Oh wait, I got it.
Turn them into modular arithmetic expressions such as \(a\equiv 5\bmod{9}\) and \(b\equiv 33\bmod{36}\).
Then, convert them into mod 3 and mod 9 expressions in order to get the units and tens digits of the base 3 equivalents of \(a\) and \(b\).
\(a\equiv 2\bmod{3}\) --> The first digit of a in base 3 is 2.
\(a\equiv 5\bmod{9}\) --> The value of the two last digits of a in base 3 is 5. Therefore, the second digit of a in base 3 is 1.
\(b\equiv 0\bmod{3}\) --> The first digit of b in base 3 is 0.
\(b\equiv 6\bmod{9}\) --> The value of the two last digits of b in base 3 is 6. Therefore, the second digit of b in base 3 is 2.
So, we end up with 12 as the last two digits of a and 20 as the last two digits of b (both in base 3).
To subtract, we do 2 - 0 = 2 for the units digit. Since the second digit of a is less than the second digit of b, we carry over 3 from the third digit, which we must assume has a value. Then we do 4 - 2 = 2 for the second digit.
The last two digits of a - b in base 3 is 22.
