Cube & Dice

medium importance~0-1 Q in Tier 16 formulas⚡ 7 shortcuts3 subtopics

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Painted cube cut into small cubes

⏱ 5 min read🧩 8 question types🎯 17 practice Q

The idea in one minute

A big cube of side n is painted on the outside and cut into n cubed small cubes. A small cube keeps paint only on the faces that were on the surface. Corners have 3 painted faces, edges have 2, face centres have 1, and the hidden core has none. Always use n minus 2 for edges, faces and the core.

The idea

Take a big cube of side $n$ . Paint all six faces. Cut it into $n \times n \times n$ small cubes.

A small cube keeps paint only on faces that were on the outside. Its position decides how many faces are painted.

The four counts

Painted faces	Where the cube is	How many
3	corner	$8$
2	on an edge, not a corner	$12(n-2)$
1	middle of a face	$6(n-2)^2$
0	inside	$(n-2)^3$

For $n = 4$ : $8$ , $24$ , $24$ and $8$ . They add to $64$ .

Rule: Corners are always 8, whatever the size. Every other count uses $(n-2)$ .

Why n minus 2

Each edge has $n$ small cubes. Two of them are corners. That leaves $n-2$ .

Each face has $n \times n$ small cubes. Remove the border ring and $(n-2) \times (n-2)$ are left.

Peel one layer off every side of the big cube. A core of $(n-2)^3$ cubes is left.

The four counts must add up to $n^3$ . Try $n = 3$ : $8 + 12 + 6 + 1 = 27$ .

Tip: If your counts do not add to $n^3$ , one of them is wrong.

At least questions

At least one face painted: all cubes except the core, $n^3 - (n-2)^3$ .
At least two faces painted: corners plus edges, $8 + 12(n-2)$ .

A cube of side 6 has $216 - 64 = 152$ cubes with at least one painted face.

Finding n

Sometimes a count is given and $n$ is asked. Divide, then add 2.

If 24 cubes have exactly two painted faces, then $12(n-2) = 24$ . So $n - 2 = 2$ and $n = 4$ .

A big cube of side 12 cm is cut into small cubes of side 3 cm. Then $n = 12 \div 3 = 4$ .

Cuboids

For an $a \times b \times c$ block, each side keeps its own $(\text{side} - 2)$ .

Corners: $8$ .
Edges: $4[(a-2)+(b-2)+(c-2)]$ .
Faces: $2[(a-2)(b-2)+(b-2)(c-2)+(c-2)(a-2)]$ .
Core: $(a-2)(b-2)(c-2)$ .

For $5 \times 4 \times 3$ : edges $4(3+2+1) = 24$ and core $3 \times 2 \times 1 = 6$ .

Worked example

A cube of side 5 is painted and cut into 125 small cubes.

Corners: $8$ .
Edges: $12 \times 3 = 36$ .
Face centres: $6 \times 3^2 = 54$ .
Core: $3^3 = 27$ .

Check: $8 + 36 + 54 + 27 = 125$ . All counts are right.

Only some faces painted

If only some faces are painted, count face by face. Each painted face has $n \times n$ small cubes. A cube on the edge shared by two painted faces is counted twice, so remove it.

Top and front painted, $n = 4$ : $16 + 16 = 32$ . The shared edge has 4 cubes with two painted faces. So exactly one painted face is $32 - 2 \times 4 = 24$ .

Watch: Two opposite painted faces share no cubes, so nothing is counted twice.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Three faces painted (corners)

How to spot it:

A painted cube is cut up. The question asks for cubes with three painted faces.

Method

Recall that only corner cubes have three painted faces.
A cube has 8 corners.

Why it works:

A small cube touches three big faces only at a corner of the big cube.

Try this

A cube of side 7 is painted on all faces and cut into small cubes of side 1. How many small cubes have three painted faces?

Show solution

Only the corner cubes have three painted faces.
A cube has 8 corners.

Answer

Type 2very common2 practice Q

Two faces painted (edges)

How to spot it:

The question asks for cubes with exactly two painted faces.

12(n-2)

Method

Find n, the number of small cubes along one edge.
Each edge has n - 2 cubes that are not corners.
Multiply by 12 edges.

Why it works:

Each of the 12 edges has n cubes, and 2 of them are corners.

Try this

A cube of side 5 is painted and cut into 125 small cubes. How many have exactly two painted faces?

Show solution

$n = 5$ , so each edge has $5 - 2 = 3$ such cubes.
$12 \times 3 = 36$ .

Answer

Type 3very common2 practice Q

One face painted (face centres)

How to spot it:

The question asks for cubes with exactly one painted face.

6(n-2)^{2}

Method

Find n.
On each face, remove the border ring. A block of (n - 2) by (n - 2) is left.
Multiply by 6 faces.

Why it works:

Cubes in the middle of a face touch no edge, so they have only that one face painted.

Try this

A cube of side 6 is painted and cut into small cubes of side 1. How many have exactly one painted face?

Show solution

$n - 2 = 4$ .
One face: $4 \times 4 = 16$ .
$6 \times 16 = 96$ .

Answer

Type 4common2 practice Q

No face painted (core)

How to spot it:

The question asks for cubes with no paint at all.

