Difference between Kaprekar Number and Square is Multiple of Repunit

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Theorem

Let $n$ be a Kaprekar number of $m$ digits.

Then:

$n^2 - n = k R_m$

where:

$R_m$ is the $m$-digit repunit
$k$ is an integer.


Proof

Since $n$ is a Kaprekar number of $m$ digits:

$\begin {cases} n^2 = a \times 10^m + b \\ n = a + b \end {cases}$

for some positive integers $a$ and $b$.


Hence:

\(\ds n^2 - n\) \(=\) \(\ds a \times 10^m + b - a - b\)
\(\ds \) \(=\) \(\ds a \paren {10^m - 1}\)
\(\ds \) \(=\) \(\ds 9 a R_m\)

$\blacksquare$

Sources