Digits of Permutable Prime

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Theorem

Let $p$ be a permutable prime with more than $1$ digit.

Then $p$ contains only digits from the set:

$\left\{ {1, 3, 7, 9}\right\}$


Proof

First note that from 3-Digit Permutable Primes, both $337$ and $199$ are permutable primes.

Hence it follows that all the elements of $\left\{ {1, 3, 7, 9}\right\}$ appear in at least one permutable prime.


Let $p$ contain an even digit $d$.

Then at least one anagram $p'$ of $p$ has $d$ at the end.

From Divisibility by 2, it follows that $p'$ is even.

As $p$ has more than $1$ digit it follows that $p'$ is composite.


Let $p$ contain the digit $5$.

Then at least one anagram $p'$ of $p$ has $5$ at the end.

From Divisibility by 5, it follows that $p'$ is divisible by $5$.

Except for $5$ itself, all integers which are divisible by $5$ are composite.

As $p$ has more than $1$ digit it follows that $p'$ is composite.


The result follows by definition of permutable prime.

$\blacksquare$