KaprekarNumber 2006-06-14 16:53:19 2007-07-22 17:15:36 Definition % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let $n$ be a $k$-digit integer in base $b$. Then $n$ is said to be a {\em Kaprekar number} in base $b$ if $n^2$ has the following property: when you add the number formed by its right hand digits to that formed by its left hand digits, you get $n$. Or to put it algebraically, an integer $n$ such that in a given base $b$ has $$n^2 = \sum_{i = 0}^{k - 1} d_ib^i$$ (where $d_x$ are digits, with $d_0$ the least significant digit and $d_{k - 1}$ the most significant) such that $$\sum_{i = {k \over 2} + 1}^k d_ib^{i - {k \over 2} - 1} + \sum_{i = 1}^{k \over 2} d_ib^{i - 1} = n$$ if $k$ is even or $$\sum_{i = \lceil {k \over 2} \rceil}^k d_ib^{i - \lfloor {k \over 2} \rfloor - 1} + \sum_{i = 1}^{k \over 2} d_ib^{i - 1} = n$$ if $k$ is odd. $b^x - 1$ for a natural $x$ is always a Kaprekar number in base $b$. \begin{thebibliography}{1} \bibitem{dk} D. R. Kaprekar, ``On Kaprekar numbers" {\it J. Rec. Math.} 13 (1980-1981), 81 - 82. \end{thebibliography}