, 2010

As the prime numbers under consideration have at least two digits, the last digit can be only $1$, $3$, $7$, or $9$. Thus $n$ is prime-prone if and only if, among numbers $10n + 1$, $10n + 3$, $10n + 7$, and $10n + 9$, at least three are primes.

If $n = 3k$, then $10n + 3 = 30k + 3$ and $10n + 9 = 30k + 9$ are divisible by $3$ and hence composite. If $n = 3k + 2$, then $10n + 1 = 30k + 21$ and $10n + 7 = 30k + 27$ are divisible by $3$ and hence composite again. Consequently, all prime-prone integers are congruent to $1$, and hence to each other, modulo $3$. Thus they differ by a multiple of $3$, i.e., by at least $3$.