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)

I was reading the expressions, but the implication just fazed me. I had a mental block about the 1 case, even when it was explained more explicitely. Gosh, I'm not getting senile? er, I'll change that ? to ! --Paddy3118 09:05, 8 June 2011 (UTC)

Change it === Which sounds better? ===

What should we say, "split into parts" or "split once into whole number components made of groups of neighbouring digits from within the number"?

We are splitting into the number into numeric values the belong to the set of "whole numbers" aren't we? Or is there a misunderstanding here?

"split into parts" sounds too vague to me.

--Markhobley 16:06, 8 June 2011 (UTC)

If I had to change it from what we have now I'd say "split into two parts". The way it is now seems fine to me especially with the examples in the description. Since the number is specified as an integer, I don't think we need to add "whole number" noise. --Mwn3d 16:16, 8 June 2011 (UTC)