Not just the special numbers.

Kaprekar's operation always results in a multiple of 9.

Given a > b > c > d. Kaprekar's operation gives:

    1000a + 100b + 10c + d - (1000d + 100c + 10b + a)
    =999(a - d) + 90(b - c)
    =9(111(a - d) + 10(b - c))

And so on for integers with other numbers of digits.

Any number subtracted from a re-arrangement of the digits results in a multiple of 9.

Why?

A number is the sum of its digits, raised to the appropriate powers

    Sum(i=0..n, a[i]*10^i)

123, for example is 3 + 2 * 10 + 1 * 100.

A power of 10 (10^n) can be re written as

    (1 + 9 * Sum(i=0..n-1, 10^i). 

For example: 1000 = 1 + 999 = 1 + 900 + 90 + 9

When you subtract two numbers with the same digits, you end up being able to factor a nine out of these sums fairly easily.

I wrote a proof of the number - reverse(number) a while back. It can be found on archive.org

http://web.archive.org/web/20050314023901/http://jimfl.tense...