The Cube Cutting Framework
Cubes Reasoning is a branch of spatial-analytical reasoning where a solid large cube has some or all of its exterior faces painted with specific colors and is then cut by equidistant planar slices along three perpendicular axes into \(n^3\) identical smaller cubes.
The goal is to track the spatial position of each smaller cube—corner, edge, face-interior, or central core—to determine exactly how many small cubes possess 3, 2, 1, or 0 painted faces.
External Shell vs. Internal Core
Imagine peeling an onion in three dimensions. Only the outermost 1-cube-thick surface layer touches exterior paint.
Why It Matters
Tests 3D spatial visualization without drawing tools. Candidates who rely on rote formulas falter when entrance exams alter paint configurations or slice cuboids.
Where Used
A staple in MAH MBA CET (3–5 questions per slot), SNAP (speed spatial reasoning), NMAT, and multi-step set variations in CAT DILR.
Prerequisites
Basic solid geometry (a cube has 6 faces, 12 edges, 8 vertices), integer division, cubes of numbers 1–10 ($n^3$), and Cartesian coordinate awareness.
CRITICAL DISTINCTIONS Rigorous Geometric Definitions
The Large Cube has edge length $L$. It is sliced into $n^3$ Smaller Cubes of edge length $s = L/n$.
Every single small cube has exactly 6 square faces. "Painted face count" counts how many of those 6 are colored.
External faces lie on the outer boundary of the large cube. Internal faces are created by cuts and are completely raw/unpainted.
To divide an edge into $n$ segments requires exactly $(n - 1)$ parallel cuts. Dividing all 3 axes requires $3(n - 1)$ total cuts.
Cube Fundamentals & The $n^3$ Slicing Model
A regular hexahedron (cube) is a 3-dimensional Platonic solid bounded by 6 identical congruent square faces, meeting at right angles.
3 opposite pairs: Top-Bottom, Left-Right, Front-Back.
Where exactly two perpendicular faces intersect.
Where exactly three perpendicular faces and edges meet.
Opposite Faces Never Touch Each Other
Notice how folding this 2D cross-net along the internal seams reconstructs the 3D cube. In any open net, opposite faces are always separated by exactly one intervening square and share the same color code below.
The "1-Step Gap" Test for Open Nets: When folded, two faces are opposite if and only if they are in the same straight sequence separated by exactly one intervening square. Notice that Left (2) & Right (4) are separated by Front (3), Front (3) & Back (5) are separated by Right (4), and Top (1) & Bottom (6) are separated by Front (3).
How to Determine the Value of $n$
Method 1 (From Dimensions): $n = \frac{\text{Side of Large Cube } (L)}{\text{Side of Small Cube } (s)}$
Method 2 (From Total Small Cubes): $n = \sqrt[3]{\text{Total Smaller Cubes } (N)}$
Method 3 (From Slicing Cuts): If $C$ total equidistant cuts are made equally across 3 axes: $n = \frac{C}{3} + 1$
Understanding the Paint Model
In management entrance questions, the paint applied to the exterior creates an exposure matrix. Before applying any formula, the student must explicitly inspect the painting configuration.
Every outer face of the original cube receives paint. In this setup, every corner small cube touches 3 painted faces, every edge small cube touches 2 painted faces, and every face-center touches 1 painted face.
Only $n^2$ cubes receive any paint at all! Zero cubes can have 2 or 3 painted faces because no two painted faces exist to intersect. All other $n^3 - n^2$ cubes have zero paint.
The two painted faces meet along exactly 1 shared edge of length $n$. Exactly $n$ smaller cubes possess 2 painted faces. Zero cubes have 3 painted faces!
Opposite faces never share an edge or corner. Hence, zero smaller cubes have 2 or 3 painted faces. Exactly $2n^2$ cubes have 1 painted face, and $n^2(n - 2)$ have zero paint.
A small cube receives paint on a face if and only if that face was part of an externally painted face on the original large cube. The moment a slice cut is made, the freshly exposed interior faces are 100% RAW & UNPAINTED.
