- Thread starter RaffiO
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If all you need is the answer, some calculators (such as the one in Windows) can do it for you.

\(\displaystyle

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)

\)

\(\displaystyle \text{Dividing the first term by $19$ we obtain $(10^{18}-1)$}\)

\(\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 $19$}

\)

\(\displaystyle \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.}\)

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What is the actual question? Do you just need to show that it is divisible by 19 or do you actually want the answer to the division. If so, I think you will have to use brute force. Doesn't take that long.