AMC 8 · 2003 · #13

Grade 4 geometry-3d
spatial-visualizationface-adjacencysystematic-enumerationcombinations-basic caseworksystematic-enumeration ↑ Prerequisites: spatial-visualizationsystematic-enumeration
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Problem
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?

Pick an answer.

(A)
4
(B)
6
(C)
8
(D)
10
(E)
12

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Create a Physical Representation

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 - neighbors." Three groups, three subtractions, then add the group with 4 red faces.

1STEP 1

Build the figure: a hollow rectangular ring of 10 floor cubes plus 1 cube on each corner — 10 + 4 = 14 cubes, matching the problem.

10 (floor ring) + 4 (corner stacks) = 14
2STEP 2

Key rule: red faces = 6 - neighbors. Since the bottom is painted too, only glued faces are hidden — the ground hides nothing.

red faces = 6 - neighbors
3STEP 3

Sort the 14 cubes by position: floor corners have 3 neighbors, floor-edge middles 2 neighbors, top corners 1 neighbor — 4 + 6 + 4 = 14.

4 + 6 + 4 = 14 ✓
4STEP 4

Apply 6 - neighbors: type A gets 3 red, type B gets 4 red, type C gets 5 red. Only type B hits exactly four, and there are 6 such cubes.

Type A: 6-3=3, Type B: 6-2=4, Type C: 6-1=5 → (B) 6
Answer
6
Cross-check by counting total red faces two ways. Type A contributes 4 × 3 = 12 red faces, Type B contributes 6 × 4 = 24, Type C contributes 4 × 5 = 20, for a grand total of 12 + 24 + 20 = 56 red faces. Independently, the painted figure has a 3 × 4 = 12-square bottom, a 12-square top of the floor ring (with the inner 1 × 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.
💡Key takeaway

Every face that hugs a neighbor stays white, so red faces = 6 - 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.