Talk:Kaprekar numbers

Java count missing

Just add (and show), the count of how many there are <1million and you will have completed the stretch goal! --Paddy3118 14:07, 7 June 2011 (UTC)

I had actually added it as a test because I saw the other examples do it, but for some reason I didn't read it as a requirement so I took it out. Mornings are hard. --Mwn3d 14:47, 7 June 2011 (UTC)

:-)
--Paddy3118 17:29, 7 June 2011 (UTC)

Why the complexity?

The wikipedia page says "Let X be a non-negative integer. X is a Kaprekar number for base b if there exist non-negative integers n, A, and positive number B satisfying ...". In other words A can be zero. Why do we have to have a bunch of text claiming that A cannot be zero but A can be an empty string and that it is meaningful to add an empty string to a number? (Can't we just take advantage of the fact that leading zero digits do not change the value of a number?) --Rdm 20:56, 7 June 2011 (UTC)

To me the text "However a conceptual single split at the very end or before the first digit that produces one empty string does have the empty string counted" means that A can be zero. Unless I'm confused about what A and B are. --Mwn3d 22:30, 7 June 2011 (UTC)

Consider 100*100 = 10000 which could be split as 100 + 00. Now the 00 is disallowed. Compare that with 1*1 = 1 which is a K-number. For this to be a k-number then it must be expressed as either 1 plus no digists to the right or no digits to the left + 1. Either way no digits is treated as a positive integer of value zero but any string of one or more noughts is expressly forbidden.

Having 1 in the series seemed to add that complexity that needed explaining in my mind. --Paddy3118 23:14, 7 June 2011 (UTC)

It has to do with the way that the splitting is done mathematically, i.e., ${\mathit {N}}=A+B$ and $N^{2}=A\times 10^{x}+B$ where $10^{x}>B>0$; in the case of $\mathrm {1}$, it becomes possible to use ${\mathit {A}}=0$. This only works for that value; nothing larger is equal to itself when squared and zero is inadmissible by the constraint on ${\mathit {B}}$. Trust a mathematician to fluff the operation of splitting a number-string and introduce an unwanted special case! –Donal Fellows 23:48, 7 June 2011 (UTC)