2021 AMC Fall 10A Problem 24

1. How many 1s? Each face adds to $2$, and there are $6$ faces, so the face totals add to $12$. Every edge lies on exactly $2$ faces, so each $1$ is counted twice in that $12$. So there are exactly $12 \div 2 = 6$ edges labeled $1$.
2. Split the edges. Stand the cube up. There are $4$ edges around the top square, $4$ around the bottom square, and $4$ upright edges joining them. The top face needs exactly two $1$s, all on its own edges; the same for the bottom. That uses $4$ of the six $1$s, so exactly $2$ of the $4$ upright edges are $1$s.
3. Name things. Number the upright edges $1, 2, 3, 4$ going around. Each side face contains two neighboring upright edges, one top edge and one bottom edge. Call the side faces $12$, $23$, $34$, $41$ by the uprights they contain.

Case A: the two upright 1s are opposite (uprights $1$ and $3$, or $2$ and $4$). There are $2$ such choices.

1. Take uprights $1$ and $3$. Every side face contains exactly one of them. So each side face needs exactly one more $1$: its top edge or its bottom edge, not both.
2. The top face needs two $1$s among its $4$ edges: $\binom42 = 6$ ways. Each side face whose top edge is $0$ must then have its bottom edge be $1$, so the bottom edges are fixed, and the bottom face also gets exactly two $1$s. Everything works.
3. Case A total: $2 \times 6 = 12$.

Case B: the two upright 1s are neighbors (like uprights $1$ and $2$). There are $4$ such choices.

1. Face $12$ already has two $1$s, so its top and bottom edges are $0$.
2. Face $34$ has no upright $1$s, so its top and bottom edges must both be $1$.
3. The top face now has one $1$ (from face $34$) and needs one more, from the top edge of face $23$ or of face $41$. Say it is face $23$. Then face $23$ is full, so its bottom edge is $0$. Face $41$ has top edge $0$, so its bottom edge is $1$. Check the bottom face: $1$ (from face $34$) $+ 1$ (from face $41$) $= 2$. It works. The other choice (face $41$) works the same way.
4. So there are $2$ ways for each choice of uprights: $4 \times 2 = 8$.

Total. $12 + 8 = 20$.

The answer is $\boxed{\textbf{(E)}\ 20}$.

First a counting fact for the whole cube (six $1$s, because each edge sits on two faces), then a smart place to start (the four upright edges, which every side face uses). Once the uprights are chosen, every other edge is either forced or a simple two-way choice. Also note the sentence about rotations and reflections: it tells you not to divide by symmetries.

How many ones? Each edge lies on exactly two faces, so adding the six face totals counts every edge twice: $6 \cdot 2 = 2 \cdot (\text{number of ones})$, so exactly $6$ edges are labeled $1$.

Pivot on the vertical edges. The top and bottom faces each contain $2$ ones among their horizontal edges, accounting for $4$ ones, so exactly $2$ of the $4$ vertical edges are $1$. Number the top corners $1, 2, 3, 4$ in order; vertical edge $v_i$ hangs from corner $i$, and side face $F_{i,i+1}$ contains $v_i$, $v_{i+1}$, top edge $t_{i,i+1}$ and bottom edge $b_{i,i+1}$.

Case A: diagonal pair ($v_1, v_3$; $2$ such pairs). Each side face has one vertical $1$ and needs exactly one of its $t, b$: $t_{i,i+1} + b_{i,i+1} = 1$ on all four sides. The top face needs two of the four $t$'s equal to $1$: $\binom42 = 6$ ways. The $b$'s are then forced and the bottom face automatically totals $2$. Total $2 \cdot 6 = 12$.

Case B: adjacent pair ($v_1, v_2$; $4$ such pairs). Face $F_{12}$ is full, so $t_{12} = b_{12} = 0$; face $F_{34}$ has no vertical ones, so $t_{34} = b_{34} = 1$. Faces $F_{23}$ and $F_{41}$ each need exactly one of $t, b$, while the top face needs one more $1$ from $t_{23}, t_{41}$ and the bottom one more from $b_{23}, b_{41}$. Choosing $t_{23} = 1$ forces $b_{23} = 0$, $b_{41} = 1$, $t_{41} = 0$; the mirror choice also works. So $2$ completions, total $4 \cdot 2 = 8$.

Altogether $12 + 8 = 20$.

The answer is $\boxed{\textbf{(E)}\ 20}$.