$25 \cdot 4$ in the $xy$-plane passes through the point $(4,3)$, and the $x$-intercept is a (positive) prime number, and the $y$-intercept is a positive integer. The number of such lines is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(45th American High School Mathematics Examination, 1994)
[Solution] The equation of a line with intercepts at $(a, 0)$ and $(0, b)$ is
$$
\frac{x}{a}+\frac{y}{b}=1,
$$

Since the line passes through the point $(4,3)$, we have $\frac{4}{a}+\frac{3}{b}=1$, which gives
$$
(a-4)(b-3)=12,
$$

Noting that the factors of 12 are $1,2,3,4,6,12$, substituting them into the above equation, and considering that $a$ is a prime number and $b$ is a positive integer, we have only
$$
(a, b)=(5,15) \text { or }(7,7)
$$

as solutions, meaning there are two lines that satisfy the conditions.
Therefore, the answer is $(C)$.