9 How many integer solutions $(x, y, z)$ are there to the equation $x y z=2008$ ?
(A) 30
(B) 60
(C) 90
(D) 120
(E) 150
9. Answer: (D)
$2008=2^{3} \times 251$. Consider $|x|=2^{p_{1}} \times 251^{q_{1}},|y|=2^{p_{2}} \times 251^{q_{2}}$ and $|z|=2^{p_{3}} \times$ $251^{q_{3}}$. Then $p_{1}+p_{2}+p_{3}=3$ and $q_{1}+q_{2}+q_{3}=1$, so the number of positive integer solutions is ${ }^{5} \mathrm{C}_{2} \times{ }^{3} \mathrm{C}_{2}=30$. Together with the 4 possible distributions of the signs: $(+,+,+),(+,-,-),(-,+,-),(-,-,+)$, the equation has $30 \times 4=120$ integer solutions.