Cubes with Painted Faces (Standard 6-Face Model)
When all six exterior faces of the large cube are painted and each edge is cut into $n$ equal segments ($n \ge 3$), the smaller cubes partition into four mutually exclusive groups:
Always strictly equal to the 8 corner vertices of the cube for any $n \ge 2$. Independent of the value of $n$!
Located along the 12 edges of the cube, strictly excluding the 2 corner endpoints on each edge.
Located on the interior area of the 6 faces, excluding the perimeter strip of corner and edge cubes.
The completely shielded internal sub-cube left when 1 layer of cubes is removed from all 6 exterior faces.
Boundary Cases & Numerical Breakdown Matrix
| Value of $n$ | Total $n^3$ | 3 Faces (Corners) | 2 Faces ($12(n-2)$) | 1 Face ($6(n-2)^2$) | 0 Faces ($(n-2)^3$) | Total Check |
|---|---|---|---|---|---|---|
| $n = 1$ | 1 | 0 (Has all 6 painted) | 0 | 0 | 0 | 1 = 1 |
| $n = 2$ | 8 | 8 (All cubes) | 0 | 0 | 0 | 8 = 8 |
| $n = 3$ | 27 | 8 | 12 | 6 | 1 | 8+12+6+1 = 27 |
| $n = 4$ | 64 | 8 | 24 | 24 | 8 | 8+24+24+8 = 64 |
| $n = 5$ | 125 | 8 | 36 | 54 | 27 | 8+36+54+27 = 125 |
| $n = 6$ | 216 | 8 | 48 | 96 | 64 | 8+48+96+64 = 216 |
Location-Based Counting & The Partition Theorem
Why do these four formulas never double-count and never miss a single small cube? Because they form a rigorous mathematical partition of 3D space.
Let $x = n - 2$. Then $n = x + 2$.
Using the binomial cube expansion: $n^3 = (x + 2)^3 = x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3$
Simplifying the terms: $n^3 = x^3 + 6x^2 + 12x + 8$
Substituting back $x = n - 2$:
$$\underbrace{(n - 2)^3}_{\text{Inner Core}} + \underbrace{6(n - 2)^2}_{\text{Face Centers}} + \underbrace{12(n - 2)}_{\text{Edges}} + \underbrace{8}_{\text{Corners}} = n^3$$
Q.E.D. Every point in the $n \times n \times n$ discrete lattice belongs to exactly one of the four geometric sets.
"At Least 2 Faces Painted"
"At Most 1 Face Painted"
Paint Configurations Other Than Six Painted Faces
CAT and XAT test scenarios where only some faces are painted. To solve partial-paint problems without rote memorization, follow this 5-step spatial protocol:
e.g., Top and Right faces painted.
- • 3 faces painted: 0
- • 2 faces painted: $n$ (along the shared edge)
- • 1 face painted: $2n(n - 1)$ ($n^2 - n$ per face)
- • 0 faces painted: $n(n - 1)^2$
- • Sum: $n + 2n(n - 1) + n(n - 1)^2 = n^3$
e.g., Top and Bottom faces painted.
- • 3 faces painted: 0
- • 2 faces painted: 0 (Opposite faces never touch)
- • 1 face painted: $2n^2$ ($n^2$ on each face)
- • 0 faces painted: $n^2(n - 2)$
- • Sum: $2n^2 + n^2(n - 2) = n^3$
e.g., Top, Front, Right meet at 1 corner.
- • 3 faces painted: 1 (the common corner vertex)
- • 2 faces painted: $3(n - 1)$ ($n-1$ per shared edge)
- • 1 face painted: $3(n - 1)^2$ ($(n-1)^2$ per face)
- • 0 faces painted: $(n - 1)^3$ (unpainted sub-cube)
- • Sum: $1 + 3(n-1) + 3(n-1)^2 + (n-1)^3 = n^3$
e.g., Top, Bottom (opposite) + Front (adjacent).
- • 3 faces painted: 0 (Top & Bottom never meet)
- • 2 faces painted: $2n$ ($n$ on Top-Front + $n$ on Bottom-Front)
- • 1 face painted: $n(3n - 4)$
- • 0 faces painted: $n(n - 1)(n - 2)$
- • Sum: $2n + n(3n-4) + n(n-1)(n-2) = n^3$
Repainting, Recutting & Cut Sequence Variations
Exams frequently test conceptual depth by altering the sequence of operations or asking about cuboids.
