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!!
For any $n \geq 1$, prove that there exists a prime $p$ with at least n of its digits equal to $0$
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number-theory
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0This 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? – 2012-06-23
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3See http://math.stackexchange.com/questions/60825/proof-that-there-are-infinitely-many-prime-numbers-starting-with-a-given-digit-s - start your prime numbers with the string $10^n$. – 2012-06-23
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0(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.) – 2012-06-23
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0@Cocopuffs:you really simplified my problem(some others too).Thanks for the help. – 2012-06-23
2 Answers
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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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1This seems to be the only solution that requires no technology beyond the 18th century! – 2012-06-23
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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.