Convert $22202_3$ (in base $3$) into base $11$.
I know I can just convert into base $10$ and divide by $11$.
$2\times 3^4+2\times3^3+2\times3^2+2=236_{10}$
Then repeated division by $11$ yields $(1A5)_{11}$, which is the correct answer. However, I would like to do it directly through long division and I'm getting stuck.
$$ \begin{array}{c|cc} 11&2&2&2&0&2 \\ &&&2&1& -5\\ \hline \end{array} $$
Explanation for what I'm doing: A single $2$ in base $3$ is not divisible by $11$. Neither is $22 \ (2\times3+2=8)$. So take $222 \ (2\times3^2+2\times3+2 = 26)$. $11$ goes into $26$ two times with $4$ remainder. Write down the $2$. The $4$ remainder is carried to the next digit, $0$, making $40 \ (4\times3+0=12)$. $11$ goes into $12$ one time, so write down $1$. The remainder, also $1$, is carried to the next digit, $2$. The next position $12 \ (1\times3+2=5)$ is not divisible by $11$, hence the remainder is $5$, which is written as $-5$.
Now I can't go on, because $21$ in base $3$ is equivalent to $2\times3+2=8$, which isn't divisible by $11$. Something is obviously wrong, but I can't see it, because I did get the digit $5$, which is in the correct position and is part of the correct answer.
To get the units digit of $1A5_{11}$, you tried to perform division with remainder of $22202_3$ by $11_{10} = 102_3$.
Your last step was to divide $12_3$ by $102_3$, and you didn't record the quotient $0$. The overall quotient $210_3$ is the one for the next step to calculate the "elevens" digit. The remainder $5_{11}$ matches the units digit of $1A5_{11}$.
Explicitly in long division:
$$ \begin{array}{ r|ccccccc } &&& 2&1&0 \\ \hline 102& 2&2&2&0&2 \\ & 2&1&1\\ \hline && 1&1&0 \\ && 1&0&2 \\ \hline &&&& 1&2 \\ &&&&& 0 \\ \hline &&&& 1&2 & (= 5_{11}) \\ \end{array} $$