The congruence $(34\times 10^{(6n-1)}-43)/99 \equiv -1~\text{ or }~5 \pmod 6$ Trying to prove this congruence:
$$ \frac{34\times 10^{(6n-1)}-43}{99} \equiv-1~\text{ or }~5 \pmod 6,\quad
n\in\mathbb{N}$$
Progress
Brought it to the form
$$34\times 10^{6n-1}-43\equiv -99\pmod{6\cdot 99}$$
but how to proceed further? $6\cdot 99$ is not a small number.

${\ 6\mid f_{n+1}-f_n =\, \dfrac{34\cdot(\color{#c00}{10^6\!-1}) 10^{6n\!-1}}{99}\ }$ by ${\ 3\cdot 99\mid \color{#c00}{10^6\!-1} = 999999 = 99\cdot\!\!\!\!\!\!\!\!\!\!\!\underbrace{10101}_{\equiv\, 1+1+1\pmod 3}}$
So $\,{\rm mod}\ 6\!:\ f_{n+1}\!\equiv f_n\,\overset{\rm induct}\Rightarrow\,f_n\equiv f_1\equiv -1$
