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My main question is the generalization, though one can answer the first one and it will get accepted.

  • Are there infinitely many primes involving $3,7$ only?

Generalization: For what sets of given $k$ distinct digits (not all even) from $\{0,1,...,9\}$ where $1\leq k \leq 9,$ there are infinitely many prime numbers involving only these $k$ digits?

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Do you want ALL the digits to be used, or possibly just some of them? –  tomasz Jun 23 '12 at 14:51
    
@tomasz: all the given digits. –  Ehsan M. Kermani Jun 23 '12 at 14:53
    
You probably want $1<k$, because e.g., $333\ldots 333$ is always a multiple of $3$. –  Aaron Jun 23 '12 at 14:54
    
@Aaron: any number written only with $0,3,6,9$ is a multiple of $3$, much the same as with $0,2,4,8$ and $2$, or just $a>1$ and $a$ itself. But I think there are infinitely many primes of the form $1111\ldots1$. In any case, it does not really depend on $k$, rather, on the digits picked. –  tomasz Jun 23 '12 at 14:55
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There are many sets that allow at most one-digit primes: any subset of $\{0,2,4,5,6,8\}$, any subset of $\{0,3,6,9\}$, and any singleton except $\{1\}$. It is not known whether there are infinitely many primes using only the digit $1$, though it is conjectured that this is the case. As @tomasz says, a better question would be which subsets allow the construction of infinitely many primes, though in at least some cases the answer simply isn’t known. –  Brian M. Scott Jun 23 '12 at 15:03
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I note that Primes that contain digits 3 and 7 only are tabulated at the Online Encyclopedia of Integer Sequences. My understanding is that there is no set $D$ of fewer than 10 digits for which it has been proved that there are infinitely many primes which use only the digits in $D$.

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