$33 \cdot 36$ If $\{a\}$ represents the smallest integer greater than $a$, such as $\{2.5\}=3, \{4\}=5, \{-1.5\}=-1$; and $[a]$ represents the largest integer not greater than $a$, such as $[2.5]=2, [4]=4, [-1.5]=-2$. Then, the solution to the system of equations
$$
\left\{\begin{array}{l}
3[x]+2\{y\}=18 \\
3\{x\}-[y]=4 .
\end{array}\right.
$$

is
(A) $\left\{\begin{array}{l}2<x \leqslant 3, \\ 5 \leqslant y<6\end{array}\right.$
(B) $\left\{\begin{array}{l}2 \leqslant x<3, \\ 5<y \leqslant 6\end{array}\right.$
(C) $\left\{\begin{array}{l}2 \leqslant x<3 \\ 5 \leqslant y<6\end{array}\right.$.
(D) $\left\{\begin{array}{l}2<x \leqslant 3, \\ 5<y \leqslant 6 .\end{array}\right.$
(8th "Jinyun Cup" Junior High School Mathematics Invitational Competition, 1991)
[Solution] From the given, we know $\{x\}=[x]+1$, and $\{y\}=[y]+1$. Therefore, the given system of equations can be transformed into
$$
\begin{array}{ll} 
& \left\{\begin{array}{l}
3[x]+2[y]=16, \\
3[x]-[y]=1 .
\end{array}\right. \\
\text { Solving, we get } \\
\therefore \quad 2 \leqslant x<3, \quad \text { and } 5 \leqslant y<6 .
\end{array}
$$

Solving, we get

Therefore, the correct choice is $(C)$.