Walter rolls four standard six-sided dice and finds that the product of the numbers of the upper faces is $144$. Which of he following could not  be the sum of the upper four faces? 
$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18$
We have $144 = 2^4 3^2$.
As the numbers on the dice are less than $9$, the two $3$s must come from different dice. This leaves us with three cases: $(6,6,a,b)$, 
$(6,3,a,b)$, and $(3,3,a,b)$.
In the first case we have $ab=2^2$, leading to the solutions $(6,6,4,1)$ and $(6,6,2,2)$.
In the second case we have $ab=2^3$, leading to the only solution $(6,3,4,2)$.
In the third case we have $ab=2^4$, leading to the only solution $(3,3,4,4)$.
We found all four possibilities for the numbers on the upper faces of the dice. The sums of these numbers are $17$, $16$, $15$, and $14$.
Therefore the answer is $\mathrm{(E)}$.