1. Painting BEFORE Cutting vs. AFTER Cutting
Painting Before Cutting: Only the $6n^2$ external square units receive paint. The inner cut faces remain raw. Exactly $(n-2)^3$ cubes have 0 paint.
Painting After Cutting: The $n^3$ small cubes are disassembled and dipped in paint. Every small cube now has all 6 of its faces painted!
New faces painted = $6n^3 - 6n^2 = 6n^2(n - 1)$.
2. Rectangular Block (Cuboid) Cutting
When a block of dimensions $L \times W \times H$ is cut into unit cubes ($n_1 = L, n_2 = W, n_3 = H$):
- • Total Cubes $= n_1 \times n_2 \times n_3$
- • 3 Faces Painted $= 8$ (Corner vertices)
- • 2 Faces Painted $= 4(n_1 - 2) + 4(n_2 - 2) + 4(n_3 - 2)$
- • 1 Face Painted $= 2(n_1-2)(n_2-2) + 2(n_2-2)(n_3-2) + 2(n_3-2)(n_1-2)$
- • 0 Faces Painted $= (n_1 - 2)(n_2 - 2)(n_3 - 2)$
8 Fully Worked Solved Examples
A solid wooden cube is painted black on all 6 exterior faces and then cut into 8 identical smaller cubes. How many small cubes have (i) 3 painted faces, (ii) 2 painted faces, (iii) 1 painted face, and (iv) 0 painted faces?
1. Identify $n$: Total cubes $N = 8 \implies n = \sqrt[3]{8} = 2$.
2. Geometric Analysis: When $n = 2$, each edge is divided into 2 segments. Every single small cube touches 3 mutually perpendicular exterior faces meeting at a corner vertex.
3. Calculations:
• 3 painted faces: All 8 corner vertices $= \mathbf{8}$.
• 2 painted faces: $12(n - 2) = 12(2 - 2) = \mathbf{0}$.
• 1 painted face: $6(n - 2)^2 = 6(2 - 2)^2 = \mathbf{0}$.
• 0 painted faces: $(n - 2)^3 = (2 - 2)^3 = \mathbf{0}$.
Verification: $8 + 0 + 0 + 0 = 8 = n^3$. Consistent.
A cube with all 6 faces painted red is cut into 27 equal smaller cubes. How many cuts were made, and how many cubes have exactly 2 red faces?
1. Identify $n$: $N = 27 \implies n = \sqrt[3]{27} = 3$.
2. Slicing Cuts: To create 3 segments along each axis requires $(3 - 1) = 2$ cuts. For 3 dimensions: Total Cuts $= 3 \times (n - 1) = 3 \times 2 = \mathbf{6\text{ cuts}}$.
3. Two Painted Faces: These reside exclusively on the 12 edges, excluding corners.
$$\text{Cubes with 2 painted faces} = 12(n - 2) = 12(3 - 2) = 12 \times 1 = \mathbf{12}$$
Verification: Total partition $= 8\text{ (corners)} + 12\text{ (edges)} + 6(1)^2\text{ (faces)} + (1)^3\text{ (core)} = 8 + 12 + 6 + 1 = 27$. Consistent.
A 16 cm cube is painted green on all exterior faces and cut into 4 cm smaller cubes. How many smaller cubes have at least 2 green faces?
1. Determine $n$: $n = \frac{L}{s} = \frac{16\text{ cm}}{4\text{ cm}} = 4$. Total smaller cubes $N = 4^3 = 64$.
2. Interpret Qualifying Phrase: "At least 2 faces" means $\ge 2$ painted faces, which includes:
$$\text{At least 2} = (\text{Exactly 2 faces}) + (\text{Exactly 3 faces})$$
• Exactly 3 painted faces (Corners) $= 8$.
