That is, is it the case that for every natural number $n$, there is a prime number of $n$ digits? Or, is there some $n$ such that no primes of $n$-digits exist?
I am wondering this because of this Project Euler problem: https://projecteuler.net/problem=37. I find it very surprising that there are only a finite number of truncatable primes (and even more surprising that there are only 11)! However, I was thinking that result would make total sense if there is an $n$ such that there are no $n$-digit primes, since any $k$-digit truncatable prime implies the existence of at least one $n$-digit prime for every $n\leq k$.
If not, does anyone have insight into an intuitive reason why there are finitely many trunctable primes (and such a small number at that)? Thanks!
Best Answer
Yes, there is always such a prime. Bertrand's postulate states that for any $k>3$, there is a prime between $k$ and $2k-2$. This specifically means that there is a prime between $10^n$ and $10\cdot 10^n$.
To commemorate $50$ upvotes, here are some additional details: Bertrand's postulate has been proven, so what I've written here is not just conjecture. Also, the result can be strengthened in the following sense (by the prime number theorem): For any $\epsilon > 0$, there is a $K$ such that for any $k > K$, there is a prime between $k$ and $(1+\epsilon)k$. For instance, for $\epsilon = 1/5$, we have $K = 24$ and for $\epsilon = \frac{1}{16597}$ the value of $K$ is $2010759$ (numbers gotten from Wikipedia).