10. Let $M$ denote the sum of positive integers $n$ that satisfy the following conditions: $n$ divides $2016^{2}$, and 2016 divides $n^{2}$, then the number of all distinct positive divisors of $M$ is $\qquad$ .
Solve:
$2016=2^{5} \times 3^{2} \times 7$, let $n=2^{\alpha} \times 3^{\beta} \times 7^{\gamma}$, then
$$
\left\{\begin{array} { l } 
{ \alpha \leqslant 10 , 2 \alpha \geqslant 5 , } \\
{ \beta \leqslant 4 , 2 \beta \geqslant 2 , } \\
{ \gamma \leqslant 2 , 2 \gamma \geqslant 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
3 \leqslant \alpha \leqslant 10, \\
1 \leqslant \beta \leqslant 4, \\
1 \leqslant \gamma \leqslant 2 .
\end{array} \Rightarrow M=\left(2^{3}+2^{4}+\cdots+2^{10}\right)\left(3+3^{2}+3^{3}+3^{4}\right)\left(7+7^{2}\right)\right.\right.
$$
$=2^{3} \cdot 255 \cdot 3 \cdot 40 \cdot 7 \cdot 8=2^{9} \cdot 3^{2} \cdot 5^{2} \cdot 7 \cdot 17$, so the number of positive divisors of $M$ is $10 \cdot 3 \cdot 3 \cdot 2 \cdot 2=360$.