# Prove that there are infinitely many primes with $666$ in their decimal representation without Dirichlet's theorem.

A satanic prime is a prime number with $666$ in the decimal representation.

The smallest satanic prime is $6661$.

Prove that there are infinitely many satanic primes.

I used Dirichlet's theorem for the progression $10000n+6661$ and it is done.

I'm interested in solutions without Dirichlet's theorem.

• Dirichlet is a big hammer, but sometimes you need a big hammer :) – Thomas Andrews Sep 22 '17 at 14:28
• I suspect that, in a reasonable sense, almost all primes are satanic – Henry Sep 22 '17 at 14:43
• I'd like to throw out here a (I think much harder) question: Are there infinitely many non-satanic primes? – Nate Sep 22 '17 at 15:33
• @Nate Yes. Any prime with no 6 cannot be satanic. Maynard proved in 2016 that there are infinitely many primes with no 6 in their decimal expansions. – Parcly Taxel Sep 22 '17 at 17:16
• Is this true in general for any natural number? – jdods Sep 26 '17 at 19:04

## 2 Answers

Let $x=666\cdot10^n$; it has $n+3$ digits. Consider the interval $(x,(1+1/666)x)=(666\cdot10^n,667\cdot10^n)$. Then the prime number theorem says that there is at least one prime in this interval for sufficiently large $x$; such a prime must begin with 666 and is thus satanic.

Concretely, use Schoenfeld's 1976 result that says for every $x\ge2010760$ there is a prime in $(x,(1+1/16597)x)$; we extend this interval to the $1+1/666$ interval above. So there is at least one satanic prime with $n$ digits for $n\ge7$, and the result is proved.

• Thanks! If I understand correctly, your argument can prove that for any fixed digits there is a prime with these digits at the beginning, am I right? – tong_nor Sep 22 '17 at 15:15
• @tong_nor Yes. For any $\epsilon>0$ and sufficiently large $x$ the prime number theorem guarantees at least one prime in $(x,(1+\epsilon)x)$. – Parcly Taxel Sep 22 '17 at 15:17
• I suspect Schoenfeld's result is a bigger hammer than Dirichlet, in a way. – wythagoras Sep 23 '17 at 11:06
• @wythagoras These are Bertrand's postulate-type arguments. I happened to think of them first because I used them in my 14/21 interesting numbers answer... – Parcly Taxel Sep 23 '17 at 11:16

Consider the set $S$ of all numbers without 666 in their base 10 expression. Here's a fun fact: the sum $\sum_{s\in S} \frac{1}{s}$ converges. It's actually pretty easy to prove, so I'll leave it as an exercise (or google "Kempner series").

On the other hand, a famous result of Euler says the sum of the reciprocals of the prime numbers diverges.

• Most Elementary answer yet. Also consider numbers to base 1000 with the omitted digit being 666. – marty cohen Sep 22 '17 at 15:29
• "Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away." —Antoine de Saint-Exupéry – Erick Wong Sep 22 '17 at 15:54
• Current Wikipedia articles with details: Kempner series and divergence of the sum of the reciprocals of the primes – Jeppe Stig Nielsen Sep 23 '17 at 8:00