jmc

For each $i$th switch (designated by $x_i,y_i,z_i$), it advances itself only one time at the $i$th step; thereafter, only a switch with larger $x_j,y_j,z_j$ values will advance the $i$th switch by one step provided $d_i=2^{x_i}{3}^{y_i}{5}^{z_i}$ divides $d_j=2^{x_j}{3}^{y_j}{5}^{z_j}$. Let $N=2^9{3}^9{5}^9$ be the max switch label. To find the divisor multiples in the range of $d_i$ to $N$, we consider the exponents of the number $\frac{N}{d_i}=2^{9-x_i}{3}^{9-y_i}{5}^{9-z_i}$. In general, the divisor-count of $\frac{N}{d}$ must be a multiple of 4 to ensure that a switch is in position A: $4n=[(9-x)+1][(9-y)+1][(9-z)+1]=(10-x)(10-y)(10-z)$, where $0\le x,y,z\le 9.$ We consider the cases where the 3 factors above do not contribute multiples of 4. Case of no 2's: The switches must be $(\mathrm{odd})(\mathrm{odd})(\mathrm{odd})$. There are $5$ odd integers in $0$ to $9$, so we have $5\times 5\times 5=125$ ways. Case of a single 2: The switches must be one of $(2\cdot \mathrm{odd})(\mathrm{odd})(\mathrm{odd})$ or $(\mathrm{odd})(2\cdot \mathrm{odd})(\mathrm{odd})$ or $(\mathrm{odd})(\mathrm{odd})(2\cdot \mathrm{odd})$. Since $0\le x,y,z\le 9,$ the terms $2\cdot 1,2\cdot 3,$ and $2\cdot 5$ are three valid choices for the $(2\cdot odd)$ factor above. We have $\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot 3\cdot 5^2=225$ ways. The number of switches in position A is $1000-125-225=\boxed{650}$.