Let $a_1$ be any number from the set {$0, 1, 2, \ldots, 9$}. For each $n \in \mathbb{N}$, denote by $a_{n+1}$ the last digit of the number $19a_n + 98$ in decimal notation. Prove that the number $0.a_1a_2a_3\ldots$ is a rational number.
Attempt: I know that the number $a_{n+1}$ is congruent to $-a_n-2 \pmod{10}$. Then I looked at all the possibilities for $a_1$, for example, for $a_1=2$ we get $a_1=2$, $a_2=-2-2=-4$, which is congruent to $6 \pmod{10}$, $a_3=-6-2=-8$, which is congruent to $2 \pmod{10}$, then $a_4=a_2$, $\ldots$, so we get $0.262626\ldots$ I would do similarly for other cases. Is there perhaps a faster and more elegant solution (way, to deal with that exercise)?