## Kaprekar constant

### Problem 414

Published on Sunday, 10th February 2013, 07:00 am; Solved by 172; Difficulty rating: 60%
6174 is a remarkable number; if we sort its digits in increasing order and subtract that number from the number you get when you sort the digits in decreasing order, we get 7641-1467=6174.

Even more remarkable is that if we start from any 4 digit number and repeat this process of sorting and subtracting, we'll eventually end up with 6174 or immediately with 0 if all digits are equal.

This also works with numbers that have less than 4 digits if we pad the number with leading zeroes until we have 4 digits.

E.g. let's start with the number 0837:

8730-0378=8352

8532-2358=6174

6174 is called the **Kaprekar constant**. The process of sorting and subtracting and repeating this until either 0 or the Kaprekar constant is reached is called the **Kaprekar routine**.

We can consider the Kaprekar routine for other bases and number of digits.

Unfortunately, it is not guaranteed a Kaprekar constant exists in all cases; either the routine can end up in a cycle for some input numbers or the constant the routine arrives at can be different for different input numbers.

However, it can be shown that for 5 digits and a base b = 6t+3≠9, a Kaprekar constant exists.

E.g. base 15: (10,4,14,9,5)_{15}

base 21: (14,6,20,13,7)_{21}

Define `C _{b}` to be the Kaprekar constant in base

`b`for 5 digits. Define the function

`sb(i)`to be

- 0 if i =
`C`or if_{b}`i`written in base`b`consists of 5 identical digits - the number of iterations it takes the Kaprekar routine in base
`b`to arrive at`C`, otherwise_{b}

`sb(i)`for all integers

`i`<

`b`

^{5}. If

`i`written in base

`b`takes less than 5 digits, the number is padded with leading zero digits until we have 5 digits before applying the Kaprekar routine.

Define `S(b)` as the sum of `sb(i)` for 0 < `i` < `b`^{5}.

E.g. S(15) = 5274369

S(111) = 400668930299

Find the sum of S(6k+3) for 2 ≤ k ≤ 300.

Give the last 18 digits as your answer.