• Exactly 2 painted faces (Edges) $= 12(n - 2) = 12(4 - 2) = 12 \times 2 = 24$.
$$\text{Total} = 8 + 24 = \mathbf{32}$$
Verification: 1-face $= 6(4-2)^2 = 24$, 0-face $= (4-2)^3 = 8$. Sum: $32 + 24 + 8 = 64$. Consistent.
A cube painted blue on all 6 faces is sliced into 125 small cubes. How many small cubes have at most one face painted?
1. Identify $n$: $N = 125 \implies n = \sqrt[3]{125} = 5$.
2. Decode "At Most 1": "At most 1" means $\le 1$ painted face:
$$\text{At most 1} = (\text{Exactly 0 painted}) + (\text{Exactly 1 painted})$$
• 0 faces painted (Inner Core) $= (n - 2)^3 = (5 - 2)^3 = 3^3 = \mathbf{27}$.
• 1 face painted (Face Centers) $= 6(n - 2)^2 = 6(5 - 2)^2 = 6 \times 9 = \mathbf{54}$.
$$\text{Total} = 27 + 54 = \mathbf{81}$$
Verification: 2-face $= 12(3) = 36$, 3-face $= 8$. Total $= 81 + 36 + 8 = 125$. Consistent.
A wooden cube is painted yellow on its Top face only and then cut into 64 identical small cubes. How many small cubes have (i) 2 or 3 painted faces, (ii) exactly 1 painted face, and (iii) 0 painted faces?
1. Identify $n$: $N = 64 \implies n = \sqrt[3]{64} = 4$.
2. Geometric Deduction: Only 1 face (Top) is painted. For any small cube to have 2 or 3 painted faces, it must touch 2 or 3 externally painted faces. Since only 1 external face was painted in total:
• Cubes with 2 or 3 painted faces $= \mathbf{0}$.
• Cubes with exactly 1 painted face $= \text{All cubes in the top layer} = n^2 = 4^2 = \mathbf{16}$.
• Cubes with 0 painted faces $= \text{All cubes in the remaining } (n - 1) \text{ layers} = n^2(n - 1) = 16 \times 3 = \mathbf{48}$.
Verification: $0 + 16 + 48 = 64$. Consistent.
A cube is painted red on two adjacent faces (Top and Right), while the remaining 4 faces remain unpainted. It is cut into 64 small cubes ($n = 4$). How many small cubes have (i) 2 painted faces, (ii) 1 painted face, and (iii) 0 painted faces?
1. Identify Shared Boundary: Top and Right faces meet along exactly 1 shared edge of length $n = 4$.
2. Two Painted Faces: Only small cubes located along this shared edge touch both painted faces:
$$\text{Cubes with 2 painted faces} = n = \mathbf{4}$$
3. One Painted Face: On Top face, there are $n^2 = 16$ cubes, minus 4 along shared edge $= 12$. On Right face, there are $n^2 = 16$ cubes, minus 4 along shared edge $= 12$:
$$\text{Cubes with 1 painted face} = 12 + 12 = 2n(n - 1) = 2(4)(3) = \mathbf{24}$$
4. Zero Painted Faces:
$$\text{Cubes with 0 painted faces} = 64 - (4 + 24) = 64 - 28 = \mathbf{36} \quad [= n(n - 1)^2 = 4 \times 3^2 = 36]$$
Verification: $4 + 24 + 36 = 64 = n^3$. Consistent.
A cube is painted black on its Top and Bottom faces, leaving the 4 side faces unpainted. It is cut into 125 small cubes ($n = 5$). How many small cubes have (i) 2 or 3 painted faces, (ii) 1 painted face, and (iii) 0 painted faces?
1. Identify Spatial Geometry: Top and Bottom faces are opposite planes separated by $(n - 2) = 3$ layers of interior cubes. They never touch or share any edge.
2. Two or Three Painted Faces: Since no small cube can touch both Top and Bottom simultaneously for $n \ge 2$, cubes with $\ge 2$ painted faces $= \mathbf{0}$.
