# 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:

 $\displaystyle 10x + y - 10 y - x$ $=$ $\displaystyle 9 \left({x - y}\right)$ 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:

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

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

Hence the result.

$\blacksquare$