$31 \cdot 1$ A merchant buys $n$ radios for $d$ dollars, where $d$ is a positive integer. Two of them he sells to a charity sale at half cost, and the rest he sells at a profit of 8 dollars each. If the total profit is 72 dollars, then the smallest possible value of $n$ is
(A) 18 .
(B) 16 .
(C) 15 .
(D) 12 .
(E) 11
(18th American High School Mathematics Examination, 1967)
[Solution] Given that the cost of each radio is $\frac{d}{n}$ yuan, according to the problem, we have
$$
(n-2)\left(\frac{d}{n}+8\right)+2 \cdot \frac{d}{2 n}=d+72 \text {, }
$$

The equation can be simplified to $n(n-11)=\frac{d}{8}$, where $n, d$ are positive integers.
If $n \leqslant 11$, then $n(n-11) \leqslant 0$, which is not valid;
If $n=12$, then $d=8 n(n-11)=96$, clearly 12 is the smallest $n$. Therefore, the answer is $(D)$.