Prime-indexed door togglers on 100 doors

A hallway has $100$ doors, numbered $1$ through $100$, and every one of them starts the day open. A sequence of workers then passes through. Worker number $k$ is assigned the $k$-th prime ($2, 3, 5, 7, \dots$) and flips the state of every door whose number is a multiple of that prime: an open door gets shut and a shut door gets reopened. Once every prime up to $100$ has had its worker, how many doors remain open?

57. Door $n$ is flipped once for each distinct prime dividing it, so it ends open exactly when $n$ has an even number of distinct prime factors. The doors that end shut are the $25$ primes, the $10$ higher prime powers ($4,8,9,16,25,27,32,49,64,81$) and the $8$ numbers with three distinct primes ($30,42,60,66,70,78,84,90$): $43$ in all, leaving $100-43=57$ open.