Repdigit Number consisting of Instances of 9 is Kaprekar

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Theorem

A repdigit number that consists entirely of the digit $9$ is a Kaprekar number.


Proof

We note as examples:

\(\ds 9^2\) \(=\) \(\ds 81\)
\(\ds 8 + 1\) \(=\) \(\ds 9\)
\(\ds 99^2\) \(=\) \(\ds 9801\)
\(\ds 98 + 01\) \(=\) \(\ds 99\)
\(\ds 999^2\) \(=\) \(\ds 998001\)
\(\ds 998 + 001\) \(=\) \(\ds 999\)


Now we consider:

\(\ds \paren {\sum_{k \mathop = 0}^n 9 \times 10^k}^2\) \(=\) \(\ds \paren {10^{n + 1} - 1}^2\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds 10^{n + 1} \paren {10^{n + 1} - 2} + 1\)

So $\paren {10^{n + 1} - 1}^2$ can be split as:

$\paren {10^{n + 1} - 2} + 1 = 10^{n + 1} - 1 = \ds \sum_{k \mathop = 0}^n 9 \times 10^k$


Thus $\ds \sum_{k \mathop = 0}^n 9 \times 10^k$ is a Kaprekar number.

$\blacksquare$


Sources