The proof

--> ]> The proof

All even-digit Kaprekar Series numbers are multiples of nine.

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

n_{1} = 10^{(2 m - 1)} x_{1} + 10^{(2 m - 2)} x_{2} + ... + 10^{(2 m - i)} x_{i} + ... + 10 x_{(2 m - 1)} + x_{2 m}

sorted in ascending order, the number would be

n_{2} = 10^{(2 m - 1)} x_{2 m} + 10^{(2 m - 2)} x_{(2 m - 1)} + ... + 10^{(i - 1)} x_{i} + ... + 10 x_{2} + x_{1}

The difference between the two numbers

{Diff}_{2 m} = n_{1} - n_{2} = \sum_{i = 1}^{2 m} 10^{(2 m - i)} x_{i} - \sum_{i = 1}^{2 m} 10^{(i - 1)} x_{i} = \sum_{i = 1}^{2 m} {10^{(2 m - i)} x_{i} - 10^{(i - 1)} x_{i}} = \sum_{i = 1}^{m} 10^{(i - 1)} {10^{(2 m - 2 i + 1)} - 1} x_{i} - \sum_{i = m + 1}^{2 m} 10^{(2 m - i)} {10^{(2 i - 2 m - 1)} - 1} x_{i}

Lets take a closer look at part of the above expression

(2 m - 2 i + 1) is a positive number here when 1 \leq i \leq m

(2 i - 2 m - 1) is a positive number for (m + 1) \leq i \leq 2 m

So our task is reduced to proving that $(10^{j} - 1)$ is divisible by 9 for a positive j.

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

i.e. $(10^{j} - 1)$ is divisible by 9 for a positive j.

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

(10^{(j + 1)} - 1)

= 10 (10^{j} - 1) + 9

Now since $(10^{j} - 1)$ is divisible by 9, $(10^{(j + 1)} - 1)$ would also be divisible by nine.

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

(10^{j} - 1) = (10^{1} - 1) = 9

which obviously is divisible by 9.

Hence, by mathematical induction, $(10^{j} - 1)$ is divisible by 9 for all positive integers.

Hence $(n_{2} - n_{1})$ is divisible by nine 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 is divisible by 9.

All odd-digit Kaprekar Series numbers are multiples of ninety nine.

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

n_{1} = 10^{2 m} x_{1} + 10^{(2 m - 1)} x_{2} + ... + 10^{(2 m + 1 - i)} x_{i} + ... + 10 x_{2 m} + x_{2 m + 1}

sorted in ascending order, the number would be

n_{2} = 10^{2 m} x_{2 m + 1} + 10^{(2 m - 1)} x_{2 m} + ... + 10^{(i - 1)} x_{i} + ... + 10 x_{2} + x_{1}

The difference between the two numbers

{Diff}_{2 m + 1} = n_{1} - n_{2} = \sum_{i = 1}^{2 m + 1} 10^{(2 m + 1 - i)} x_{i} - \sum_{i = 2 m + 1}^{1} 10^{(i - 1)} x_{i} = \sum_{i = 1}^{2 m + 1} {10^{(2 m + 1 - i)} - 10^{(i - 1)}} x_{i} = \sum_{i = 1}^{m} 10^{(i - 1)} {10^{(2 m - 2 i + 2)} - 1} x_{i} - \sum_{i = m + 2}^{2 m + 1} 10^{(2 m + 1 - i)} {10^{(2 i - 2 m - 2)} - 1} x_{i} = \sum_{i = 1}^{m} 10^{(i - 1)} {10^{2 (m - i + 1)} - 1} x_{i} - \sum_{i = m + 2}^{2 m + 1} 10^{(2 m + 1 - i)} {10^{2 (i - m - 1)} - 1} x_{i}

Notice that $(m - i + 1) \geq 0$ for $1 \leq i \leq m$ And $(i - m - 1) > 0$ for $m + 2 \leq i \leq 2 m + 1$ And that the expression is zero for $i = m + 1$

So, the problem is reduced to proving that $(10^{2 j} - 1)$ is divisible by 99.

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

10^{2 (j + 1)} - 1 = 100 (10^{2 j} - 1) + 99

Since $(10^{2 j} - 1)$ is divisible by 99, so the expression is divisible by 99 for (j + 1) too.

Lets consider the case j = 1. The expression for this value is
$(10^{2 j} - 1) = (10^{2} - 1) = 99$
Which is divisible by 99. Hence by mathematical induction, the expression is divisible by 99 for all positive integer j.

Hence $(n_{2} - n_{1})$ 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 is divisible by 99.