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# Problem 414

Kaprekar constant

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.
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 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.

6174是个神奇的数；如果我们用将它的数字按降序排列组成的数减去将它的数字按升序排列组成的数，我们得到7641-1467=6174。

8730-0378=8352
8532-2358=6174

6174被称为卡布列克常数。这个排序和相减直至到达0或卡布列克常数的过程称为卡布列克路径

• 0，如果i = Cb，或者i在b进制下5个数字全部相同。
• 否则为在b进制下从这个数出发经过卡布列克路径到达Cb的迭代次数。

S(111) = 400668930299