13.5 Let $a, b, c, d$ be integers, and the roots of the following four equations in $x$
$$
\begin{array}{rr}
(a-2 b) x=1, & (b-3 c) x=1, \\
(c-4 d) x=1, & x+100=d,
\end{array}
$$

are all positive. Then the smallest possible value of $a$ is
(A) 2433 .
(B) 2425 .
(C) 2401 .
(D) 2400 .
(China Guangzhou, Wuhan, Fuzhou, etc. Five Cities Junior High School Mathematics League, 1990)
[Solution] From the given conditions, we have
$$
\left\{\begin{array}{l}
\frac{1}{a-2b} > 0, \\
\frac{1}{b-3c} > 0, \\
\frac{1}{c-4d} > 0, \\
d - 100 > 0.
\end{array} \text { i.e., } \quad \left\{\begin{array}{l}
a-2b > 0, \\
b-3c > 0, \\
c-4d > 0, \\
d > 100.
\end{array}\right.\right.
$$

Given that \(a, b, c, d\) are all integers, we know that \(d\) is at least 101.
Also, \(c > 4d \geq 4 \cdot 101 = 404\), so \(c\) is at least 405.
And \(b > 3c \geq 3 \cdot 405 = 1215\), so \(b\) is at least 1216.
Furthermore, \(a > 2b \geq 2 \cdot 1216 = 2432\), so \(a\) is at least 2433. Therefore, the answer is \((A)\).