The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the 2005th term of the sequence?

$ \textbf{(A)}\ 29\qquad
\textbf{(B)}\ 55\qquad
\textbf{(C)}\ 85\qquad
\textbf{(D)}\ 133\qquad
\textbf{(E)}\ 250$
1. The first term of the sequence is given as \(2005\).
2. To find the next term, we need to calculate the sum of the cubes of the digits of \(2005\):
   \[
   2^3 + 0^3 + 0^3 + 5^3 = 8 + 0 + 0 + 125 = 133
   \]
   So, the second term is \(133\).

3. Next, we calculate the sum of the cubes of the digits of \(133\):
   \[
   1^3 + 3^3 + 3^3 = 1 + 27 + 27 = 55
   \]
   So, the third term is \(55\).

4. Next, we calculate the sum of the cubes of the digits of \(55\):
   \[
   5^3 + 5^3 = 125 + 125 = 250
   \]
   So, the fourth term is \(250\).

5. Next, we calculate the sum of the cubes of the digits of \(250\):
   \[
   2^3 + 5^3 + 0^3 = 8 + 125 + 0 = 133
   \]
   We see that the sequence has started to repeat, as \(250\) leads back to \(133\).

6. The sequence is now observed to be periodic with a cycle length of 3: \(133 \rightarrow 55 \rightarrow 250 \rightarrow 133 \rightarrow \ldots\).

7. To find the 2005th term, we note that the sequence repeats every 3 terms. We need to find the position of the 2005th term within the cycle:
   \[
   2005 \mod 3 = 1
   \]
   This means the 2005th term corresponds to the first term in the cycle, which is \(133\).

The final answer is \(\boxed{133}\).