5. (10 points) Class 1 of Grade 4 used the class fund to purchase three types of stationery, A, B, and C, with unit prices of 3 yuan, 2 yuan, and 1 yuan, respectively. It is known that the number of B stationery purchased is 2 less than the number of A stationery purchased, and the cost of A stationery does not exceed half of the total cost. If the total cost for the three types of stationery is exactly 66 yuan, then the maximum number of A stationery that can be purchased is $\qquad$ pieces.
【Analysis】Let the cost of purchasing stationery A be $x$ yuan, then B is $(x-2)$ yuan. Since the total cost for A, B, and C stationery is exactly 66 yuan, the cost of purchasing A and B stationery $\leqslant 66$. Also, the cost of purchasing stationery A does not exceed half of the total cost. Solve the inequality accordingly.

【Solution】9 Solution: Let the quantity of stationery A purchased be $x$, and the quantity of stationery C be $y$, then the quantity of stationery B is $(x-2)$, so $3 x+2(x-2)+y=66$,

which is $y=70-5 x$,
and $3 x \leqslant \frac{66}{2}$,
solving gives $x \leqslant 11$,
so the maximum number of stationery A that can be purchased is 11,
Answer: The maximum number of stationery A that can be purchased is 11.
Therefore, the answer is: 11.