(5) From the set $M=\{1,2,3, \cdots, 2009\}$, after removing all multiples of 3 and all multiples of 5, the number of elements remaining in $M$ is $\qquad$.
(5) 1072 Hint: In set $M$, the multiples of 3 are $\left[\frac{2009}{3}\right]=669$ (numbers), the multiples of 5 are $\left[\frac{2009}{5}\right]=401$ (numbers), and the multiples of 15 are $\left[\frac{2009}{15}\right]=133$ (numbers). Therefore, the number of remaining elements is
$$
2009-(669+401-133)=1072 \text { (numbers). }
$$