2015 AMC 10B Problem 20

Color the cube's corners black and white so that every edge joins different colors (this works: color a corner by the parity of its coordinate sum). Each move flips the color, so after $7$ moves Erin is on the opposite color from her start. The opposite-colored corners are the three neighbors of the start and the single antipodal corner. "Unable to return along an edge" rules out the neighbors, so every valid path must end at the antipode.

Label the top face $ABCD$ and the bottom face $EFGH$ with $E$ below $A$, $F$ below $B$, etc. Start at $A$; the antipode is $G$. By symmetry the three first moves are equivalent, so count paths beginning $A \to B$ and multiply by $3$.

From $B$ the next corner is $C$ or $F$ (not $A$).

• $A, B, C$: from $C$ the unvisited neighbors are $D$ and $G$, but $G$ must be last, so $D$. Then $H$ is forced, then $E$ (again saving $G$), then $F$, then $G$. One path.
• $A, B, F$: from $F$ go to $E$ (saving $G$), then $H$ is forced, then $D$ (saving $G$), then $C$, then $G$. One path.

So there are $2$ paths per first move and $3 \cdot 2 = 6$ in all.

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

Parity coloring is the standard first move on any grid or cube walk: it tells you which endpoints are even possible before you count anything. Once the endpoint is forced, the rule "never step onto the finish early" eliminates a branch at almost every corner, so a tree search that looks like $3 \cdot 2 \cdot 2 \cdots$ collapses to a few forced lines. Use the cube's symmetry to count one branch and multiply.

• Counting all 18 Hamiltonian paths from a corner (or all paths ending at a non-adjacent vertex without the parity restriction) instead of only those ending at the antipode.
• Forgetting to multiply by $3$ for the first move, or double counting by treating the two branches from $B$ as symmetric and then also enumerating both.