3. One Painted Face: All cubes on the Top face ($n^2 = 25$) and all cubes on the Bottom face ($n^2 = 25$):
$$\text{Cubes with 1 painted face} = 2n^2 = 2 \times 25 = \mathbf{50}$$
4. Zero Painted Faces: All cubes in the intermediate $(n - 2)$ vertical slices:
$$\text{Cubes with 0 painted faces} = n^2(n - 2) = 25 \times 3 = \mathbf{75}$$
Verification: $0 + 50 + 75 = 125 = n^3$. Consistent.
A solid cube is painted Red on Top, Blue on Front, and Green on Right. The other 3 faces remain unpainted. It is cut into 64 small cubes ($n = 4$). Determine the number of small cubes that have:
(a) Exactly 3 painted faces (Red, Blue, Green)
(b) Exactly 2 painted faces
(c) Exactly 1 painted face
(d) Zero painted faces
1. Corner Analysis (3 Faces): The three painted faces meet at exactly 1 common corner vertex. Therefore, exactly $\mathbf{1}$ small cube has all 3 colors (Red, Blue, Green).
2. Edge Analysis (2 Faces): They share 3 edges (Top-Front, Front-Right, Top-Right). Each edge has length $n = 4$, minus 1 corner cube $= 3$ cubes per edge:
$$\text{Cubes with 2 painted faces} = 3 \times (n - 1) = 3 \times 3 = \mathbf{9}$$
(3 Red-Blue, 3 Blue-Green, 3 Red-Green).
3. Face-Interior Analysis (1 Face): On each of the 3 painted faces, $n^2 = 16$ cubes exist. Deducting the 1 corner and the $2(n - 1) = 6$ edge cubes on that face leaves $(n - 1)^2 = 3^2 = 9$ cubes:
$$\text{Cubes with 1 painted face} = 3 \times (n - 1)^2 = 3 \times 9 = \mathbf{27}$$
(9 Red only, 9 Blue only, 9 Green only).
4. Unpainted Cubes (0 Faces): Removing 1 layer from the 3 painted sides leaves an unpainted solid block of $(n - 1) \times (n - 1) \times (n - 1) = 3^3 = \mathbf{27}$ cubes.
Verification: $1\text{ (3-face)} + 9\text{ (2-face)} + 27\text{ (1-face)} + 27\text{ (0-face)} = 64 = n^3$. Complete.
10 Critical Pitfalls & Exam Traps
Each edge has $n$ small cubes, but the 2 ends are corner vertices. Failing to subtract 2 results in counting corners twice, inflating edge cubes from $12(n - 2)$ to $12n$.
If a cube is painted on only 2 faces, students automatically plug into $12(n - 2)$. This is completely wrong! The standard formula assumes all 12 edges are painted. In 2-adjacent paint, only 1 edge is painted.
To get 4 pieces along an edge requires only 3 cuts. Students often multiply $n \times 3$ instead of $3(n - 1)$, producing incorrect cuts counts.
When cutting occurs after painting, only the exterior faces receive paint. Internal faces created by the saw are 100% bare wood.
For $n = 2$, formula $(2 - 2)^3 = 0$ unpainted, but students erroneously think 1 cube must be inside. At $n = 2$, all 8 cubes are external corners!
"At least 2 faces painted" means 2 or 3 faces painted. "At most 1 face painted" means 0 or 1 face painted. Never confuse inequality directions.
Top and Bottom faces never share an edge or corner. If only opposite faces are painted, 2-painted and 3-painted cubes are strictly 0.
When dimensions are $4 \times 5 \times 6$, $n$ is not a single number. You must use $(n_1 - 2)(n_2 - 2)(n_3 - 2)$ for the inner core.
If cubes are separated first and then painted, every small cube gets 6 painted faces. Reversing this sequence completely changes the answer.
Always check that $\text{3-face} + \text{2-face} + \text{1-face} + \text{0-face} = n^3$. If your sum does not match $n^3$, an arithmetic or counting error exists.
