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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $297$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $297$