# Kaprekar's Process for 2-Digit Numbers

## Theorem

Kaprekar's process, when applied to a non-repdigit $2$-digit positive integer leads to the cycle:

$09 \to 81 \to 63 \to 27 \to 45 \to 09$

Note that it is important to retain the leading zero on the $9$, or the process trivially terminates in $0$.

## Proof

Let $n \in \Z_{>0}$ be a $2$-digit positive integer.

Without loss of generality, let $n$ be expressed in decimal notation as:

$n = 10 x + y$

where:

$x > y$
$0 \le x \le 9, 0 \le y \le 9$

The reversal of $n$ is $10 y + x$.

We have:

 $\ds 10x + y - 10 y - x$ $=$ $\ds 9 \left({x - y}\right)$ where $1 \le x - y \le 9$

Thus after the first iteration of Kaprekar's process, $n$ is one of:

$09$
$18$
$27$
$36$
$45$

or their reversals.

Applying Kaprekar's process to $09$ gives:

 $\ds 09$ $=$ $\ds 90 - 09$ $\ds$ $=$ $\ds 81$ $\ds$ $\to$ $\ds 81 - 18$ $\ds$ $=$ $\ds 63$ $\ds$ $\to$ $\ds 63 - 36$ $\ds$ $=$ $\ds 27$ $\ds$ $\to$ $\ds 72 - 27$ $\ds$ $=$ $\ds 45$ $\ds$ $\to$ $\ds 45 - 54$ $\ds$ $=$ $\ds 09$

All those multiples of $9$ can be seen to be in (or end up in) this loop.

Hence the result.

$\blacksquare$