Master Shortcuts & Exam Speed Strategy
| Question Asks For | All 6 Faces Painted | 2 Adjacent Faces Painted | 2 Opposite Faces Painted | 3 Adjacent Faces Painted |
|---|---|---|---|---|
| 3 Faces Painted | $8$ | $0$ | $0$ | $1$ |
| 2 Faces Painted | $12(n - 2)$ | $n$ | $0$ | $3(n - 1)$ |
| 1 Face Painted | $6(n - 2)^2$ | $2n(n - 1)$ | $2n^2$ | $3(n - 1)^2$ |
| 0 Faces (Unpainted) | $(n - 2)^3$ | $n(n - 1)^2$ | $n^2(n - 2)$ | $(n - 1)^3$ |
| Sum Verification | $n^3$ | $n^3$ | $n^3$ | $n^3$ |
1. Find $n$ first: Take cube root of total smaller cubes, or side length ratio $L/s$.
2. Inspect paint condition: Are all 6 faces painted? If YES, use the 8, 12, 6, 1 multiplier rule. If NO, identify shared edges.
3. Count Corners first: In standard setups, it is always 8. In partial setups, it is either 0 or 1.
4. Reconcile: Mentally verify your chosen categories add up to $n^3$.
15 Verified Practice Questions
A cube with all 6 faces painted red is cut into 27 identical small cubes. How many small cubes have exactly 2 red faces?
Step 1: Total cubes $N = 27 \implies n = \sqrt[3]{27} = 3$.
Step 2: Cubes with 2 painted faces lie on the 12 edges, excluding the 2 corner endpoints:
$$\text{Cubes} = 12(n - 2) = 12(3 - 2) = 12 \times 1 = \mathbf{12}$$
Correct Answer: 12 cubes.
A solid cube is painted on all 6 faces and divided into 64 small identical cubes. How many of the small cubes have zero painted faces?
Step 1: $n = \sqrt[3]{64} = 4$.
Step 2: Zero painted faces correspond to the completely shielded inner core:
$$\text{Unpainted cubes} = (n - 2)^3 = (4 - 2)^3 = 2^3 = \mathbf{8}$$
Correct Answer: 8 cubes.
A large cube of side 12 cm is painted green on all exterior faces and cut into small cubes of side 3 cm each. How many small cubes have exactly one face painted?
Step 1: Find $n = \frac{L}{s} = \frac{12}{3} = 4$.
Step 2: Cubes with exactly 1 painted face reside at the center of each of the 6 faces:
$$\text{1-painted cubes} = 6(n - 2)^2 = 6(4 - 2)^2 = 6 \times 4 = \mathbf{24}$$
Correct Answer: 24 cubes.
A solid cube is painted blue on all 6 faces and cut into 125 smaller cubes. What is the ratio of the number of cubes with exactly 2 painted faces to the number of cubes with zero painted faces?
Step 1: $n = \sqrt[3]{125} = 5$.
Step 2: Cubes with 2 painted faces $= 12(n - 2) = 12(5 - 2) = 12 \times 3 = 36$.
Step 3: Cubes with 0 painted faces $= (n - 2)^3 = (5 - 2)^3 = 3^3 = 27$.
Step 4: Ratio $= \frac{36}{27} = \frac{4}{3} = \mathbf{4 : 3}$.
Correct Answer: 4 : 3.
What is the minimum number of cuts required to divide a solid wooden cube into 216 identical smaller cubes?
Step 1: $N = 216 \implies n = \sqrt[3]{216} = 6$.
Step 2: To get $n$ segments along an edge requires $(n - 1)$ parallel cuts.
Step 3: Across length, width, and height, total cuts $= 3(n - 1) = 3(6 - 1) = 3 \times 5 = \mathbf{15\text{ cuts}}$.
Correct Answer: 15 cuts.
A cube of side 20 cm is painted red on all 6 faces and cut into 5 cm cubes. How many of the resulting smaller cubes have at least 2 painted faces?
Step 1: $n = \frac{20}{5} = 4$. Total cubes $= 4^3 = 64$.
Step 2: "At least 2 faces" $= (\text{Exactly 2 faces}) + (\text{Exactly 3 faces})$.
• 3 painted faces $= 8$ (corners).
• 2 painted faces $= 12(4 - 2) = 24$ (edges).
$$\text{Total} = 8 + 24 = \mathbf{32}$$
Correct Answer: 32 cubes.
