I noted a conjecture down a while back after doing a fair few calculations on repdigits and noticing patterns. Define some digit repetition syntax for a 'string of integers' $x$ and an integer $y$, let $x;y$ be $x$ repeated and concatenated $y$ times. For example $$ 123;3 = 123123123\\ 9;10 = 9999999999 $$ then it seemed that for all strings I have tried the following was true:
For a prime $b$, if $\frac{s;a}{b}$ is an integer then $\frac{s;k}{b}$ is also an integer for any $k$ such that $k \bmod (b-1) = a$
This is best shown with some examples, for $s=165$, $a=3$, $b=5$, $$ \frac{s;a}{b} = \frac{165165165}{5} = 33033033 \in \mathbb{Z}\\ \{k:k\bmod(b-1)=a\} = \{k:k\bmod 4 =3\} = \{3,7,11,\cdots\} $$ so we can predict that $$ \frac{165;7}{5}=\frac{165165165165165165165}{5} \in \mathbb{Z} \;\;(true)\\ \frac{165;11}{5}=\frac{165165165165165165165165165165165}{5} \in \mathbb{Z} \;\;(true)\\ $$ and so on.
Moreover, it seems we can concatenate (denoted by the $|$ symbol) some other strings around this pattern take for example \begin{equation} \frac{11222221}{97}= \frac{11|2;5|1}{97} \in \mathbb{Z} \end{equation}
then we have $a=5,b=97,k=5,101,197,\cdots$ \begin{equation} \frac{11|2;101|1}{97} \in \mathbb{Z} \end{equation} (which appears to be true)
This leads to a modified conjecture
For a prime $b$ and strings $A$ and $B$, if $\frac{A|s;a|B}{b}$ is an integer then $\frac{A|s;k|B}{b}$ is also an integer for any $k$ such that $k \bmod (b-1) = a$
This seems to work if the number is a repdigit in any integer base for example in binary $427=110101011_2=(11|01;3|1)_2$ we have $$ \frac{(11|01;3|1)_2}{7} = \frac{427}{7} = 61 \in \mathbb{Z}\\ k = \{3,9,15,\cdots\}\\ \frac{(11|01;9|1)_2}{7} = \frac{1747627}{7} \in \mathbb{Z} \\ \frac{(11|01;15|1)_2}{7} = \frac{7158278827}{7} \in \mathbb{Z} $$
This makes me suspect it is something to do with the pattern rather than the numeric content per se.
Questions:
Can the conjecture be proved true or false? Can a counter example be found?
Why does this work/not work?
Cool ideas!
OK, so your initial conjecture is true. Here's why. Using your notation, $s;a$ is $\sum_{i=0}^as(10^n)^i$ for some $n$ ($n$ being the number of digits in $s$). If $b$ is either $2$ or $5$, this is a multiple of $b$ if and only if $s$ is a multiple of $b$, and the conjecture isn't terribly interesting. So let's concentrate on $b$ not $2$ or $5$ - and $b$ not a factor of $s$, because again, not terribly interesting.
For any $k$, $s;k - s;a = \sum_{i = a + 1}^ks(10^n)^i$. A consequence of Fermat's Little Theorem states that if $c$ is relatively prime to a prime $p$, then $c^n \mod p$ takes every value from $1$ to $p - 1$ as $n$ ranges from $1$ to $p - 1$. In particular, if $k = a + (b - 1)$, then $(10^n)^i$ takes every value from $1$ to $b - 1$ as $i$ ranges from $a + 1$ to $k$. Now, mod $b$, every nonzero value has an inverse - so multiplying by $s$ just permutes these numbers. So $\sum_{i = a+1}^ks(10^n)^i$ is, modulo $b$, just $\sum_{i = 1}^{b - 1}i$. This is $\frac{b(b - 1)}{2}$. Since $b$ is prime and not $2$, $\frac{b - 1}{2}$ is an integer, so $\frac{b(b-1)}{2}$ is a multiple of $b$. In other words, $\sum_{i = a+1}^ks(10^n)^i \equiv \sum_{i=1}^{b-1}i \equiv 0\mod b$. Since $s;k - s;a \equiv 0 \mod b$, if $s;a$ is divisible by $b$ then so is $s;k$.
This doesn't prove your modified conjecture - but it seems to me that you should be able to modify this proof to do the trick.