Multiple of Repdigit Base minus 1/Generalization

Theorem

Let $b \in \Z_{>1}$ be an integer greater than $1$.

Let $n$ be a repdigit number of $k$ instances of the digit $b - 1$ for some integer $k$ such that $k \ge 1$.

Let $m \in \N$ be an integer such that $1 \le m \le b^k$.

Then $m \times n$, when expressed in base $b$, is the concatenation of $m - 1$ with $b^k - m$, that is:

$m n = \sqbrk {\paren {m - 1} \paren {b^k - m} }_b$

If $b^k - m$ has less than $k$ digits, leading zeros are added to it until it has $k$ digits.

$\ds n$

$=$

$\ds \sum_{j \mathop = 0}^{k - 1} \paren {b - 1} b^j$

$\ds $

$=$

$\ds b^k - 1$

$\ds \leadsto \ \ $

$\ds m n$

$=$

$\ds m \paren {b^k - 1}$

$\ds $

$=$

$\ds \paren {m - 1} b^k + b^k - m$

which is exactly the representation $\sqbrk {\paren {m - 1} \paren {b^k - m} }_b$ with leading zeroes for $b^k - m$, as defined.

$\blacksquare$

In base $10$, $972 \times 999 = 971 \, 028$.

Hence $971 = 972 - 1$ and $028 = 10^3 - 972$.