A wooden cube is painted on two adjacent faces and cut into 27 identical small cubes ($n = 3$). How many smaller cubes have zero painted faces?
Step 1: $n = 3$. Total cubes $= 27$.
Step 2: The two painted faces meet at 1 edge. Cubes touching the painted faces:
• 2-face painted $= n = 3$ (along shared edge).
• 1-face painted $= 2n(n - 1) = 2(3)(2) = 12$.
• Total painted cubes $= 3 + 12 = 15$.
Step 3: Unpainted cubes $= 27 - 15 = \mathbf{12} \quad [= n(n - 1)^2 = 3(2^2) = 12]$.
Correct Answer: 12 cubes.
A cube is painted on two opposite faces only and then cut into 64 small cubes ($n = 4$). How many small cubes have at least one face painted?
Step 1: $n = 4$. Opposite faces never intersect.
Step 2: Each painted face contains $n^2 = 4^2 = 16$ small cubes. Since they do not overlap, every cube touching a painted face has exactly 1 painted face.
$$\text{Cubes with at least 1 painted face} = 2n^2 = 2 \times 16 = \mathbf{32}$$
Correct Answer: 32 cubes.
Three mutually adjacent faces of a 15 cm cube are painted red. The cube is cut into 3 cm smaller cubes. How many smaller cubes have exactly two painted faces?
Step 1: $n = \frac{15}{3} = 5$.
Step 2: Three mutually adjacent faces share exactly 3 edges meeting at 1 corner vertex.
Step 3: Along each of the 3 shared edges, there are $n$ cubes, but the common corner vertex has 3 painted faces. Thus, each edge contributes $(n - 1)$ cubes with 2 painted faces:
$$\text{Total 2-painted} = 3(n - 1) = 3(5 - 1) = 3 \times 4 = \mathbf{12}$$
Correct Answer: 12 cubes.
A cube with all 6 faces painted has exactly 96 smaller cubes with only one painted face. What is the total number of smaller cubes?
Step 1: Formula for 1-face painted: $6(n - 2)^2 = 96$.
Step 2: Solve for $n$:
$$(n - 2)^2 = \frac{96}{6} = 16 \implies n - 2 = 4 \implies n = 6$$
Step 3: Total smaller cubes $N = n^3 = 6^3 = \mathbf{216}$.
Correct Answer: 216 smaller cubes.
A cube is painted on three faces: Top, Bottom (opposite), and Front (adjacent to both). It is cut into 64 small cubes ($n = 4$). How many small cubes have zero painted faces?
Step 1: $n = 4$, Total $= 64$.
Step 2: Along the height (Top-Bottom), only $(n - 2) = 2$ interior layers are unpainted.
Step 3: Along the depth (Front-Back), only the $(n - 1) = 3$ layers behind Front are unpainted.
Step 4: Along the width (Left-Right), all $n = 4$ layers remain unpainted on exterior sides.
$$\text{Unpainted block} = n \times (n - 1) \times (n - 2) = 4 \times 3 \times 2 = \mathbf{24}$$
Correct Answer: 24 cubes.
A large cube was painted red and cut into 27 small cubes ($n = 3$). The 27 cubes are then separated and all their unpainted faces are painted green. How many total small faces were painted green?
Step 1: Total faces on 27 small cubes $= 27 \times 6 = 162\text{ faces}$.
Step 2: Before cutting, only the exterior faces were red: $6 \times n^2 = 6 \times 3^2 = 54\text{ red faces}$.
Step 3: Green faces $= \text{Total faces} - \text{Red faces} = 162 - 54 = \mathbf{108\text{ faces}}$.
Correct Answer: 108 faces.
A solid rectangular block of dimensions 6 cm × 8 cm × 10 cm is painted black on all exterior faces and cut into 2 cm × 2 cm × 2 cm cubes. How many cubes have zero painted faces?
Step 1: Find segments along each dimension:
$$n_1 = \frac{6}{2} = 3, \quad n_2 = \frac{8}{2} = 4, \quad n_3 = \frac{10}{2} = 5$$
Total cubes $= 3 \times 4 \times 5 = 60$.
