Cyclic Permutation of Kaprekar Number

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Theorem

Let $n$ be a Kaprekar number of $k$ digits.

Let $m$ be an integer formed from a cyclic permutation of the digits of $n$.

Let $m$ be squared and the result split into $2$ parts, where the $2$nd part is of $k$ digits.

Let these two parts be added, in the way of operating on a Kaprekar number.

If the result is more than $k$ digits long, split that into $2$ parts, where the $2$nd part is of $k$ digits, and add the parts.

The result will be another cyclic permutation of the digits of $n$.


Proof


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Examples

972

$972$ is a cyclic permutation of the $3$-digit Kaprekar number $297$.

Thus we have:

\(\ds 972^2\) \(=\) \(\ds 944 \, 784\)
\(\ds 944 + 784\) \(=\) \(\ds 1728\)
\(\ds 1 + 728\) \(=\) \(\ds 729\)

and it is seen that $729$ is another cyclic permutation of $297$.

$\blacksquare$


2727

$2727$ is a cyclic permutation of the $4$-digit Kaprekar number $7272$.

Thus we have:

\(\ds 2727^2\) \(=\) \(\ds 7 \, 436 \, 529\)
\(\ds 743 + 6529\) \(=\) \(\ds 7272\)

and it is seen that $7272$ is a (trivial) cyclic permutation of $7272$.

$\blacksquare$


Sources

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