--> ]> 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 = i = 1 2 m 10 ( 2 m - i ) x i - i = 1 2 m 10 ( i - 1 ) x i = i = 1 2 m { 10 ( 2 m - i ) x i - 10 ( i - 1 ) x i } = i = 1 m 10 ( i - 1 ) { 10 ( 2 m - 2 i + 1 ) - 1 } x i - 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 i m

( 2 i - 2 m - 1 ) is a positive number for ( m + 1 ) i 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 = i = 1 2 m + 1 10 ( 2 m + 1 - i ) x i - i = 2 m + 1 1 10 ( i - 1 ) x i = i = 1 2 m + 1 { 10 ( 2 m + 1 - i ) - 10 ( i - 1 ) } x i = i = 1 m 10 ( i - 1 ) { 10 ( 2 m - 2 i + 2 ) - 1 } x i - i = m + 2 2 m + 1 10 ( 2 m + 1 - i ) { 10 ( 2 i - 2 m - 2 ) - 1 } x i = i = 1 m 10 ( i - 1 ) { 10 2 ( m - i + 1 ) - 1 } x i - i = m + 2 2 m + 1 10 ( 2 m + 1 - i ) { 10 2 ( i - m - 1 ) - 1 } x i

Notice that ( m - i + 1 ) 0 for 1 i m And ( i - m - 1 ) > 0 for m + 2 i 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.