2015 AMC 10B Problem 25

The condition is $$abc=2(ab+ac+bc).$$ Since $abc\le6bc$, we have $a\le6$. Also $a=1$ and $a=2$ give no positive solutions, so test $a=3,4,5,6$. For $a=3$, $(b-6)(c-6)=36$, giving $(b,c)=(7,42),$ $(8,24),$ $(9,18),$ $(10,15),$ $(12,12)$. For $a=4$, $(b-4)(c-4)=16$, giving $(5,20),(6,12),(8,8)$. For $a=5$, $(3b-10)(3c-10)=100$, and the only solution with $a\le b\le c$ is $(b,c)=(5,10)$. For $a=6$, $(b-3)(c-3)=9$, and the only solution with $a\le b\le c$ is $(6,6)$. The total number of triples is $5+3+1+1=10$. Thus, the correct answer is B.