6. On the circumference of the circle, there are 3000 chess pieces placed, numbered clockwise as $1,2,3, \cdots$, 2999,3000. First, remove the 3rd chess piece, then proceed clockwise, removing 1 chess piece every 2 chess pieces, $\cdots$, until the 1st chess piece is removed. Question: (1) How many chess pieces are left on the circumference? (2) Among the remaining chess pieces on the circumference, starting from the chess piece with the smallest number, what is the number of the 181st chess piece?
6. Solution: In the first round, all numbers divisible by 3 are taken away, which means the last one taken in the first round is number 3000, a total of 1000 coins are taken, leaving 2000 coins, with number 1 still being the first. From these 2000 coins, every third one is taken away, the last one taken in the second round is the 1998th of the 2000 coins, a total of 666 coins are taken, and the 1999th and 2000th coins are not taken. When number 1 is taken, a total of $1000+666+1=1667$ coins have been taken, leaving 1333 coins.

The first round of taken coins is marked in green, and the second round of taken coins is marked in red:
$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18$, $19,20,21,22,23,24$, $\qquad$
It can be seen that every 18 numbers form a cycle, 10 numbers are removed from 18, leaving 8. After taking number 1, the smallest remaining number is 2, counting the 181st from 2 is the same as counting the 182nd from 1. $182 \div 8=22$ remainder $6,22 \times 18=396$.

List the numbers after 396 and mark them according to the above analysis:
$$
397,398,399,400,401,402,403,404,405,406,407,408,409, \cdots \cdots
$$

It can be seen that the 6th remaining number is 407, which means after taking number 1, the 181st coin from the smallest remaining number is 407.