Cyclic Permutations of 5-Digit Multiples of 41

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Theorem

Let $n$ be a multiple of $41$ with $5$ digits.

Let $m$ be an integer formed by cyclically permuting the digits of $n$.


Then $m$ is a multiple of $41$.


Proof

First we note that $10^5 - 1 = 41 \times 271 \times 9$

$10$ generates exactly $5$ elements in $\Z_{41}$ Subgroup Generated by One Element is Cyclic


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\(\ds \forall k \in \N: \, \) \(\ds 10^{0 + 5 k}\) \(\equiv\) \(\ds 1\) \(\ds \pmod {41}\)
\(\ds 10^{1 + 5 k}\) \(\equiv\) \(\ds 10\) \(\ds \pmod {41}\)
\(\ds 10^{2 + 5 k}\) \(\equiv\) \(\ds 18\) \(\ds \pmod {41}\)
\(\ds 10^{3 + 5 k}\) \(\equiv\) \(\ds 16\) \(\ds \pmod {41}\)
\(\ds 10^{4 + 5 k}\) \(\equiv\) \(\ds 37\) \(\ds \pmod {41}\)

In $\Z_{41}$: $ \set {10^1, 10^2, 10^3, 10^4, 10^5} \leadsto \set {10, 18, 16, 37, 1}$


Let $n$ be a multiple of $41$ with $5$ digits.

Then we have:

\(\ds 41 \times c\) \(=\) \(\ds a_4 \times 10^4 + a_3 \times 10^3 + a_2 \times 10^2 + a_1 \times 10^1 + a_0\)


Let $m$ be an integer formed by cyclically permuting the digits of $n$.

Then we have:

\(\ds 10 \times 41 \times c\) \(=\) \(\ds 10 \times \paren {a_4 \times 10^4 + a_3 \times 10^3 + a_2 \times 10^2 + a_1 \times 10^1 + a_0}\) multiply original number by $10$
\(\ds \) \(=\) \(\ds a_4 \times 10^5 + a_3 \times 10^4 + a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1\)
\(\ds \) \(=\) \(\ds a_3 \times 10^4 + a_2 \times 10^3 + a_1 \times 10^2 + a_0 \times 10^1 + a_4 \times 10^0\) $10^5$ and $10^0 \equiv 1 \pmod {41}$


Therefore $m$ is a multiple of $41$.

$\blacksquare$


Also see


Sources

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