$4 \cdot 12$ If the system of inequalities $\left\{\begin{array}{l}9 x-a \geqslant 0, \\ 8 x-b<0\end{array}\right.$ has only the integer solutions $1,2,3$, then the number of ordered pairs $(a, b)$ of integers that satisfy this system of inequalities is
(A) 17.
(B) 64.
(C) 72.
(D) 81.
(China Junior High School Mathematics League, 1998)
[Solution] From the original system of inequalities, we get $\frac{a}{9} \leqslant x<\frac{b}{8}$.
Plotting the possible interval of the solution set of this system of inequalities on the number line, as shown in the figure below:

It is easy to see that
$$
0<\frac{a}{9} \leqslant 1, \quad 3<\frac{b}{8} \leqslant 4 .
$$

From $0<\frac{a}{9} \leqslant 1$, we get $0<a \leqslant 9$.
$\therefore \quad a=1,2,3, \cdots, 9$, a total of 9.
From $3<\frac{b}{8} \leqslant 4$, we get $3 \times 8<b \leqslant 4 \times 8$.
$\therefore b=3 \times 8+1,3 \times 8+2,3 \times 8+3, \cdots, 3 \times 8+8$. A total of 8. $\therefore \quad 9 \times 8=72$ (cases).
Therefore, the answer is (C).