Can a repunit number be a perfect number?

Question

Can a repunit number be a perfect number?

I think a repunit number can be a perfect number, since the last digit of an odd perfect number can be $1$.

Am I correct?

Answer 1

The answer is NO, a repunit number cannot be a perfect number. (Edit: The following assumes base-$10$. If base-$5$ is considered, then as pointed out in a comment by Oscar Lanzi, ${11}_{5} = {6}_{10}$ is perfect and a repdigit.)

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To see why, consider an odd perfect number $N$, assuming such a number exists.

Euler showed that $N$ must have the so-called Eulerian form $$N = q^k n^2$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$.

Since $N > {10}^{1500}$ as shown by Ochem and Rao (2012), then in particular $N$ cannot be equal to $1$. This means that, under the assumption that $N$ is a repunit number, then $N$ has the form $$N = 100\overline{N} + 11$$ whence $$N \equiv 3 \pmod 4$$ holds. This contradicts the Eulerian form, since $N$ must be congruent to $1$ modulo $4$.