[This has been cross-posted to MO.]
A positive integer $N$ is said to be a perfect number if
$$\sigma(N) = 2N,$$
where $\sigma(x)$ is the sum of the divisors of $x$. For example, $6$ is perfect since the divisors of $6$ are $1$, $2$, $3$ and $6$, and
$$\sigma(6) = 1 + 2 + 3 + 6 = 12 = 2 \cdot 6.$$
It is currently unknown whether there are any odd perfect numbers. Euler derived the general form
$$M = {p^k}{m^2}$$
that an odd perfect number must take, where $p$ is a prime with $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$. (Even perfect numbers have a similar form, but with $k = 1$ and $p \equiv 3 \pmod 4$.)
Since the sum-of-divisors function is weakly multiplicative, we have
$$\sigma(M) = 2M = \sigma({p^k}{m^2}) = \sigma(p^k)\sigma(m^2) = 2{p^k}{m^2}.$$
Note that $\gcd(p^k, \sigma(p^k)) = 1$ (since prime powers are solitary). Therefore, $p^k \mid \sigma(m^2)$. Consequently, $\sigma(m^2)/p^k \in \mathbb{N}$, where $\mathbb{N}$ is the set of positive integers. Moreover, $\sigma(m^2)/p^k \mid 2M$, so that the odd divisors of $\sigma(m^2)/p^k$ are also the divisors of $M$ (except for $p^k$).
Now my question is this:
Is it possible to prove (or is it evident) that such odd divisors of $M$ coming from $\sigma(m^2)/p^k$ are repdigits (in an odd prime base, other than $10$)? Or (for a stronger statement) can they be repunits (in an odd prime base, other than $10$)?