8 The set $A$ is a 20-element subset of the set $M=\{1,2,3, \cdots, 2012\}$, and any two elements in $A$ differ by a multiple of 12. The number of such subsets $A$ is $\qquad$ .
$88 \mathrm{C}_{168}^{20}+4 \mathrm{C}_{167}^{20}$ Hint: For $x, y \in M$, if $12 \mid(x-y)$, then $x$ and $y$ are said to be of the same type. Thus, when $n \in\{1,2,3,4,5,6,7,8\}$, $n$ has 168 numbers of the same type, and for each $n$, the 168-element set
$$
T_{n}=\{x \mid x=n+12 k, k=0,1,2, \cdots, 167\}
$$

has any 20-element subset that meets the condition, yielding a total of $8 \mathrm{C}_{168}^{20}$ subsets;
When $n \in\{9,10,11,12\}$, $n$ has 167 numbers of the same type, and for each $n$, the 167-element set $T_{n}=\{x \mid x=n+12 k, k=0,1,2, \cdots, 166\}$ also has any 20-element subset that meets the condition, yielding a total of $4 \mathrm{C}_{167}^{20}$ subsets;
Therefore, the number of required sets is $8 \mathrm{C}_{168}^{20}+4 \mathrm{C}_{167}^{20}$.