2015 AMC 10B Problem 20

Step 1: where can the path end?

1. Color the cube's corners black and white so that every edge joins a black corner to a white one. (Think of the top face as a checkerboard, and the bottom face with the colors swapped.)
2. Every move along an edge changes color. After $7$ moves, an odd number, Erin is on the opposite color from where she started.
3. The corners with the opposite color are the $3$ corners next to the start and the $1$ corner directly opposite the start (the far end of the long diagonal through the cube).
4. She cannot end next to the start, because then she could walk back. So every good path ends at the opposite corner.

Step 2: count the paths.

5. Name the top corners $A, B, C, D$ going around, and the bottom corners $E, F, G, H$ with $E$ under $A$, $F$ under $B$, $G$ under $C$, $H$ under $D$. Start at $A$. The opposite corner is $G$, so $G$ must be the last corner visited, and she must avoid it until then.
6. From $A$ there are $3$ first moves, and the cube looks the same whichever one she takes. So count the paths that start $A \to B$, then multiply by $3$.
7. From $B$ she can go to $C$ or $F$.
- $A, B, C$: next is $D$ or $G$. She must save $G$, so $D$. From $D$ only $H$ is new. From $H$ she must save $G$, so $E$. From $E$ only $F$ is new, then $G$. One path: $A, B, C, D, H, E, F, G$.
- $A, B, F$: next is $E$ or $G$. Save $G$, so $E$. From $E$ only $H$ is new. From $H$ save $G$, so $D$. From $D$ only $C$ is new, then $G$. One path: $A, B, F, E, H, D, C, G$.
8. That is $2$ paths for each first move, so $3 \cdot 2 = 6$ paths.

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

Two-coloring the corners is a standard first step for walks on a grid or a cube: it tells you where the walk can possibly end before you count anything. Once the end is known, the rule "do not step on the finish early" forces almost every move, so the counting is short. The cube's symmetry lets you count one branch and multiply.