Need more info. Does person B open the ones that are closed, and THEN close the doors that are multiples of 3? Or close the multiples-of-three doors, and THEN open all the closed doors?
If the former, then each person undoes the previous person's work (i.e. returns all doors to "open") and then does their own thing. Which means person 100 opens EVERYTHING that's closed (which is just Door 99, but that's irrelevant), and closes Door 100.
If the latter, then everybody closes some lockers and then OPENS ALL THE LOCKERS BACK UP AGAIN. In which case, all the lockers are open after Person 100 is done.
Unless I am misreading the problem somehow . . . ?
