How to Solve Cube Cutting and Painted-Faces Problems
Solve painted-cube cutting problems with corner, edge, face and internal formulas, a worked example, and practice questions with answers.
Expected Interview Answer
A cube of side n units, painted on all outer faces and cut into n³ unit cubes, splits into four countable categories by position — corner cubes always have 3 painted faces, edge cubes have 2, face cubes have 1, and internal cubes have 0 — and each category’s count follows a fixed formula in n.
Corner cubes are always exactly 8, regardless of n, since a cube always has 8 corners. Edge cubes (excluding corners) number 12×(n−2), one strip per edge minus the two corner cubes on that edge. Face cubes (excluding edges) number 6×(n−2)², one central patch per face. Internal cubes with zero painted faces number (n−2)³, the smaller cube hidden entirely inside. These four counts must always sum to n³ as a sanity check. When only some faces are painted, or the cube is cut unevenly, re-derive position-by-position rather than forcing the standard formulas.
- Four simple formulas cover the entire standard painted-cube problem type
- The sum-to-n³ check catches arithmetic errors immediately
- The same positional logic extends to non-standard painting patterns
AI Mentor Explanation
Think of a Rubik’s-cube-sized trophy case with 3×3×3 = 27 small display cubes: the 8 corner cubes each show three painted faces just like a cube corner sits at the intersection of three outer faces, the edge cubes show two painted faces, and the single center cube is completely hidden, showing zero. This is exactly the counting structure of painted-cube problems: 8 corners always, 12(n−2) edges, 6(n−2)² faces, and (n−2)³ hidden internal cubes, all summing back to n³. A scorer double-checking these counts against the total cube count catches errors the same way a scorecard total must match the sum of individual scores. Getting n wrong — miscounting how many times each edge was sliced — throws off every category at once.
Worked example — 4×4×4 painted cube
3 painted faces (corners)
- always 8
2 painted faces (edges)
- 12×(4−2) = 24
1 painted face (faces)
- 6×(4−2)² = 24
0 painted faces (internal)
- (4−2)³ = 8
- Total: 8+24+24+8 = 64
Step-by-Step Explanation
Step 1
Determine n
Find how many unit cubes the original cube is cut into along one edge.
Step 2
Apply the four category formulas
Corners = 8, edges = 12(n−2), faces = 6(n−2)², internal = (n−2)³.
Step 3
Sanity-check the sum
All four counts must add up to n³ — recheck n if they do not.
Step 4
Adjust for non-standard painting
If not all faces are painted, recount positionally rather than using the standard formulas.
What Interviewer Expects
- Recall of all four position-count formulas (corner, edge, face, internal)
- Understanding why corners are always exactly 8 regardless of n
- Using the sum-to-n³ check to validate the answer
- Adapting the method when not all six faces are painted
Common Mistakes
- Forgetting to subtract corners when counting edge cubes, i.e. using 12n instead of 12(n−2)
- Using n instead of (n−2) for the internal cube dimension
- Applying the standard formulas when the problem paints fewer than all six faces
- Confusing n (cubes per edge) with the total number of cuts made (n−1)
Best Answer (HR Friendly)
“I would classify every small cube by its position in the larger cube: the 8 corners always have three painted faces, the edges (excluding corners) have two, the face centers have one, and the fully interior cubes, forming an (n−2) cube inside, have none. I would apply the formulas 12(n−2), 6(n−2)², and (n−2)³ respectively, and always verify the four counts sum to n³ as a final check.”
Follow-up Questions
- How does the answer change if only 4 of the 6 faces are painted?
- How would you count cubes with exactly one painted face if the cube is cut unevenly along each axis?
- How many unit cubes have at least one painted face, in terms of n?
- How would you solve the same problem for a cuboid rather than a cube?
MCQ Practice
1. A cube is painted on all faces and cut into 125 unit cubes. How many have exactly one painted face?
n = 5 (since 5³=125). Face cubes = 6×(5−2)² = 6×9 = 54.
2. A cube is painted and cut into 64 unit cubes. How many have exactly two painted faces?
n = 4. Edge cubes = 12×(4−2) = 24.
3. A cube is painted and cut into n³ unit cubes. The number with zero painted faces is always?
The fully interior region forms a smaller cube of side (n−2), giving (n−2)³ unpainted cubes.
Flash Cards
Corner cubes (3 painted faces)? — Always exactly 8, regardless of n.
Edge cubes (2 painted faces)? — 12 × (n − 2).
Face cubes (1 painted face)? — 6 × (n − 2)².
Internal cubes (0 painted faces)? — (n − 2)³ — forms a smaller cube hidden inside.