AMC 8 · 2003 · #13
Easy mode Grade 4Problem
Imagine 14 small white cubes stacked together to make one shape. The whole outside of the shape, including the bottom, is dipped in red paint.
Then the shape is taken apart, back into the 14 separate cubes.
Some of these cubes will have paint on exactly 4 of their faces. How many of them?
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Fourteen identical white cubes are glued together into a single figure. The entire outside surface of the figure — including the bottom — is painted red. The figure is then taken apart into the original $14$ cubes. How many of those cubes end up with exactly $4$ red faces?
Givens: $14$ unit cubes assembled into one figure; The figure forms a rectangular ring of $10$ cubes on the floor ($3 \times 4$ outline with the middle hollow) plus $4$ corner cubes stacked one high on top of the ring's corners; Every exterior face of the assembled figure is painted red, including the bottom; Touching faces between two cubes are NOT painted; After painting, the cubes are separated again; Answer choices: (A) $4$, (B) $6$, (C) $8$, (D) $10$, (E) $12$
Unknowns: The number of individual cubes that have exactly $4$ red faces
Understand
Restated: Fourteen identical white cubes are glued together into a single figure. The entire outside surface of the figure — including the bottom — is painted red. The figure is then taken apart into the original $14$ cubes. How many of those cubes end up with exactly $4$ red faces?
Givens: $14$ unit cubes assembled into one figure; The figure forms a rectangular ring of $10$ cubes on the floor ($3 \times 4$ outline with the middle hollow) plus $4$ corner cubes stacked one high on top of the ring's corners; Every exterior face of the assembled figure is painted red, including the bottom; Touching faces between two cubes are NOT painted; After painting, the cubes are separated again; Answer choices: (A) $4$, (B) $6$, (C) $8$, (D) $10$, (E) $12$
Plan
Primary tool: #10 Create a Physical Representation
Secondary: #2 Make a Systematic List by Cube Type
The figure is a 3D arrangement of $14$ cubes, which is exactly when Tool #10 (Create a Physical Representation) earns its keep — stack real blocks (or sketch the floor plan with a separate top layer) so you can see which faces of each cube touch a neighbor. Tool #2 (Make a Systematic List) is the bookkeeping partner: group the $14$ cubes by position type (bottom-corner, bottom-edge-middle, top-corner), count the neighbors for one cube in each group, and use the rule "red faces $= 6 - \text{neighbors}$." Three groups, three subtractions, then add the group with $4$ red faces.
Execute — Answer: B
1.G.A.2 Step 1 - Build the figure with blocks (or sketch it) to see the structure.
- The bottom layer is a hollow rectangle: $3$ cubes along the front, $3$ along the back, and $2$ on each side, for $10$ cubes total.
- On top of each of the $4$ floor corners sits one extra cube, adding $4$ more.
- Total: $10 + 4 = 14$ cubes — matches the problem.
💡 Composing 3D shapes from unit cubes is a Grade 1 spatial skill. The split into "floor ring + corner stacks" gives a clean count and lines up the three cube types.
1.G.A.2 Step 2 - Use the rule: red faces per cube $= 6 - \text{(neighbors that cube touches)}$.
- The bottom of the figure is painted too, so floor cubes get NO hidden face from the ground.
- The only hidden faces are the ones glued to another cube.
💡 Each touch covers exactly one face on each cube. "Bottom painted too" is the line that makes this clean — no extra subtractions for cubes sitting on the floor.
4.OA.A.3 Step 3 - Sort the $14$ cubes into three types and count neighbors of one cube in each type.
- Type A — floor corner ($4$ cubes at the four floor corners): neighbors are $1$ cube along the front/back edge, $1$ cube along the side edge, and $1$ cube stacked on top.
- That is $3$ neighbors.
- Type B — floor edge middle ($6$ cubes: the middle of each side of the ring): each has $2$ neighbors along the ring and nothing stacked above.
- Type C — top corner ($4$ stacked cubes): each sits on $1$ cube below and has no other neighbors, so $1$ neighbor.
💡 Cubes in the same position type behave identically by symmetry, so one count per type covers all $14$ cubes.
4.OA.A.3 Step 4 - Apply $6 - \text{neighbors}$ to each type.
- Type A: $6 - 3 = 3$ red faces.
- Type B: $6 - 2 = 4$ red faces.
- Type C: $6 - 1 = 5$ red faces.
- Only Type B hits exactly $4$ red faces, and there are $6$ cubes of that type.
💡 "Exactly $4$ red faces" means "exactly $2$ neighbors," which singles out the middles of the four ring sides.
1.G.A.2 Build the figure with blocks (or sketch it) to see the structure. The bottom lay 1.G.A.2 Use the rule: red faces per cube $= 6 - \text{(neighbors that cube touches)}$. T 4.OA.A.3 Sort the $14$ cubes into three types and count neighbors of one cube in each typ 4.OA.A.3 Apply $6 - \text{neighbors}$ to each type. Type A: $6 - 3 = 3$ red faces. Type B Review
Reasonableness: Cross-check by counting total red faces two ways. Type A contributes $4 \times 3 = 12$ red faces, Type B contributes $6 \times 4 = 24$, Type C contributes $4 \times 5 = 20$, for a grand total of $12 + 24 + 20 = 56$ red faces. Independently, the painted figure has a $3 \times 4 = 12$-square bottom, a $12$-square top of the floor ring (with the inner $1 \times 2$ hole and the four corner squares now hidden under the stacked cubes), a top surface on each of the $4$ corner stacks ($4$ squares), and outer/inner side strips around the ring plus the four sides of each stacked cube. Carefully adding the exposed surface yields the same $56$ squares, confirming the neighbor count is correct. Also, the answer $6$ is the middle choice — a reasonable AMC 8 sanity signal that the count is not at either extreme.
Alternative: Tool #17 (Visualize Spatial Relationships): translate "exactly $4$ red faces" directly into "exactly $2$ hidden faces," i.e., the cube has exactly $2$ neighbors. In the assembled figure, every cube on the floor ring lies along one of the four straight ring sides. The cubes at the four floor corners are bends with $3$ touches (two ring neighbors plus the stacked cube above), and the cubes on top each touch only $1$ cube (the floor corner beneath). That leaves exactly the interior cubes of the four ring sides — $1$ in the front side, $1$ in the back side, $2$ in the left side, $2$ in the right side — which is $1 + 1 + 2 + 2 = 6$. Same answer (B).
CCSS standards used (min grade 4)
1.G.A.2Compose two-dimensional shapes or three-dimensional shapes to create a composite shape (Visualizing the $14$ unit cubes as a composite figure — a floor ring of $10$ plus $4$ stacked corner cubes — so neighbors of each cube can be read off the build.)4.OA.A.3Solve multistep word problems with whole numbers using the four operations (Grouping the $14$ cubes into three position types, applying the rule $\text{red faces} = 6 - \text{neighbors}$ to each type, and selecting the type whose count equals $4$.)
⭐ Every face that hugs a neighbor stays white, so red faces $= 6 - \text{neighbors}$. Sort the $14$ cubes by where they sit — floor corner, floor edge, top corner — and the $6$ cubes in the middles of the ring's sides are the ones with exactly $4$ red faces.
⭐ Every face that hugs a neighbor stays white, so red faces $= 6 - \text{neighbors}$. Sort the $14$ cubes by where they sit — floor corner, floor edge, top corner — and the $6$ cubes in the middles of the ring's sides are the ones with exactly $4$ red faces.
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