12. Let $M=\{1,2,3, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A$, then $15 x \notin A$. Then the maximum number of elements in $A$ is $\qquad$.
Since $\frac{1995}{15}=133,\left[\frac{133}{15}\right]=8$, construct the set $A=\{1,2, \cdots, 8,134,135, \cdots, 1995\}$ to meet the requirement. On the other hand, since $(9,135),(10,150), \cdots,(133,1995)$, in each pair of numbers, only one can be taken, so at least $133-8=125$ numbers must be removed from $A$. Therefore, the maximum number of elements in $A$ is $1995-133+8=1870$