Talk:Kaprekar number

message: Following the description on this page, how can 1 be a Kaprekar number? One squared is one, which cannot be split in to two one-digit numbers.
Therefore, either one is not a Kaprekar number, or the description on this page needs a special case for the number one. --Dan Huby 14:55, 6 Apr 2004 (UTC)

message: What is the usefulness of a Kaprekar number? - Zepheus (ツィフィアス) 17:17, 3 August 2006 (UTC)
thread_title: Usefulness

message: Since the two parts don't need to be of equal length, a Kaprekar number can have an odd number of digits. 2607:FB90:2113:EA80:B5CB:1FF3:4674:FBF5 (talk) 20:06, 31 May 2015 (UTC)
thread_title:  Odd number of digits 

message: Given that no sources have been provided for any of the lists in the Other bases section of the article, I have written a script for a symbolic algebra package to check them.  The output of this script agrees with the article's lists for decimal and hexadecimal, but uncovered problems with the duodecimal and base 7 lists.
My script confirms that all the numbers given in the list of duodecimal Kaprekar numbers are indeed Kaprekar, but found two more, 4873012 and 7239212, which were missing from the list.  That these are also Kaprekar is easy to check with a calculator:
:: 48730122 = 1X3004690012   and   4873012 = 04690012 + 1X3012
:: 72392122 = 438X06X00412   and   7239212 = 06X00412 + 438X12.
There were many more problems with the base 7 list. According to my script, more than two thirds of the numbers given in this list are not Kaprekar, and there are numerous Kaprekar numbers missing from it.  I have not checked any more than a handful of these with an explicit hand calculation, but for all those I have checked, the calculation has vindicated the output of my script.  One example is the number 347, which had been given in the article's list.   But   3472 = 15527. There are only 3 ways to split this latter number into two parts and add them, 1557 + 27, 157 + 527   and  17 + 5527, and none of these sums is equal to 347.  Therefore 347 cannot be Kaprekar.
On the other hand, 227, which was not given in the list of base 7 Kaprekar numbers, is easily verified to be Kaprekar, since
: 2272 = 5147, and   227 = 57 + 147
David Wilson (talk · cont) 11:22, 25 September 2017 (UTC)
thread_title:  Corrections 

message: This page seems to be the only one that accepts 0 and any powers of 10 (resp. any powers of base) as Kaprekar numbers. In any other page dealing with K numbers, these numbers are excluded.
CKWG  — Preceding unsigned comment added by 185.22.143.114 (talk) 15:12, 15 November 2019 (UTC)
thread_title:  0 a Kaprekar number? 

message: Hello, I think the proof for finiteness of p-Kaprekar numbers may be incorrect (located under the Definition and properties section). 
The proof claims that \beta = m^2 for the given example; however, if you compute \beta, it's actually 
: 
\begin{align}
\beta & = n^2 \bmod b^p \\ 
 & = ((b^p + 2m)b^p + m^2) \bmod b^p \\
 & = m^2 \bmod b^p \\
\end{align}
Note that this is an important claim in the proof because it may be the case that m^2 \ge b^p . For instance, consider n=19, b = 10, and  p =1. In this case, b^p = 10  and m = 9. Which means that m^2 > b^p because 81 > 10. 
The proof is still structurally correct. Below is the fix I believe is necessary for the proof. 
Along with the re-computing \beta, we need to recompute \alpha. If you let  \beta = m^2 \bmod b^p  then  \alpha = b^p + 2m + \frac{m^2-(m^2\bmod b^p)}{b^p}. Therefore, 
:  \alpha + \beta = n + m + \frac{m^2-(m^2\bmod b^p)}{b^p} + m^2\bmod b^p 
:  > n 
Please let me know if I'm missing anything. Thank you for reviewing this!
KamyarGhiam (talk) 00:03, 10 August 2023 (UTC)
thread_title:  Proof of finite p-Kaprekar numbers may be incorrect