Divisibility by 41 and 5-digit number How to prove that if a $5$-digit number is divisible by $41$,then all the numbers generated from it by cyclic shift are also divisible by $41$

The reason is that $10^5 \equiv 1 \bmod 41$:
$$
0 \equiv 10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0  \bmod 41
\implies\\
0 \equiv 10(10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0) \equiv 10^4 a_3 + 10^3 a_2 + 10^2 a_1 + 10 a_0 + a_4 \bmod 41
$$
