Six dice are stacked on the floor as shown. On each die, the 1 is opposite the 6 , the 2 is opposite the 5 , and the 3 is opposite the 4 . What is the maximum possible sum of numbers on the 21 visible faces?
(A) 69
(B) 88
(C) 89
(D) 91
(E) 96

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-088.jpg?height=398&width=483&top_left_y=587&top_left_x=1252)
Label the six dice as shown:

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-167.jpg?height=398&width=501&top_left_y=2010&top_left_x=866)

The maximum overall exposed sum occurs when the sum of the exposed faces on each die is maximized.

Die $\mathrm{P}$ has 5 exposed faces. The sum of these faces is a maximum when the 1 is hidden, so the
maximum exposed sum on die $\mathrm{P}$ is $2+3+4+5+6=20$.

Dice $\mathrm{Q}$ and $\mathrm{S}$ each have 3 exposed faces. Two of these are opposite to each other, so have a sum of 7 . Thus, to maximize the exposed sum on these dice, we position them with the 6 as the unpaired exposed face. (This is on the left face of the stack.) Each of these dice has a maximum exposed sum of $6+7=13$.

Dice $\mathrm{R}$ and $\mathrm{U}$ each have 4 exposed faces. Two of these are opposite to each other, so have a sum of 7 . Thus, to maximize the exposed sum on these dice, we position them with the 6 and the 5 as the unpaired exposed faces (on the top and right of the stack). Each of these dice have a maximum exposed sum of $5+6+7=18$.

Die $\mathrm{T}$ has 2 exposed faces, which are opposite each other, so have a sum of 7.

Therefore, the maximum possible sum of the exposed faces is $20+13+13+18+18+7=89$.

ANswer: (C)