Step 2: The unpainted core subtracts 2 units from each dimension:
$$\text{Unpainted} = (n_1 - 2)(n_2 - 2)(n_3 - 2) = (3 - 2)(4 - 2)(5 - 2) = 1 \times 2 \times 3 = \mathbf{6}$$
Correct Answer: 6 cubes.
A large cube has opposite pairs of faces painted Red, Green, and Blue. It is cut into 125 small cubes ($n = 5$). How many small cubes have exactly two painted faces with different colors, where one of the colors is Red?
Step 1: $n = 5$. Cubes with 2 painted faces lie on the 12 edges, each having $(n - 2) = 3$ cubes.
Step 2: The 2 Red faces are opposite. A cube has 12 edges. Edges meet 2 perpendicular faces:
• 4 edges connect Top Red with the 4 sides (2 Green, 2 Blue).
• 4 edges connect Bottom Red with the 4 sides (2 Green, 2 Blue).
• 4 edges connect Green with Blue sides (no Red face).
Step 3: Exactly $4 + 4 = 8$ edges connect a Red face with another color.
$$\text{Total cubes} = 8 \times (n - 2) = 8 \times 3 = \mathbf{24}$$
Correct Answer: 24 cubes.
An $n \times n \times n$ cube is painted on all 6 faces and cut into $n^3$ small cubes. If the number of small cubes with exactly 2 painted faces is equal to the number of small cubes with exactly 1 painted face, what is the value of $n$?
Step 1: Equate the two formulas:
$$12(n - 2) = 6(n - 2)^2$$
Step 2: For $n \ge 3$, $(n - 2) \neq 0$. Divide both sides by $6(n - 2)$:
$$2 = n - 2 \implies n = \mathbf{4}$$
Step 3: Verification check at $n = 4$:
• 2-painted faces: $12(4 - 2) = 24$.
• 1-painted face: $6(4 - 2)^2 = 6(4) = 24$.
Correct Answer: n = 4 (Total 64 cubes).
Frequently Asked Questions
How many small cubes are created when each edge is divided into n equal parts?
When each of the 3 dimensions (length, width, height) of a large cube is divided into $n$ equal segments, the total number of identical smaller cubes created is exactly $n^3$.
How many small cubes have three painted faces in a standard cube?
When all six exterior faces of a cube are painted, exactly 8 smaller cubes have three painted faces for any $n \ge 2$. These 8 cubes are strictly the 8 corner vertices of the original cube.
Why do edge counts exclude corners in cube problems?
Corners already touch three perpendicular exterior faces. Edge-interior cubes touch only two exterior faces. Including corners in edge counts would double-count vertices and corrupt the partition of small cubes.
Why are face-interior cubes counted using the formula 6(n - 2)²?
A cube has 6 faces. On each face of $n \times n$ small cubes, removing the 1-cube perimeter border of corners and edges leaves an interior square of $(n - 2) \times (n - 2)$ cubes that touch only that single exterior face.
Why are there (n - 2)³ unpainted interior cubes?
Stripping away the 1-cube exterior boundary layer from all six faces leaves an internal, completely unexposed solid cube of dimension $(n - 2)$ along length, width, and height, yielding exactly $(n - 2)^3$ unpainted cubes.
What changes when only some original faces are painted?
When fewer than six faces are painted, the standard $8, 12(n-2), 6(n-2)^2,$ and $(n-2)^3$ formulas cannot be used. Instead, you must map the specific painted faces and count intersections along shared edges and corners.
Does painting before cutting give the same result as painting after cutting?
No. Painting before cutting leaves all internal cut faces completely unpainted. Painting after cutting coats every single face of all $n^3$ small cubes, resulting in $6n^3$ painted small faces and zero unpainted cubes.
How can I verify that my cube counts are correct?
Verify using the partition theorem: the sum of cubes with 3 painted faces, 2 painted faces, 1 painted face, and 0 painted faces must equal the total number of smaller cubes $n^3$.
Related Concepts & What's Next
Advance next to Analogies — master word relationships, functional associations, semantic hierarchies, and numerical ratio analogies for SNAP, NMAT, and MBA CET.