$31 \cdot 11-$ a two-digit number, when the tens and units digits are swapped, the resulting two-digit number is $\frac{7}{4}$ times the original number. How many such two-digit numbers are there?
(A) 1.
(B) 2.
(C) 4.
(D) Infinitely many.
(E) 0.
(China Tianjin Junior High School Mathematics Competition, 1984)
［Solution］Let the two-digit number be $10a + b$. According to the problem, we have
$$
10b + a = \frac{7}{4}(10a + b),
$$

which simplifies to $66a - 33b = 0$, or $2a - b = 0$.
Since $a, b$ are both single digits, then $\left\{\begin{array}{l}a=1,2,3,4, \\ b=2,4,6,8 .\end{array}\right.$ Therefore, the answer is $(C)$.