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Why cannot I use Dirichlet's theorem on primes in arithmetic progressions to compute the density of the set of prime whose first digit is 1?

Thank you very much :)

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  • $\begingroup$ Why would you expect to be able to? Numbers whose first digit is '1' don't lie in any sort of arithmetic progression, or arithmetic progressions (in fact, you should be able to easily show that every arithmetic progression takes on all non-zero leading digits infinitely often - try this!). $\endgroup$ – Steven Stadnicki Nov 22 '13 at 18:17
  • $\begingroup$ Having only the condition of the first digit being $1$, you have a union of finite subsets of arithmetic progressions: $10k+j,100k+j,1000k+j,\dots$ for $j\in [0,9],j\in[0,99],j\in[0,999]\dots$. $\endgroup$ – abiessu Nov 22 '13 at 18:18
  • $\begingroup$ (Incidentally, it is possible to prove that there are infinitely many primes with first digit 1 - in any base - but it's a consequence of a much more complicated result and I don't know of a good elementary proof for it.) $\endgroup$ – Steven Stadnicki Nov 22 '13 at 19:03
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Since $\gcd(10,1)=1$, Dirichlet's theorem on primes in arithmetic progression would tell you that there are infinitely many primes of the form $10n+1$, that is, with last digit $1$. However, as Steven Stadnicki said, you cannot characterize a number $n$ as having first digit $1$ by a single linear congruence condition, which is what Dirichlet's theorem gives us to work with.

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