8. If positive integers $a, b, c$ satisfy $2017 \geq 10 a \geq 100 b \geq 1000 c$, then the number of tuples $(a, b, c)$ is . $\qquad$
Answer: 574.
Solution: From the condition, we know that $c \leq\left[\frac{2017}{1000}\right]=2$.
When $c=1$, we have $10 \leq b \leq 20$. For each such positive integer $b$, from $10 b \leq a \leq 201$, we know the number of corresponding $a$ is $202-10 b$. Therefore, the number of such positive integer tuples is
$$
\sum_{b=10}^{20}(202-10 b)=\frac{(102+2) \times 11}{2}=572 .
$$

When $c=2$, from $20 \leq b \leq\left[\frac{2017}{100}\right]$, we know $b=20$. Then $200 \leq a \leq\left[\frac{2017}{10}\right]=201$, so $a=200,201$. There are 2 tuples $(a, b, c)$ in this case.
In summary, the number of positive integer tuples that satisfy the condition is $572+2=574$.