$ Thank you for pointing out this very negative result. $\endgroup$ – Peter Nov 24 '15 at 15:44
  • $\begingroup$ I recommend you set up an Internet alert on this page: oeis.org/A020469 $\endgroup$ – Robert Soupe Nov 27 '15 at 18:24

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