5. 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 integer pairs $(a, b)$ that satisfy this system of inequalities is $(\quad)$.
(A) 17
(B) 64
(C) 72
(D) 81
5. (C).

From the original system of inequalities, we get $\frac{a}{9} \leqslant x<\frac{b}{8}$.
Drawing 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 $0<\frac{a}{9} \leqslant 1, 3<\frac{b}{8} \leqslant 4$.
From $0<\frac{a}{9} \leqslant 1$, we get $0<a \leqslant 9$.
$\therefore a=1,2,3, \cdots, 9$, 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.
$\therefore 9 \times 8=72$ (cases). Hence, the answer is (C).