The original question Let $S_b(n)$ be the sum of the digits of $n$ when $n$ is expressed in base $b$. was asked by Anson Chan on Jan. 24, 2019. Since it was deemed to not have enough context, it was closed & deleted. As I thought it was an interesting question (& had provided the only answer), I unsuccessfully tried to reopen it. I am presenting an adjusted version of it here, along with my updated answer.
For a fixed integer $b \ge 2$ and for any positive integer $n$, let $S_b(n)$ be the sum of the digits of $n$ when expressed in base $b$. For example, $S_4(26) = S_4(1 \cdot 4^2 + 2 \cdot 4 + 2 \cdot 1) = 1 + 2 + 2 = 5$.
Determine all positive integers $m$ with the property that for all $n$, whenever $m$ is a factor of $S_b(n)$, them $m$ is also a factor of $S_b(n + 1) - 1$.
This indicates, to me, a property where the remainder of an integer and the sum of its digits are the same. In particular, with base $10$, a number, when divided by $3$ or $9$, has the same remainder as the sum of the digits. This occurs due to $3$ and $9$ dividing $10 - 1 = 9$. Thus, in general, it seems it will work with any factors of $b - 1$ since $S_b(n + 1) - 1$ would have the same remainder as $S_b(n) + 1 - 1 = S_b(n)$. How do you prove this, plus also find any other cases which may work, or show there are no other such values?