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# Kaprekar number

Let $n$ be a $k$-digit integer in base $b$. Then $n$ is said to be a 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

$n^{2}=\sum_{{i=0}}^{{k-1}}d_{i}b^{i}$ |

(where $d_{x}$ are digits, with $d_{0}$ the least significant digit and $d_{{k-1}}$ the most significant) such that

$\sum_{{i={k\over 2}+1}}^{k}d_{i}b^{{i-{k\over 2}-1}}+\sum_{{i=1}}^{{k\over 2}}% d_{i}b^{{i-1}}=n$ |

if $k$ is even or

$\sum_{{i=\lceil{k\over 2}\rceil}}^{k}d_{i}b^{{i-\lfloor{k\over 2}\rfloor-1}}+% \sum_{{i=1}}^{{k\over 2}}d_{i}b^{{i-1}}=n$ |

if $k$ is odd.

$b^{x}-1$ for a natural $x$ is always a Kaprekar number in base $b$.

# References

- 1 D. R. Kaprekar, “On Kaprekar numbers” J. Rec. Math. 13 (1980-1981), 81 - 82.

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## Mathematics Subject Classification

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