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

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

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 $41$.

$\blacksquare$