Reversal formed by Repdigits of Base minus 1 by Addition and Multiplication

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Theorem

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

Let $n = b^k - 1$ for some integer $k$ such that $k \ge 1$.


Then:

$n + n$ is the reversal of $\paren {b - 1} n$

when both are expressed in base $b$ representation.


Proof

By Power of Base minus 1 is Repdigit Base minus 1, $n$ is a repdigit number consisting of $k$ occurrences of $b - 1$.

Let $a = b - 1$.

Thus $n$ can be expressed in base $b$ as:

$n = {\overbrace {\sqbrk {aaa \cdots a} }^k}_b$


We have that:

\(\ds n + n\) \(=\) \(\ds 2 n\)
\(\ds \) \(=\) \(\ds \sqbrk {1aa \cdots ac}_b\) Multiple of Repdigit Base minus 1: $k - 1$ occurrences of $a$, and $c = b - 2$

where there are:

$k - 1$ occurrences of $a$
$c = b - 2$.


Then:

\(\ds \paren {b - 1} \times n\) \(=\) \(\ds a n\)
\(\ds \) \(=\) \(\ds \sqbrk {caa \cdots a1}_b\) Multiple of Repdigit Base minus 1

where there are:

$k - 1$ occurrences of $a$
$c = b - 2$.

Hence the result.

$\blacksquare$


Examples

Reversal of $9 + 9$

\(\ds 9 + 9\) \(=\) \(\ds 18\)
\(\ds 9 \times 9\) \(=\) \(\ds 81\)


Reversal of $99 + 99$

\(\ds 99 + 99\) \(=\) \(\ds 198\)
\(\ds 9 \times 99\) \(=\) \(\ds 891\)


Reversal of $7 + 7$ Base $8$

\(\ds 7 + 7\) \(=\) \(\ds 16_8\)
\(\ds 7 \times 7\) \(=\) \(\ds 61_8\)


Historical Note

David Wells reports in Curious and Interesting Numbers of $1986$ that this result was discussed in Volume $10$ of Journal of Recreational Mathematics by D.Y. Hsu, but further corroborating evidence for this has not yet been found.


Sources

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