All even-digit Kaprekar Series numbers (base r) are multiples of (r - 1).

Let use consider a 2m-digit number(base r), whose digits are sorted in descending order:

n1 = r(2m - 1) x1 + r(2m - 2) x2 + ... + r(2m - i) xi + ... + r * x2m-1 + x2m

sorted in ascending order, the number would be

n2 = r(2m - 1) x2m + r(2m - 2) x2m-1 + ... + r(i - 1) xi + ... + r * x2 + x1

The difference between the two numbers

Diff2m = n1 - n2

       = Σ(i = 1 to 2m) r(2m - i) xi - Σ(i = 2m to 1) r(i - 1) xi

       = Σ(i = 1 to 2m) {r(2m - i) - r(i - 1)} xi

       = Σ(i = 1 to m) r(i - 1) {r(2m - 2i + 1) - 1} xi

- Σ(i = m+1 to 2m) r(2m - i) {r(2i - 2m - 1) - 1} xi

Where Σ stands for Sigma - the summation notation

Lets take a closer look at part of the above expression

(2m - 2i + 1) is a positive number here when 1 <= i <= m

(2i - 2m - 1) is a positive number for m + 1 <= i <= 2m

So our task is reduced to proving that (r^j - 1) is divisible by (r - 1) for a positive j.

Let us assume that this is true for a positive j.

i.e. (r^j - 1) is divisible by (r - 1) for a positive j.

Consider the next number (j + 1). For this number, the expression changes to

(r(j+1) - 1)
= r (rj - 1) + (r - 1)

Now since (r^j - 1) is divisible by (r - 1), (r^(j+1) - 1) would also be divisible by (r - 1).

Lets consider the case of j = 1. For this value of j, the expression reduces to

r1 - 1 = (r - 1)

which obviously is divisible by (r - 1).

Hence, by mathematical induction, (r^j - 1) is divisible by (r - 1) for all positive integers.

Hence n2 - n1 is divisible by (r - 1) for all even-digit numbers. Since every number in the Kaprekar series is obtained by subtracting two numbers, with the same digits, sorted in descending and ascending orders; we come to the conclusion that all numbers in an even-digit Kaprekar series for a number in radix r is divisible by (r - 1).

All odd-digit Kaprekar Series numbers (base r) are multiples of (r * r - 1).

Let use consider a 2m+1-digit number (base r), whose digits are sorted in descending order:

n1 = r2m x1 + r(2m - 1) x2 + ... + r(2m + 1 - i) xi + ... + r * x2m + x2m+1

sorted in ascending order, the number would be

n2 = r2m x2m+1 + r(2m - 1) x2m + ... + r(i - 1) xi + ... + r * x2 + x1

The difference between the two numbers

Diff2m+1 = n1 - n2

        = Σ(i = 1 to 2m+1) r(2m + 1 - i) xi -  Σ(i = 2m+1 to 1) r(i - 1) xi

        = Σ(i = 1 to 2m+1) {r(2m + 1 - i) - r(i - 1)} xi

        = Σ(i = 1 to m) r(i - 1) {r(2m - 2i + 2) - 1} xi
             - Σ(i = m+2 to 2m+1) r(2m + 1 - i) {r(2i - 2m - 2) - 1} xi

        = Σ(i = 1 to m) r(i - 1) {r2(m - i + 1) - 1} xi
             - Σ(i = m+2 to 2m+1) r(2m + 1 - i) {r2(i - m - 1) - 1} xi

Where Σ stands for Sigma - the summation notation

Notice that (m - i + 1) >= 0 for 1 <= i <= m
And (i - m - 1) > 0 for m + 2 <= i <= 2m + 1
And that the expression is zero for i = m + 1

Diff2m+1 = n1 - n2

So, the problem is reduced to proving that (r^2j - 1) is divisible by (r * r - 1).

Lets assume that this is true for a positive j.
Lets consider the next number (j + 1). The expression for (j + 1) would be

r2(j+1) - 1
= r2 (102j - 1) + r2 - 1

Since (r^2j - 1) is divisible by (r * r - 1), so the expression is divisible by (r * r - 1) for (j + 1) too.

Lets consider the case j = 1. The expression for this value is r² - 1 = (r * r - 1). Which is obviously divisible by (r * r - 1). Hence by mathematical induction, the expression is divisible by (r * r - 1) for all positive integer j.

Hence n2 - n1 is divisible by 99 for all odd-digit numbers. Since every number in the Kaprekar series is obtained by subtracting two numbers, with the same digits, sorted in descending and ascending orders; we come to the conclusion that all numbers in an odd-digit Kaprekar series in radix r are divisible by (r * r - 1).

Kaprekar Series Generator at Sourceforge.