$32 \cdot 54$ can make the integers $m$ such that $2 m-1,4 m+5,20-m$ can be the three sides of a triangle are
(A) 2.
(B) 6.
(C) 12
(D) 18.
(Jiangxi Province Nanchang City Junior High School Mathematics Competition, 1990)
［Solution］From the triangle inequality, we have
$$
\left\{\begin{array}{l}
(2 m-1)+(20-m)>4 m+5 \\
(2 m-1)+(4 m+5)>20-m .
\end{array}\right.
$$

Simplifying, we get \( m > \frac{16}{7} \) and \( m < \frac{14}{3} \), so \( \frac{16}{7} < m < \frac{14}{3} \). The values of \( m \) that satisfy these two inequalities can only be 3 or 4.
Therefore, the answer is (A).