# Is there a method to determine a prime number containing the first n digits?

For example, the number $10243$ is prime and contains the digits '0,' '1,' '2,' '3,' and '4.' Similarly, the number $20143$ is prime. Is there a method to determine whether a prime number exists that contains the first, say, $8$ digits? Or whether a number exists that contains an arbitrary number of digits in order starting from an arbitrary number (say, $3, 4, 5, 6, 7$)?

• Do you want the prime to contain only those digits, each exactly once, or something more relaxed? – JimmyK4542 Jul 30 '15 at 6:35

The number $10235647$ is prime, and contains each of the digits $0$ through $7$ exactly once.
For any number with the digits $3,4,5,6,7$ arranged in some order, the sum of its digits will be $3+4+5+6+7 = 15$, and so, that number will be divisible by $3$, and thus, will not be prime.
Let $d_1,d_2,\dots,d_k$ be an arbitrary sequence of $k$ digits, with $d_k$ odd and not equal to $5$. Let $a=10^k$ and let $b$ be the number with decimal expansion $d_1\dots d_k$. By Dirichlet's Theorem, there are infinitely many primes in the arithmetic progression of numbers of the form $an+b$, so there are infinitely many primes whose decimal expansion ends in $d_1\dots d_k$.
Added: We now look at the "opposite" problem. Let $d_1,d_2,\dots,d_k$ be any sequence of digits, with $d_1\ne 0$. Then there are infinitely many primes whose decimal expansion starts with the digits $d_1,d_2,\dots,d_k$, in that order. The proof is in principle simple, but takes a while to write out. It uses the Prime Number Theorem.