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Kaprekar constant

Problem 414 Published on Sunday, 10th February 2013, 07:00 am; Solved by 210;
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:

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 Cb to be the Kaprekar constant in base b for 5 digits. Define the function sb(i) to be

  • 0 if i = Cb or if 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 Cb, otherwise
Note that we can define sb(i) for all integers i < b5. 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 < b5.
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.