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Let $\{P_n\}, n\in \mathbb{N}$ be the sequence prime numbers such that $P_1=2, P_2=3\dots$.

Define a new sequence $\{M_n\}$, $n\in \mathbb{N}$, such that $M_n=$Product of the digits of the $nth$ prime in its decimal representation.

Now the question is: Does there exist an $N\in \mathbb{N}$, such that $M_n=0$ $ \forall n\geq N$ ?

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  • $\begingroup$ This is equivalent to asking there exist a minimum $n$ such that every prime after that contains either a 2,4,6,8 and a 5 in its decimal representation $\endgroup$ – frogeyedpeas May 25 '17 at 17:28
  • $\begingroup$ Ohi misread the quesiton. I thought it was asking if there exists a point such that the last digit is always 0. ( $\endgroup$ – frogeyedpeas May 25 '17 at 23:35
  • $\begingroup$ "at least one 2 or 4 or 6 or 8 And a 5" is what i meant to say to be precise (noting that the products of these digits is 0). $\endgroup$ – frogeyedpeas May 25 '17 at 23:36
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    $\begingroup$ Look at oeis.org/A038618 In particular: "Maynard proves that this sequence is infinite" $\endgroup$ – Lisa May 26 '17 at 21:29
  • $\begingroup$ If you look at other bases, the question is not settled. For example, with binary, the question is equivalent to whether there are infinitely many Mersenne primes, and closely related to whether there are infinitely many even perfect numbers. $\endgroup$ – Robert Soupe May 27 '17 at 23:37
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No. This 2016 paper by James Maynard says that there are infinitely many primes not containing the digit $0$. So $M_n \gt 0$ infinitely often.

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  • $\begingroup$ That is astonishing. Not that it's true, but that somebody was able to prove it. $\endgroup$ – TonyK May 26 '17 at 0:11

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