The switch labeled $d$ is advanced exactly once for each step $i$ whose label is a multiple of $d.$ The multiples of $2^x 3^y 5^z$ among the labels are the $2^{x'} 3^{y'} 5^{z'}$ with $x \le x' \le 9,$ etc., so that switch advances $(10 - x)(10 - y)(10 - z)$ times. It returns to position $A$ exactly when this count is a multiple of $4.$

Write $a = 10 - x,$ $b = 10 - y,$ $c = 10 - z,$ each ranging over $1$ through $10.$ We count the triples where $abc$ is not divisible by $4:$ either all three are odd, or exactly one is even but not divisible by $4.$ Among $1, \ldots, 10$ there are $5$ odd values and $3$ values ($2, 6, 10$) that are twice an odd number. That gives $5^3 = 125$ triples of the first kind and $3 \cdot 3 \cdot 5^2 = 225$ of the second, or $350$ in all.

Therefore $1000 - 350 = 650$ switches end in position $A.$