Cyclic Permutations of 5-Digit Multiples of 41

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

$\forall k \in \N: $

$10^{0 + 5k} \equiv 1 \pmod {41}$

$10^{1 + 5k} \equiv 10 \pmod {41}$

$10^{2 + 5k} \equiv 18 \pmod {41}$

$10^{3 + 5k} \equiv 16 \pmod {41}$

$10^{4 + 5k} \equiv 37 \pmod {41}$

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:

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

Then we have:

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