AMC 8 · 2020 · #9
Easy mode Grade 3Problem
Imagine a birthday cake shaped like a cube, 4 inches on each side. The cake has icing on the top and on all four side faces. The bottom has no icing.
Now suppose the cake is sliced into 64 smaller cubes, each 1×1×1 inch. Some of these little cubes are corners, some are along edges, and some are buried inside.
How many of the 64 small cubes have icing on exactly 2 of their sides?
Pick an answer.
AMC 8 2020 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: Akash's birthday cake is a $4 \times 4 \times 4$ inch cube with icing on the top face and all $4$ side faces but no icing on the bottom. The cake is sliced into $64$ unit cubes ($1 \times 1 \times 1$ inch each). How many of these small cubes end up with icing on exactly $2$ of their faces?
Givens: Big cube edge length $= 4$ inches; sliced into $4 \times 4 \times 4 = 64$ unit cubes; Icing on the top face and all $4$ side faces ($5$ faces total); No icing on the bottom face; Answer choices: (A) $12$, (B) $16$, (C) $18$, (D) $20$, (E) $24$
Unknowns: The number of unit cubes that have icing on exactly $2$ of their $6$ faces
Understand
Restated: Akash's birthday cake is a $4 \times 4 \times 4$ inch cube with icing on the top face and all $4$ side faces but no icing on the bottom. The cake is sliced into $64$ unit cubes ($1 \times 1 \times 1$ inch each). How many of these small cubes end up with icing on exactly $2$ of their faces?
Givens: Big cube edge length $= 4$ inches; sliced into $4 \times 4 \times 4 = 64$ unit cubes; Icing on the top face and all $4$ side faces ($5$ faces total); No icing on the bottom face; Answer choices: (A) $12$, (B) $16$, (C) $18$, (D) $20$, (E) $24$
Plan
Primary tool: #10 Create a Physical Representation
Secondary: #7 Identify Subproblems, #2 Make a Systematic List
Because the icing covers $5$ of the $6$ faces, the usual symmetric "corners have 3, edges have 2, faces have 1" rule does not apply — the missing bottom flips some categories. Tool #10 (build a physical cube of $4 \times 4 \times 4$ unit cubes, or sketch it) makes it visible which positions touch exactly two iced faces. Tool #7 (Identify Subproblems) splits the count into three location types — top edges, vertical edges, and bottom corners — that each contribute to the "exactly two iced" count. Tool #2 (Systematic List) then counts each type without missing or double-counting any cube.
Execute — Answer: D
K.G.B.4 Step 1 - Picture the cube and label its $6$ kinds of positions: $8$ corners, $12$ edges (each edge has $4 - 2 = 2$ "middle" unit cubes between its two corner unit cubes), $6$ face-centers, and $1$ deep interior block.
- A unit cube's iced-face count equals the number of iced big-cube faces it touches.
💡 Sorting the small cubes into corner / edge / face / interior families is the kindergarten skill of analyzing a 3D shape's parts.
K.G.B.4 Step 2 Split into three sub-questions, one per location type that could give exactly $2$ iced faces: (a) cubes on the $4$ top edges where the iced top meets an iced side, (b) cubes on the $4$ vertical edges where two iced sides meet, (c) cubes at the $4$ bottom corners where two iced sides meet the un-iced bottom.
💡 Breaking the cube's surface into pieces I can count separately is the Tool #7 sub-problems move on a 3D shape.
3.OA.A.1 Step 3 - Count case (a) — top edges.
- There are $4$ edges around the top face.
- Each edge has $2$ middle unit cubes (the $2$ end unit cubes are top corners, which touch $3$ iced faces).
- Each middle cube touches the iced top and one iced side: exactly $2$ iced faces.
💡 Multiplying "$4$ groups of $2$" is the third-grade meaning of multiplication.
3.OA.A.1 Step 4 - Count case (b) — vertical edges.
- There are $4$ vertical edges joining top corners to bottom corners.
- The top end of each is a top corner (3 iced faces) and the bottom end is a bottom corner (handled below).
- The $2$ middle cubes on each vertical edge touch two iced side faces: exactly $2$ iced faces.
💡 Same "$4$ groups of $2$" pattern as case (a) — multiplication keeps the count tidy.
3.OA.A.1 Step 5 - Count case (c) — bottom corners.
- There are $4$ unit cubes at the bottom corners of the big cube.
- Each touches two iced side faces and the un-iced bottom, so each has exactly $2$ iced faces.
- (The $4$ top corners are excluded because they touch the iced top plus two iced sides, giving $3$ iced faces.) Also, the $2$ middle cubes on each of the $4$ bottom edges touch only one iced side, so they have just $1$ iced face — not counted.
💡 Listing each of the $4$ bottom corners once and checking its iced faces is straight systematic counting.
3.OA.D.8 Step 6 Add the three cases to get the total number of unit cubes with exactly two iced faces.
💡 Combining sub-problem answers with a single addition is the wrap-up step of any multi-step word problem in Grade 3.
K.G.B.4 Picture the cube and label its $6$ kinds of positions: $8$ corners, $12$ edges ( K.G.B.4 Split into three sub-questions, one per location type that could give exactly $2 3.OA.A.1 Count case (a) — top edges. There are $4$ edges around the top face. Each edge h 3.OA.A.1 Count case (b) — vertical edges. There are $4$ vertical edges joining top corner 3.OA.A.1 Count case (c) — bottom corners. There are $4$ unit cubes at the bottom corners 3.OA.D.8 Add the three cases to get the total number of unit cubes with exactly two iced Review
Reasonableness: Sanity check the parts. The full big cube has $4 \cdot 4 \cdot 4 = 64$ unit cubes. Of these, $8$ are deep inside (no icing), $4 \cdot 4 = 16$ sit in the centers of the $4$ side faces (each iced once), $4$ centers of the un-iced bottom have $0$ iced faces, $4$ centers of the top have $1$ iced face, and the $20$ we counted have $2$ iced faces, while the $4$ top corners have $3$ iced faces. The remaining bottom-edge middles ($4 \cdot 2 = 8$ cubes) have $1$ iced face. Tallying: $8 + 16 + 4 + 4 + 20 + 4 + 8 = 64$. The bookkeeping closes, so $20$ is consistent. Also $20$ is choice (D), which is one of the offered answers.
Alternative: Tool #16 (Change Focus / Complement) is a clean alternative. If the cake had icing on all $6$ faces, the standard "2-faces-iced" count would be the $12$ edges $\times$ $2$ middle cubes $= 24$. Removing icing from the bottom changes two categories: the $8$ bottom-edge middles drop from "$2$-iced" to "$1$-iced" ($-8$), and the $4$ bottom corners drop from "$3$-iced" to "$2$-iced" ($+4$). Net change: $-8 + 4 = -4$, giving $24 - 4 = 20$ — the same answer (D).
CCSS standards used (min grade 3)
K.G.B.4Analyze and compare two- and three-dimensional shapes (Identifying the corner, edge, face-center, and interior positions of the $4 \times 4 \times 4$ cube so each unit cube can be classified by how many big-cube faces it touches.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Counting each case as "$4$ groups of $2$" (top edges and vertical edges) using the grade-3 meaning of multiplication.)3.OA.D.8Solve two-step word problems using four operations within 100 (Combining the three case counts $8 + 8 + 4 = 20$ to finish the multi-step word problem.)
⭐ This AMC 8 problem only needs Grade 3 multiplication and addition you already know — count edges and corners, group them with $\times$, then add!
⭐ This AMC 8 problem only needs Grade 3 multiplication and addition you already know — count edges and corners, group them with $\times$, then add!
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