jmc

From the second bullet point, we know that the second digit must be $3$, for a number divisible by $10$ ends in zero. Since there is a remainder of $1$ when $N$ is divided by $9$, the multiple of $9$ must end in a $2$ for it to have the desired remainder$(\mathrm{mod}10).$ We now look for this one: $$\begin{gathered} 9(1)=9 \\ 9(2)=18 \\ 9(3)=27 \\ 9(4)=36 \\ 9(5)=45 \\ 9(6)=54 \\ 9(7)=63 \\ 9(8)=72 \end{gathered}$$ The number $72+1=73$ satisfies both conditions. We subtract the biggest multiple of $11$ less than $73$ to get the remainder. Thus, $73-11(6)=73-66=\boxed{7}$.