8. The numbers from 1 to 1000 are written in a circle. Starting from the first, every 15th number is crossed out (i.e., the numbers $1, 16, 31$, etc.), and during subsequent rounds, already crossed-out numbers are also taken into account. The crossing out continues until it turns out that all the numbers to be crossed out have already been crossed out previously. How many numbers will remain uncrossed out?
8. After the first cycle, all numbers that give a remainder of 1 when divided by 15 will be crossed out; the last such number will be 991. The first number to be crossed out during the second cycle will be $991+15-1000=6$; subsequently, during the second cycle, all numbers that give a remainder of 6 when divided by 15 will be crossed out (the last such number is 996). The first number to be crossed out during the third cycle will be $996+15-1000=11$; subsequently, during the third cycle, all numbers that give a remainder of 11 when divided by 15 will be crossed out (the last such number is 986). The first number to be crossed out during the fourth cycle will be $986+15-1000=1$. Since this number has already been crossed out earlier, further counting intervals of 15 numbers will always land on already crossed out numbers, so no new numbers will be crossed out.

Thus, ultimately, all numbers that give a remainder of 1, 6, or 11 when divided by 15, and only these numbers, will be crossed out. However, numbers that give a remainder of 1, 6, or 11 when divided by 15 coincide with all numbers that give a remainder of 1 when divided by 5. There are $\frac{1000}{5}=200$ such numbers among the first 1000 (numbers $1,6,11,16, \ldots, 996=5 \cdot 199+1$). Therefore, $1000-200=$ $=800$ numbers will remain uncrossed.