2015 AMC 12B Problem 23

Numerically equal volume and surface area means $abc=2(ab+bc+ca).$ Dividing by $abc$ gives $1=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\le\frac{6}{a},$ so $a\le6.$ The cases $a=1$ and $a=2$ give no positive solutions. For $a\ne2,$ set $$\begin{gathered} u=(a-2)b-2a,\\ v=(a-2)c-2a. \end{gathered}$$ The equation then factors as $uv=4a^2.$ For $a=3,$ $(b-6)(c-6)=36$ gives $(b,c)=(7,42),$ $(8,24),$ $(9,18),$ $(10,15),$ $(12,12).$ For $a=4,$ $(b-4)(c-4)=16$ gives $(5,20),(6,12),(8,8).$ For $a=5,$ the congruence of the two factors modulo $3$ leaves only the valid pair $(b,c)=(5,10),$ and for $a=6,$ $(b-3)(c-3)=9$ leaves only $(b,c)=(6,6).$ Thus there are $5+3+1+1=10$ triples. Thus, the correct answer is B.