2016 AMC 10A Problem 18

Step 1: every face sums to $18$. The numbers $1$ to $8$ add to $36$. Each corner of a cube touches $3$ faces, so adding all six face sums counts every corner three times: $3 \cdot 36 = 108$. Six equal sums of $108$ means each face sums to $18$.

Step 2: name the corners. Call the top face $A, B, C, D$ going around, and the bottom face $E, F, G, H$, with $E$ under $A$, $F$ under $B$, $G$ under $C$, $H$ under $D$. Opposite corners (at the two ends of a long diagonal through the cube) are $A$ and $G$, $B$ and $H$, $C$ and $E$, $D$ and $F$.

Step 3: opposite edges have equal sums. The top face and the front face share the edge $AB$:
$$\text{top} = (A + B) + (C + D) = 18, \qquad \text{front} = (A + B) + (E + F) = 18.$$
So $C + D = E + F$. The same trick works for any two faces that share an edge, and it always says: two parallel edges on opposite sides of the cube have equal sums.

Step 4: opposite corners differ by the same amount. Apply step 3 to the three edges at corner $A$:
- Edge $AB$ and its opposite edge $HG$: $A + B = H + G$, so $A - G = H - B$.
- Edge $AD$ and its opposite edge $FG$: $A + D = F + G$, so $A - G = F - D$.
- Edge $AE$ and its opposite edge $CG$: $A + E = C + G$, so $A - G = C - E$.

So there is one number $d$ with
$$A - G = H - B = F - D = C - E = d.$$
The four corners $A, C, F, H$ (no two of them are next to each other) each hold a number $d$ larger than the corner opposite them.

Step 5: this pattern is enough. Any face contains two of $A, C, F, H$ and two other corners, which are the opposites of the other two. For the top face: $A + B + C + D = A + C + (H - d) + (F - d)$. Every face works out to $(A + C + F + H) - 2d$, so all faces have equal sums.

Step 6: which differences $d$ are possible? We need to split $1$ to $8$ into four pairs that all differ by $d$.
- $d = 1$: $\{1,2\}, \{3,4\}, \{5,6\}, \{7,8\}$. Works.
- $d = 2$: $\{1,3\}, \{2,4\}, \{5,7\}, \{6,8\}$. Works.
- $d = 3$: $1$ pairs with $4$, $2$ with $5$, $3$ with $6$; then $7$ would need $4$ or $10$. Fails.
- $d = 4$: $\{1,5\}, \{2,6\}, \{3,7\}, \{4,8\}$. Works.
- $d \ge 5$: the number $4$ would need a partner of $9$ or more, or below $0$. Fails.

Step 7: count placements for each $d$. Turn the cube so $8$ sits at corner $A$. Its partner goes to $G$. The larger numbers of the other three pairs go to $C$, $F$, $H$, with their partners opposite. There are $3 \cdot 2 \cdot 1 = 6$ ways to assign the three pairs to $C, F, H$. But spinning the cube about the long diagonal through $A$ and $G$ keeps $8$ in place and moves $C \to F \to H \to C$, so these $6$ ways come in groups of $3$ that are the same arrangement. That leaves $6 \div 3 = 2$ different arrangements.

Three choices of $d$, each with $2$ arrangements: $3 \cdot 2 = 6$.

The answer is $\boxed{\textbf{(C)}\ 6}$.

When many sums must all be equal, subtract two equations that share terms; the shared part cancels and a simpler fact appears (here, opposite edges have equal sums, and then opposite corners differ by a fixed amount). Once the pattern is forced, counting is a small number puzzle plus care with rotations. To count "up to rotation", fix the position of one special number first, then see which spins still keep it in place.