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 $3 s+1$. Thus, the first four terms in the sequence are $6,3,10,5$. The 100th term is
(A) 1
(B) 2
(C) 3
(D) 4
(E) 6
In order to get a better feel for this sequence we should write out the first few terms (and hope we spot a pattern!). We'll use the notation $t_{n}$ for the $n$th term in the sequence; that is, $t_{1}$ is the first term, $t_{2}$ is the second term, and so on.

So we have:

$$
\begin{aligned}
& t_{1}=6, t_{2}=\frac{1}{2} t_{1}=3, t_{3}=3 t_{2}+1=10, t_{4}=\frac{1}{2} t_{3}=5, t_{5}=16, t_{6}=8, t_{7}=4, t_{8}=2 \\
& t_{9}=1, t_{10}=4, t_{11}=2, t_{12}=1, t_{13}=4, \ldots
\end{aligned}
$$

Since each term in the sequence depends only on the previous term, then if a term repeats (as it does here), the sequence will then cycle. Here, the terms (starting with the 7th) will cycle 4, 2, 1, 4, 2, 1 etc.

We notice that the cycle has length 3 , and $t_{9}=1$, so $1=t_{9}=t_{12}=t_{15}=t_{18}=\cdots$, and so every term whose subscript is a multiple of 3 will be equal to 1 .

Thus, $t_{99}=1$ and so $t_{100}=4$.

ANSWER: (D)

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