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 exceeding the real number $x$. Then the number of elements in set $A$ is $\qquad$
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  
$$
0 \leqslant\left[\frac{5(k+1)}{6}\right]-\left[\frac{5 k}{6}\right] \leqslant 1 \text {. }
$$

Therefore, the elements in $A$ can cover all integers from 83 to 832, a total of 750 elements.