AMC10 2015 B

Answer (B): Because the volume and surface area are numerically equal, $abc=2(ab+ac+bc)$. Rewriting the equation as $ab(c-6)+ac(b-6)+bc(a-6)=0$ shows that $a\le6$. The original equation can also be written as $(a-2)bc-2ab-2ac=0$. Note that if $a=2$, this becomes $b+c=0$, and there are no solutions. Otherwise, multiplying both sides by $a-2$ and adding $4a^2$ to both sides gives $[(a-2)b-2a][(a-2)c-2a]=4a^2$. Consider the possible values of $a$. $a=1:\ (b+2)(c+2)=4$ There are no solutions in positive integers. $a=3:\ (b-6)(c-6)=36$ The 5 solutions for $(b,c)$ are $(7,42)$, $(8,24)$, $(9,18)$, $(10,15)$, and $(12,12)$. $a=4:\ (b-4)(c-4)=16$ The 3 solutions for $(b,c)$ are $(5,20)$, $(6,12)$, and $(8,8)$. $a=5:\ (3b-10)(3c-10)=100$ Each factor must be congruent to 2 modulo 3, so the possible pairs of factors are $(2,50)$ and $(5,20)$. The solutions for $(b,c)$ are $(4,20)$ and $(5,10)$, but only $(5,10)$ has $a\le b$. $a=6:\ (b-3)(c-3)=9$ The solutions for $(b,c)$ are $(4,12)$ and $(6,6)$, but only $(6,6)$ has $a\le b$. Thus in all there are 10 ordered triples $(a,b,c)$: $(3,7,42)$, $(3,8,24)$, $(3,9,18)$, $(3,10,15)$, $(3,12,12)$, $(4,5,20)$, $(4,6,12)$, $(4,8,8)$, $(5,5,10)$, and $(6,6,6)$.