【Question 6】
A bottle contains 1001 blue balls, 1000 red balls, and 1000 green balls. At the same time, Xiaoming has enough balls of these three colors in his hand. Next, Xiaoming takes out two balls from the bottle each time, and then puts one or two balls back into the bottle according to the following rules:
(1) If a blue ball and a green ball are taken out, then put back one red ball;
(2) If a red ball and a green ball are taken out, then put back one red ball;
(3) If two red balls are taken out, then put back two blue balls;
(4) If the two balls taken out do not match the above three cases, then put back one green ball.

The above operations are repeated until only one ball is left in the bottle. The last remaining ball is $\qquad$ (fill in "red", "blue", "green", or "uncertain")
【Analysis and Solution】
Operation type problem; parity.
\begin{tabular}{|c|c|c|c|c|c|}
\hline & \multicolumn{2}{|c|}{ Ball Drawing Situations } & \multicolumn{2}{|c|}{ Changes in the Number of Blue, Red, and Green Balls } \\
\hline & Balls Drawn & Balls Returned or Two Balls Returned & Blue Balls & Red Balls & Green Balls \\
\hline (1) & 1 Blue Ball, 1 Green Ball & 1 Red Ball & -1 & +1 & -1 \\
\hline (2) & 1 Red Ball, 1 Green Ball & 1 Red Ball & Unchanged & Unchanged & -1 \\
\hline (3) & 2 Red Balls & 2 Blue Balls & +2 & -2 & Unchanged \\
\hline \multirow{3}{*}{ (4) } & 2 Blue Balls & 1 Green Ball & -2 & Unchanged & +1 \\
\cline { 1 - 6 } & 2 Green Balls & 1 Green Ball & Unchanged & Unchanged & -1 \\
\cline { 1 - 6 } & 1 Blue Ball, 1 Red Ball & 1 Green Ball & -1 & -1 & +1 \\
\hline
\end{tabular}

From the table, we can observe that the parity of the number of red balls and blue balls changes simultaneously;
and initially, there are 1001 blue balls and 1000 red balls in the bottle;
that is, initially, there is an odd number of blue balls and an even number of red balls, with different parities for the number of blue and red balls;
thus, no matter how the operations are performed, the parities of the number of blue and red balls will always be different;
therefore, it is impossible to end up with "0 blue balls, 0 red balls, and 1 green ball."
Moreover, we notice that in the situations of returning balls, there is no case of "returning 1 blue ball";
thus, it is also impossible to end up with "1 blue ball, 0 red balls, and 0 green balls."
In conclusion, the remaining ball is a red ball.