Problem 68. Kaprekar Steps

The behavior at x=6174 is artifically set to 0, which taints the pure-recursive solution, but hey it was cool problem!

The K constant is 495 for 3 digits.
So the test with 691 is wrong.
Why tests with only one digit ?
Did I miss something ?

I don't understand the Test Suite for x = 3 and x = 1 too. what should I do in this case?

For x=3, the steps are 3000-0003=2997, 9972-2799=7173, etc.

Problem description is confusing as there are different Kaprekar constants depending on the number of digits. [0 9 495 6174 for 1, 2,3 4 digits respectively.

getting this error:
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The server encountered an internal error or misconfiguration and was unable to complete your request.
Reference #3.c2c1ab8.1412090777.18623697
any ideas?

The problem should specify that any number with less than four digits should be filled up to four digits with leading zeros. (e.g. 3 -> 0003)

Very nice and interesting problem!

I like the recursion aspect of this problem.

There is a small correction needed in the problem statement. Not all natural numbers, but 4 digit numbers can be reduced to Kaprekar number by the mentioned method. Similarly 3 digit numbers can be reduced to 495
https://en.wikipedia.org/wiki/D._R._Kaprekar

How it works x = 1？？？？？

For those confused with test cases 2,3 and 5, like myself before, do conversion to 4-digit integer. Here is an example:
x = 1:
1000-0001 = 999
9990-0999 = 8991
9981-1899 = 8082
8820-0288 = 8532
8532-2358 = 6174
Therefore, y_correct = 5

love it!!!