How to prove that every number of the form $245\cdot 22..11-239$ is congruent to $10^m \mod 41$? Consider Numbers:
\begin{align}
245&\cdot 211-239\\
245&\cdot 2211-239\\
245&\cdot 22211-239\\
245&\cdot 222211-239{}{}{}\\
245&\cdot 2222\ldots11-239
\end{align}
I am stuck, how to prove that every number of this form is congruent to $10^m \bmod 41$ for some $m$?

Since $1..1$ with $n$ ones is equal to $\frac{10^n-1}{9}$, we can write the general term of the sequence as
$$
a_n = 245 \cdot \Big(200 \cdot \frac{10^n-1}{9} + 11\Big) - 239.
$$
Looking modulo 41 we have $245 \equiv -1$, $200 \equiv -5$, $-239 \equiv 7$ and $9^{-1} \equiv -9$ (since $9 \times (-9) = -81 \equiv 1$). We can write
\begin{align*}
a_n
&\equiv (-1) \cdot 9^{-1} (-5) (10^{n}- 1) - 11 + 7 \\
&\equiv 9 \cdot  (-5 \cdot 10^{n} + 5) -4 \\
&\equiv -45 \cdot 10^{n} + 45 - 4 \\
&\equiv -4 \cdot 10^n \\
&\equiv -40 \cdot 10^{n-1}\\
&\equiv 10^{n-1}.
\end{align*}
