Prime Repdigit Number is Repunit

Theorem

Let $b \in \Z_{>0}$ be an integer greater than $1$.

Let $n \in \Z$ expressed in base $b$ be a repdigit number with more than $1$ digit.

Let $n$ be prime.

Then $n$ is a repunit (in base $b$).

Proof

Let $n$ be a repdigit number with $k$ digits.

Then by the Basis Representation Theorem:

$\ds n = \sum_{j \mathop = 0}^k m b^j$

for some $m$ such that $1 \le m < b$.

Let $m \ge 2$.

Then:

$\ds n = m \sum_{j \mathop = 0}^k b^j$

and so has $m$ as a divisor.

Hence $n$ is not prime.

The result follows by the Rule of Transposition.

$\blacksquare$