(7) Set $A=\left\{x \left\lvert\, x=\left[\frac{5 k}{6}\right]\right., k \in \mathbf{Z}, 100 \leqslant k \leqslant 999\right\}$, where $[x]$ denotes the greatest integer not greater than $x$. Then the number of elements in set $A$ is $\qquad$ .
7 When $k=100$, $\left[\frac{5 k}{6}\right]=83$; when $k=999$, $\left[\frac{5 k}{6}\right]=832$. It is also easy to see that for $100 \leqslant k \leqslant 999$, we have
$$
0 \leqslant\left[\frac{5(k+1)}{6}\right]-\left[\frac{5 k}{6}\right] \leqslant 1,
$$

Therefore, the elements in $A$ can cover all integers from 83 to 832, so there are 750 elements in total.