2002 AMC 10A Problem 18

Each die can be oriented independently, so minimize each die's visible total separately and add.

Sort the $27$ dice by how many faces show. The $8$ corner dice show $3$ faces, the $12$ edge dice show $2$, the $6$ dice at the centers of the big faces show $1$, and the central die shows none.

Because opposite faces sum to $7$, the three faces at any corner of a die are pairwise non-opposite; the faces $1, 2, 3$ are pairwise non-opposite (their opposites are $6, 5, 4$), so they do meet at a corner and a corner die can show $1 + 2 + 3 = 6$, which is the least possible. An edge die shows two adjacent faces, at best $1 + 2 = 3$. A face-center die shows a single face, at best $1$.

Minimum total:
$$8 \cdot 6 + 12 \cdot 3 + 6 \cdot 1 = 48 + 36 + 6 = 90.$$

The answer is $\boxed{\textbf{(D)}\ 90}$.

When the parts of a sum can be chosen independently, the minimum of the sum is the sum of the minima, so the problem reduces to knowing the $8/12/6/1$ breakdown of a $3\times3\times3$ cube and the smallest total on $1$, $2$ or $3$ adjacent faces of a die. The "opposite faces sum to $7$" clue is what guarantees the small numbers $1, 2, 3$ can all be visible at once.

• Miscounting the 8 corner, 12 edge and 6 face-center dice (or including the hidden center die), which shifts the total to 84 or 96.
• Doubting that $1$, $2$ and $3$ can all show on one corner die; they can, because none of them are opposite each other (their opposites are $6$, $5$, $4$).