TITLE: Find all two-digits $N$ for which the sum of the digits of $10^N-N$ is divisible by $170$.
QUESTION [5 upvotes]: My friend was surfing on the web for Math Questions. He found a Math Question:

Find all two-digits $N$ for which the sum of the digits of $10^N-N$ is divisible by $170$.

I tried it, and this is my attempt.
$10^N-N=(10^N-1)-(N-1)=999...999-(N-1)$
The digit sum of $(10^N-N)=9N-$ the digits sum of $(N-1)$
Then I am stuck.
Can someone help me? Any help is appreciated!

REPLY [6 votes]: The digit sum of $N-1$ is somewhere between $1$ and $17$ (inclusive). So our two-digit number must necessarily be such that $9N$ is within "reach" of a multiple of $170$. The possible values for $9N$ and thus $N$ is therefore
$$
\begin{array}{|c|c|c|}
\hline 9N & N & \text{Digit sum}\\
\hline 171&19 & 9\\
\hline 180&20 & 10\\
\hline 342&38 & 10\\
\hline 351&39 & 11\\
\hline 513&57 & 11\\
\hline 522&58 & 12\\
\hline 684&76 & 12\\
\hline 693&77 & 13\\
\hline 855&95 & 13\\
\hline 864&96 & 14\\
\hline
\end{array}
$$
We see that the $N$ for which $9N$ minus the digit sum of $N-1$ is a multiple of $170$ are the larger one in each "pair", which is to say
$$
20, 39, 58, 77, 96
$$