1. Given
$$
A=\{n \in \mathbf{N} \mid 1 \leqslant n \leqslant 2006 \text { and }(n+4,30) \neq 1\} \text {. }
$$

then $\operatorname{card}(A)=(\quad)$.
(A) 1605
(B) 1537
(C) 1471
(D) 1404
-1.C.
Since $30=2 \times 3 \times 5, (n+4,30) \neq 1$, therefore, $n+4=2k$ or $3k$ or $5k\left(k \in \mathbf{N}_{+}\right)$.

Also, $1 \leqslant n \leqslant 2006$, then $5 \leqslant n+4 \leqslant 2010$. Thus, there are $\left[\frac{2010}{2}\right]-2=1003$ numbers of the form $2k$, $\left[\frac{2010}{3}\right]-1=669$ numbers of the form $3k$, $\left[\frac{2010}{5}\right]=402$ numbers of the form $5k$, $\left[\frac{2010}{6}\right]=335$ numbers of the form $6k$, $\left[\frac{2010}{10}\right]=201$ numbers of the form $10k$, $\left[\frac{2010}{15}\right]=134$ numbers of the form $15k$, and $\left[\frac{2010}{30}\right]=67$ numbers of the form $30k$. Hence, $\operatorname{card}(A)=1003+669+402-335-201-134+67$ $=1471$.