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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $99$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $999$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $99$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $999$