Here is an alternate approach: Write 22 in base-3. So,

and

.

Now, multiply the two numbers together:

This addition should feel familiar from multiplying two numbers base-10. Only now, for each column, you record the result (mod 3) and carry each multiple of 3. So, the right-most column shows that

. Plugging that in, you have

, so

. Now,

. Divide the result by 3 and that is the amount you are "carrying" to the left. So, you are carrying

since you had one 3 from that column. So, the third column is the sum

. You want that sum (mod 3) to be zero. So,

shows that

. Plugging that in, you have

, which is the same result as the other method.