Product with Repdigit can be Split into Parts which Add to Repdigit

Theorem
Let $n$ be a positive integer with $d_1$ digits.

Let $m$ be a repdigit number with $d_2$ digits such that $d_2 > d_1$.

Let $r$ consist of the result when the rightmost $d_2$ digits of $m n$ is cut off and added to the remaining left hand portion.

Then $r$ is a repdigit number.

Proof
Let $b > 1$ be the base we are working on.

Let $m = \sqbrk {aaa \dots a}_b$.

Let $R = \dfrac m a = \sqbrk {111 \dots 1}_b = \dfrac {b^{d_2} - 1} {b - 1}$.

Let the rightmost $d_2$ digits of $m n$ be $y$ and the remaining left hand portion be $x$.

Then we have:

so $r$ is divisible by $R$.

It remains to show that:
 * $r \le b^{d_2} - 1 = \paren {b - 1} R$

because $\paren {b - 1} R$ is the largest $d_2$-digit integer.

This condition forces $r$ to have $d_2$ digits, so it remains a repdigit number.

Since $r$ is divisible by $R$, this is equivalent to:
 * $r < b R$

We have:

Hence the result.