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For any $n \geq 1$, prove that there exists a prime $p$ with at least n of its digits equal to $0$. I don't even know how to start?? Any help(even a hint) would do. Thanks in advance!!

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This is not the kind of thing that is easy to prove directly be elementary means. However there are results about the density of prime numbers that will easily prove such existence results (without producing an example). What kind of answer are you after? –  Marc van Leeuwen Jun 23 '12 at 9:16
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See math.stackexchange.com/questions/60825/… - start your prime numbers with the string $10^n$. –  Cocopuffs Jun 23 '12 at 9:17
    
(Parenthetical: Using Dirichlet's theorem, given an $n$ we can pick an $m$ with $n$ digits equal to $0$ that is coprime to $10$, and there will exist (infinitely many) primes $p$ with residue $m$ modulo some power of $10$ larger than $m$, which proves the result. Pretty high-tech though.) –  anon Jun 23 '12 at 9:19
    
@Cocopuffs:you really simplified my problem(some others too).Thanks for the help. –  Aang Jun 23 '12 at 9:25
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2 Answers

up vote 9 down vote accepted

The series $\sum_{p \in \mathbb P} \frac{1}{p}$ of primes diverges (proof). Show that the series $\sum_{n \in A} \frac{1}{n}$ converges, where $A$ is the set of integers with at most $k$ zeroes (modify this proof). Therefore $\mathbb P \not \subset A$.

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This seems to be the only solution that requires no technology beyond the 18th century! –  Erick Wong Jun 23 '12 at 9:34
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By Dirichlet's theorem on primes in arithmetic progressions there are infinitely many primes congruent to $1$ mod $10^{n+1}$. These primes have at least $n$ $0$'s in them.

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