well let's see if there's anything clever that makes this trivial
[MATH]
n = \sum \limits_{k=0}^{17} 9\cdot 10^k = \\
\sum \limits_{k=0}^{17} (19-10) \cdot 10^k=\\
19\sum \limits_{k=0}^{17}~10^k - 10\sum \limits_{k=0}^{17}~10^k = \\
19 (10^{18}-1) - 10(10^{18}-1)
[/MATH]
[MATH]\text{Dividing the first term by $19$ we obtain $(10^{18}-1)$}[/MATH]
\(\displaystyle \text{Is the second term even divisible by $19$? We can use Fermat's Little Theorem.}\\
\text{If $p$ is prime, for any integer $a$ it holds that $a^{p-1}-1 \equiv 0 \pmod{p}$}\\~\\
\text{so $10^{18}-1 \pmod{19} = 0$, so it is divisible by $19$}
\)
[MATH]\text{I'm failing to see anything clever that will let you simply evaluate the second term divided by 19.}\\
\text{The answer isn't trivial. Maybe one of the number theory gurus will see the answer.}[/MATH]