17. Find the smallest two-digit number $N$ such that the sum of digits of $10^{N}-N$ is divisible by 170 .
17. Ans: 20
Let $N=10 a+b$. When $b \neq 0,10^{N}-N$ is a $N$-digit number whose first $N-2$ digits are 9 and the last two digits are $(9-a)$ and $(10-b)$. Thus the sum of digits of $10^{N}-N$ equals $9(N-2)+19-(a+b)=89 a+8 b+1$. Thus, we need
$$
\begin{aligned}
a & \equiv 1 \quad(\bmod 2) ; \\
-a+3 b & \equiv-1 \quad(\bmod 5) ; \\
4 a+8 b & \equiv-1 \quad(\bmod 17) .
\end{aligned}
$$
$a=1$ has no solution, but $a=3, b=9$ is a solution.
When $b=0$, the sum of digits is $9(N-2)+(10-a)=89 a-8$. Thus $a$ is even. $a=2$ is a solution. Thus the answer is 20 since it is $<39$.