Question

Using dirichlet's theorem prove that for every positive integer n,there is a prime number whose decimal expansions contains n consecutive 2s and whose final digit is 3

Answer 1

Let x=666⋅10nx=666⋅10n; it has n+3n+3 digits. Consider the interval (x,(1+1/666)x)=(666⋅10n,667⋅10n)(x,(1+1/666)x)=(666⋅10n,667⋅10n). Then the prime number theorem says that there is at least one prime in this interval for sufficiently large xx; such a prime must begin with 666 and is thus satanic.

Concretely, use Schoenfeld's 1976 resultthat says for every x≥2010760x≥2010760 there is a prime in (x,(1+1/16597)x)(x,(1+1/16597)x); we extend this interval to the 1+1/6661+1/666interval above. So there is at least one satanic prime with nn digits for n≥7n≥7, and the result is proved.