2003 Fermat Problem 21

A sequence of numbers has 6 as its first term, and every term after the first is defined as follows: If a term, t, is even, the next term in the sequence is \frac{1}{2}t. If a term, s, is odd, the next term is 3s+1. Thus, the first four terms in the sequence are 6, 3, 10, 5. The 100th term is

\textbf{(A)}\ 1\quad \textbf{(B)}\ 2\quad \textbf{(C)}\ 3\quad \textbf{(D)}\ 4\quad \textbf{(E)}\ 6

If there are no answer choices shown, enter a numerical answer.

Full credit to this problem is given to the CEMC, you may view all Fermat contests here.