What is the last two digits of base-$3$ representation of $2005^{2003^{2004}+3}$?

$ 
\textbf{(A)}\ 21
\qquad\textbf{(B)}\ 01
\qquad\textbf{(C)}\ 11
\qquad\textbf{(D)}\ 02
\qquad\textbf{(E)}\ 22
$
To find the last two digits of the base-3 representation of \(2005^{2003^{2004} + 3}\), we need to follow these steps:

1. **Convert the problem to a simpler form using modular arithmetic:**
   We need to find the last two digits in base-3, which is equivalent to finding the number modulo \(3^2 = 9\).

2. **Simplify the exponent modulo 9:**
   We need to compute \(2003^{2004} + 3 \mod 9\). First, we simplify \(2003 \mod 9\):
   \[
   2003 \div 9 = 222 \text{ remainder } 5 \implies 2003 \equiv 5 \mod 9
   \]
   Therefore, we need to compute \(5^{2004} + 3 \mod 9\).

3. **Use properties of exponents modulo 9:**
   Notice that \(5^6 \equiv 1 \mod 9\) (since \(5^6 = 15625 \equiv 1 \mod 9\)). This is due to Euler's theorem, which states \(a^{\phi(n)} \equiv 1 \mod n\) for \(a\) coprime to \(n\). Here, \(\phi(9) = 6\).

4. **Reduce the exponent modulo 6:**
   \[
   2004 \mod 6 = 0 \implies 5^{2004} \equiv (5^6)^{334} \equiv 1^{334} \equiv 1 \mod 9
   \]
   Therefore,
   \[
   5^{2004} + 3 \equiv 1 + 3 \equiv 4 \mod 9
   \]

5. **Convert the result to base-3:**
   We need to convert 4 to base-3. The base-3 representation of 4 is:
   \[
   4_{10} = 11_3
   \]

Thus, the last two digits of the base-3 representation of \(2005^{2003^{2004} + 3}\) are \(11\).

The final answer is \(\boxed{11}\).