9. (10 points) As shown in the figure, a large triangular board is cut into four smaller triangular boards (first operation), and then each of the smaller triangular boards is cut into four even smaller triangular boards (second operation). This process continues, and after the 5th operation, a number of small triangular boards are obtained. Jia and Yi take turns coloring these small triangular boards, with each person coloring 1 to 10 small triangular boards at a time. The one who colors the last board wins. Given that Jia goes first, please devise a strategy for Jia to win.
【Analysis】According to the known information, the first operation results in 4 small equilateral triangles, the second operation results in $4 \times 4=16$ small equilateral triangles; the third operation results in $4 \times 4 \times 4=64$ small equilateral triangles; ... and the fifth operation results in $4 \times 4 \times 4 \times 4 \times 4=1024$ small triangles. First, understand the problem: the one who paints first wins, and Jia must paint first. Using the number of small triangles Jia paints, which is 10 minus the number of triangles Yi paints, we can solve this problem: Jia paints 1 first, and then when Yi paints $n$ triangles, Jia paints $11-n$ triangles, according to this, the problem can be solved.

【Solution】Solution: $4 \times 4 \times 4 \times 4 \times 4=1024$ (triangles)
$$
1024 \div(10+1)=93 \cdots 1
$$

Jia paints first. Since the remainder is 1, Jia paints 1 triangle first, then Yi paints $n(1 \leqslant n \leqslant 10)$ triangles, and Jia paints $(11-n)$ triangles next;

After each time Yi paints, Jia paints the number of triangles as 11 minus the number Yi painted;
In the end, 1 to 10 triangles will be left for Jia,
This way, Jia can paint all the remaining small triangles last.