Suppose $\overline{a_1a_2...a_{2009}}$ is a $2009$-digit integer such that for each $i = 1,2,...,2007$, the $2$-digit integer $\overline{a_ia_{i+1}}$ contains $3$ distinct prime factors. Find $a_{2008}$ 
(Note: $\overline{xyz...}$ denotes an integer whose digits are $x, y,z,...$.)
1. We are given a 2009-digit integer $\overline{a_1a_2 \ldots a_{2009}}$ such that for each $i = 1, 2, \ldots, 2007$, the 2-digit integer $\overline{a_ia_{i+1}}$ contains 3 distinct prime factors.
2. We need to find the value of $a_{2008}$.

Let's analyze the given condition:
- For each $i$, $\overline{a_ia_{i+1}}$ must be a 2-digit number with exactly 3 distinct prime factors.

3. The smallest 2-digit number with 3 distinct prime factors is $30 = 2 \times 3 \times 5$.
4. Other examples include:
   - $42 = 2 \times 3 \times 7$
   - $66 = 2 \times 3 \times 11$
   - $70 = 2 \times 5 \times 7$
   - $78 = 2 \times 3 \times 13$
   - $102 = 2 \times 3 \times 17$ (but this is not a 2-digit number)

5. We need to find a pattern that fits the entire 2009-digit number. Let's consider the number $66$:
   - $66 = 2 \times 3 \times 11$
   - If $a_i = 6$ and $a_{i+1} = 6$, then $\overline{a_ia_{i+1}} = 66$.

6. If we set $a_1 = a_2 = \ldots = a_{2009} = 6$, then for each $i$, $\overline{a_ia_{i+1}} = 66$, which satisfies the condition of having 3 distinct prime factors.

7. Therefore, $a_{2008} = 6$.

The final answer is $\boxed{6}$