Before anyone gives me a link to Wikipedia, I've already looked at it, and the history. The last edit by Michael Hardy was last year. Now, I don't know him personally, I don't think I've ever even communicated with him through the Internet, but to me his name carries a lot more weight than the screen name "Doctormatt" (attached to a revert back in April) and certainly more than the seemingly random IP addresses from which the article has been edited since last year.
I think I understand the decimal Kaprekar numbers well enough. For example $297^2 = 88209$ and $088 + 209 = 297$. In other bases, I frequently get confused by applying the conversions at the wrong time, so it would help me greatly if there was already some reference with a good example worked out clearly.
Let's assume for a moment that the version of the Wikipedia page was correct as of Mr. Hardy's version last year. The article lists duodecimal Kaprekar numbers, using X rather than A and E rather than B. That idiosyncrasy aside, the listing seems to be correct.
But then there's the assertion that
In binary, all even perfect numbers are Kaprekar numbers.
Let's see, $6^2 = 36$ which in binary is $100100$ (as $32 + 4 = 36$). Then $100 + 100 = 1000$, which is $8$, not $6$. Did I go wrong somewhere?
The current version of the article lists for binary the Mersenne numbers (both prime and non-prime) but no perfect numbers. This checks out at least for a few calculations of small numbers, but now the article contradicts itself because it still contains the assertion about even perfect numbers in binary.
Furthermore, someone added $2$ and $3$ to the duodecimal listing, as well as listings for base $7$ (a really odd choice, but okay, I guess) and hexadecimal, and they also include $2$ and $3$ in those listings, which seems to be patently wrong. For $b > 4$, we have $2^2 = 4$ and $0 + 4 = 4$, not $2$, right?
I headed to a far more reputable source, the OEIS, tried a search for "binary Kaprekar" and was just utterly confused by the results.
I also thought about Wolfram Alpha, but this seems to require a little more computational muscle than that website is willing to provide. Hopefully next week I'll have access to Mathematica, if they fix the license server like they said they would, but even so I could wind up confusing myself further still.
So maybe I should ask other people before my family thinks I've finally gone senile.
Where people often go wrong is assuming that the division between digits, or in this case between bits, must be balanced; thus the six bits in $100100_2$ must be divided three and three. But Kaprekar numbers allow for any division of digits/bits. If $100100_2$ has to be divided into a group of two bits and a group of four bits to get the proper sum (@quasi), so be it; its square root $=110_2$ is still a Kaprekar number.