1956 AMC 12 第 48 题

Let $q=2p-5;$ positivity forces $q$ to be a positive odd integer

Rewrite the quotient as $\frac{3+\frac{65}{q}}{2},$ so $q$ must be a positive divisor of $65$

Set $q=2p-5.$ A positive quotient requires $q>0,$ and $q$ is odd. Since $p=\frac{q+5}{2},$ $$\begin{aligned} \frac{3p+25}{2p-5} &=\frac{3q+65}{2q}\\ &=\frac{3+\frac{65}{q}}{2}. \end{aligned}$$ Thus $q$ must be a positive divisor of $65.$ The possibilities $q=1,$ $q=5,$ $q=13,$ and $q=65$ give $p=3,$ $p=5,$ $p=9,$ and $p=35,$ respectively, and each works. No other positive integer $p$ works.

Therefore the mathematically correct set is $\{3,5,9,35\},$ shown in the adjusted choice B.