smc

We can determine if our number is divisible by $3$ or $9$ by summing the digits. Looking at the one's place, we can start out with $0,1,2,3,4,5,6,7,8,9$ and continue cycling though the numbers from $0$ through $9$. For each one of these cycles, we add $0+1+...+9=45$. This is divisible by $9$, thus we can ignore the sum. However, this excludes $19$, $90$, $91$ and $92$. These remaining units digits sum up to $9+1+2=12$, which means our units sum is $3(\mathrm{mod}9)$. As for the tens digits, for $2,3,4,\cdots ,8$ we have $10$ sets of those: $$\frac{8\cdot 9}{2}-1=35,$$ which is congruent to $8(\mathrm{mod}9)$. We again have $19,90,91$ and $92$, so we must add $$1+9\cdot 3=28$$ to our total. $28$ is congruent to $1(\mathrm{mod}9)$. Thus our sum is congruent to $3(\mathrm{mod}9)$, and $k=1⟹\boxed{1}$.