Question:
Let $S$ be the set of all integers $k$ such that, if $k$ is in $S$, then $\frac{17k}{66}$ and $\frac{13k}{105}$ are terminating decimals. What is the smallest integer in $S$ that is greater than 2010?

Answer:
Let us first analyze the fraction $\frac{17k}{66}$. We can rewrite this fraction as $\frac{17k}{2 \cdot 3 \cdot 11}$. Since the denominator can only contain powers of 2 and 5, we have that $k$ must be a multiple of 33. We now continue to analyze the fraction $\frac{13k}{105}$. We rewrite this fraction as $\frac{13k}{3 \cdot 5 \cdot 7}$, and therefore deduce using similar logic that $k$ must be a multiple of 21. From here, we proceed to find the least common multiple of 21 and 33. Since $21 = 3 \cdot 7$ and $33 = 3 \cdot 11$, we conclude that the least common multiple of 21 and 33 is $3 \cdot 7 \cdot 11 = 231$.

We now know that $S$ contains exactly the multiples of 231. The smallest multiple of 231 that is greater than 2010 is $231 \cdot 9 = \boxed{2079}$.