Kaprekar's Process on 3 Digit Number ends in 495

Theorem
Let $n$ be a $3$-digit integer whose digits are not all the same.

Kaprekar's process, when applied to $n$, results in $495$ after no more than $6$ iterations.

Proof
Let $n = \left[{abc}\right]_{10}$ denote a $3$-digit integer whose digits are $a, b, c$.

If $a = b = c$ then Kaprekar's process trivially results in $0$ after the first iteration.

, let $a \ge b \ge c$ but such that $a \ne c$.

By the Basis Representation Theorem:
 * $n = 10^2 a + 10 b + c$

Let $n' = 10^2 a' + 10 b' + c'$ be the result of Kaprekar's process after the $1$st iteration on $n$.

We have:

Thus we have that:

Hence we have that:

There are in fact only the following possibilities for $n'$:

we inspect Kaprekar's process on $198$:

and the result is seen to follow.