Tricky Lists

By the time you get to list C, have have lost multiples of $3$ and $7$ , but regained those multiples of $3$ which are $1$ less than a multiple of $7$ . There are

• $333$ multiples of $3$ ,
• $142$ multiples of $7$ ,
• $47$ multiples of $21$ ,

so there are $333 + 142 - 47 = 428$ numbers which are multiples of either $3$ or $7$ . Numbers which are multiples of $3$ and also $1$ less than a multiple of $7$ are of the form $6 + 21k$ for $k \ge 0$ , and so there are $48$ such numbers. Thus a total of $428-48 = 380$ numbers are removed from list A when list C is created, and so there are $\boxed{620}$ different numbers still left in list C.