8. Let $a_{1}, a_{2}, \cdots, a_{105}$ be a permutation of $1,2, \cdots, 105$, satisfying: for any $m \in\{3,5,7\}$, for all $n$ such that $1 \leqslant n<n+m \leqslant 105$, we have $m \mid\left(a_{n+m}-a_{n}\right)$. Then the number of different permutations that meet the requirement is $\qquad$ (answer with a specific number).
8. 3628800 .

By the Chinese Remainder Theorem, for given $m, n, l$, the system of equations
$$
\left\{\begin{array}{l}
a_{i} \equiv m(\bmod 3), \\
a_{i} \equiv n(\bmod 5), \\
a_{i} \equiv l(\bmod 7)
\end{array}\right.
$$

has a unique solution in $a_{i} \in [1,105]$.
Furthermore, for any $m \in \{3,5,7\}, m \mid (a_{n+m} - a_{n})$, so $a_{1}, a_{2}, \cdots, a_{m}$ form a complete residue system modulo $m$.
Thus, the number of permutations that satisfy the conditions is
$$
(3!)(5!)(7!)=3628800
$$