smc

The first $2$ radios are sold for $\frac{d}{2}$ dollars each, for income of $2\cdot \frac{d}{2}$, or simply $d$. The remaining $n-2$ radios are sold for $d+8$ dollars each, for income of $(n-2)(d+8)$, which expands to $nd+8n-2d-16$. The amount invested in buying $n$ radios for $d$ dollars is $nd$. Thus, the total profit is $(d)+(nd-2d+8n-16)-nd$, which simplifies to $8n-d-16$. Setting this equal to $72$ gives $8n-d-16=72$, or $n=11+\frac{d}{8}$. Since $n$ must be an integer, $d$ must be a multiple of $8$. If we want to minimize $n$, we want to minimize $d$. Since $d>0$, setting $d=8$ yields $n=12$, giving the answer of $\boxed{12}$