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To prove the following fact: given any sequence of digits in any base, eg 314159265358979323 base 10, there are infinitely many primes that start with these digits,eg when expressed in decimal they start with 314159265358979323. I think using a prime number theorem will be helpful . But I am still not getting a point of how to start the proof.

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    The "Algorithms" in the title doesn't seem appropriate.2011-08-30
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    @Sri: Any suggestions for a better title? You could propose one by clicking edit, and users with sufficient rep can approve it if it's hunky-dory.2011-08-30
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    Note that "$n$ starts with $314159265358979323$" is equivalent to "there exists $k \in \mathbb N$ such that $314159265358979323 \cdot 10^k \le n < 314159265358979324 \cdot 10^k$". Can you show that the union of these intervals for all $k$ must contain infinitely many primes?2011-08-30
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    I am a little suspicious that two people asked the same question less than an hour apart. Can you give the source of the question?2011-08-30
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    @Sri: I certainly didn't read it as curt; I was merely saying that you can always suggest better titles (via the edit button) for questions whose titles are a bit short in the quality department... :)2011-08-30
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    Related: https://math.stackexchange.com/questions/24404822018-11-28

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