(n-2)^{3}

Method

Find n.
Peel one layer off every side.
Cube the side that is left: (n - 2) cubed.

Why it works:

The cubes inside never touched the outside, so they stay unpainted.

Try this

A cube of side 5 is painted and cut into small cubes of side 1. How many small cubes have no painted face?

Show solution

$n - 2 = 3$ .
$3^3 = 27$ .

Answer

Type 5common2 practice Q

At least or combined counts

How to spot it:

The question says at least one, at least two, or asks for two classes added together.

n^{3} - (n-2)^{3}

Method

At least one painted: all cubes minus the core.
At least two painted: corners plus edges.
For any other mix, add the classes you need.

Why it works:

Every class is already counted, so a combined count is a sum of classes.

Try this

A cube of side 4 is painted and cut into small cubes of side 1. How many have at least two painted faces?

Show solution

Corners: $8$ .
Edges: $12 \times 2 = 24$ .
$8 + 24 = 32$ .

Answer

Type 6occasional2 practice Q

Painted cuboid

How to spot it:

The block is a by b by c, not a cube.

Method

Write a - 2, b - 2 and c - 2.
Corners: 8.
Edges: 4 times the sum of the three (side - 2) values.
Faces and core use products of the (side - 2) values.

Why it works:

Each side of a cuboid has its own length, so each keeps its own (side - 2).

Try this

A cuboid measuring 5 by 4 by 3 is painted and cut into small cubes of side 1. How many have exactly two painted faces?

Show solution

$(5-2)+(4-2)+(3-2) = 3+2+1 = 6$ .
$4 \times 6 = 24$ .

Answer

Type 7occasional2 practice Q

Find n from a count or a size

How to spot it:

A count of painted cubes, or the edge lengths of the big and small cubes, is given. Another count is asked.

Method

Turn the given count into n: divide by 12 (edges) or by 6 and take the square root (faces).
Add 2 to get n. If edges are in cm, n = big edge divided by small edge.
Now find the count that is asked.

Why it works:

The formulas run both ways, so a count fixes n.

Try this

36 small cubes have exactly two painted faces. How many small cubes have exactly one painted face?

Show solution

$12(n-2) = 36$ , so $n - 2 = 3$ and $n = 5$ .
One face: $6 \times 3^2 = 54$ .

Answer

Type 8occasional

Only some faces painted

How to spot it:

The question says which faces are painted, such as the top and one side, or two opposite faces.

Method

Count the small cubes on each painted face: n times n.
If two painted faces touch, they share an edge of n cubes.
Exactly one painted face = sum of faces minus twice the shared cubes.

Why it works:

The cubes on a shared edge have two painted faces, so they must leave the exactly-one count.

Try this

A cube of side 4 is painted on its top face and on its front face only. It is cut into 64 small cubes. How many have exactly one painted face?

Show solution

Top face: $4 \times 4 = 16$ . Front face: $16$ .
The shared edge has 4 cubes with two painted faces.
$16 + 16 - 2 \times 4 = 24$ .

Answer

Formula sheet

Two faces painted

12(n-2)

Cubes on the edges, not the corners.

One face painted

6(n-2)^2

Cubes in the middle of each face.

No face painted

(n-2)^3

The hidden inner block.

Total check

8 + 12(n-2) + 6(n-2)^2 + (n-2)^3 = n^3

The four counts add up to all the small cubes.

Shortcuts that save time

⚡ Check the total

The four counts must add up to n cubed. Use it to catch mistakes.

Example

For a cube of side 5, check the four counts.

Show solution

Corners $8$ ; edges $12 \times 3 = 36$ ; faces $6 \times 9 = 54$ ; core $27$ .
$8 + 36 + 54 + 27 = 125 = 5^3$ .

Answer

125, so the counts are right

⚡ At least one painted

Take all the small cubes and remove the core.

Example

A cube of side 6 is painted and cut. How many small cubes have at least one painted face?

Show solution

Total: $6^3 = 216$ .
Core: $4^3 = 64$ .
$216 - 64 = 152$ .

Answer

152

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using n instead of (n - 2) in the formulas.

Edges, faces and the core all use (n - 2). Subtract 2 before you square or cube.

Mistake 02

Thinking the number of corner cubes changes with n.

A cube always has 8 corners, so 3 painted faces always means 8 cubes.

Mistake 03

Counting corners again inside the edge cubes.

Edge cubes are only the n - 2 cubes between the corners.

Mistake 04

Using the cube formulas for a cuboid.

Use each side's own (side - 2): a - 2, b - 2 and c - 2.

Mistake 05

Forgetting to subtract the shared edge when only some faces are painted.

Two painted faces that touch share one edge. Remove those cubes from the exactly-one count.

Quick revision

Read this the night before the exam.

3 painted faces: 8 corners, for any n.
2 painted faces: $12(n-2)$ .
1 painted face: $6(n-2)^2$ .
0 painted faces: $(n-2)^3$ .
The four counts add to $n^3$ .
At least one: $n^3 - (n-2)^3$ . At least two: $8 + 12(n-2)$ .
Given a count? Divide, then add 2 to get n.

Practice: 17 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 5 min · wrong answers go to your mistake notebook automatically.

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