Multiple of Repdigit Base minus 1
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 $m \in \Z_{>1}$ be an integer such that $1 < m < b$.
Then $m \times n$, when expressed in base $b$, is of the form:
- $m n = \sqbrk {r d d \cdots d s}_b$
where:
- $d = b - 1$
- $r = m - 1$
- $s = b - m$
- there are $k - 1$ occurrences of $d$.
Generalization
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.
Proof
| \(\ds n\) | \(=\) | \(\ds \sum_{j \mathop = 0}^{k - 1} \paren {b - 1} b^j\) | Basis Representation Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds b^k - 1\) | Sum of Geometric Sequence | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds m n\) | \(=\) | \(\ds m \paren {b^k - 1}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {m - 1} b^k + b^k - 1 + 1 - m\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {m - 1} b^k + \sum_{j \mathop = 0}^{k - 1} \paren {b - 1} b^j + \paren {1 - m}\) | Sum of Geometric Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {m - 1} b^k + \sum_{j \mathop = 1}^{k - 1} \paren {b - 1} b^j + \paren {b - 1} + \paren {1 - m}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {m - 1} b^k + \sum_{j \mathop = 1}^{k - 1} \paren {b - 1} b^j + \paren {b - m}\) |
which is exactly the representation $\sqbrk {r d d \cdots d s}_b$ as defined.
$\blacksquare$
Examples
$4 \times 999$ base $10$
- $4 \times 999$ (base $10$) is equal to $3996$.