11. From the sequence of positive integers $1,2,3,4,5, \cdots$, remove the multiples of 3 and 4, but retain all multiples of 5 (for example, 15 and 120 are retained). After removing, the remaining numbers form a new sequence: $1,2,5,7,10, \cdots$. Find the 2010th term of the new sequence.
11. Since the least common multiple of $3$, $4$, and $5$ is 60, the numbers in the new sequence can be expressed in the form $60m + r$, where $m \geqslant 0, m \in \mathbf{Z}$, and $r$ is one of the numbers left after removing the multiples of 3 and 4 (but not the multiples of 5) from the set $\{1, 2, 3, \cdots, 60\}$. Let $I=\{1,2,3, \cdots, 60\}$, $A=\{x|x \in I, 3| x\}$, $B=\{x|x \in I, 4| x\}$, and $C=\{x|x \in I, 5| x\}$. Then $|I|=60$, $|A|=\left[\frac{60}{3}\right]=20$, $|B|=\left[\frac{60}{4}\right]=15$, $|A \cap B|=\left[\frac{60}{3 \times 4}\right]=5$, $|B \cap C|=\left[\frac{60}{4 \times 5}\right]=3$, $|A \cap C|=\left[\frac{60}{3 \times 5}\right]=4$, and $|A \cap B \cap C|=\left[\frac{60}{3 \times 4 \times 5}\right]=1$. Therefore, the number of possible values for $r$ is $|\overline{A \cup B \cup C}| + |C| = |I| - |A| - |B| + |A \cap B| + |B \cap C| + |C \cap A| - |A \cap B \cap C| = 36$. Since $2010 = 36 \times 55 + 30$, the 30th term of the new sequence is 49, so the 2010th term is $55 \times 60 + 49 = 3349$.