3.8 Cubes, Dice & Figure Counting

Part 1: The Painted Cube

The standard question takes a solid cube, paints its outer surface, then cuts it into $n^3$ identical smaller cubes and asks how many small cubes carry a given number of painted faces. The answer depends only on position, and the four categories are exhaustive.

A useful check: the four counts must sum to $n^3$.

Worked example: n = 4

A cube is painted and cut into 64 smaller cubes, so $n = 4$ and $n - 2 = 2$.

• Three faces painted: 8
• Two faces painted: $12 \times 2 = $ 24
• One face painted: $6 \times 2^2 = $ 24
• No face painted: $2^3 = $ 8

Check: $8 + 24 + 24 + 8 = 64$. Correct.

Worked example: n = 5

Here $n - 2 = 3$:

• Three faces: 8
• Two faces: $12 \times 3 = 36$
• One face: $6 \times 9 = 54$
• None: $27$

Check: $8 + 36 + 54 + 27 = 125 = 5^3$. Correct.

Variants

If only some faces are painted, or different colours are used on different face pairs, do not reach for the formula. Instead count layer by layer, treating each painted face as contributing its own $n \times n$ layer and subtracting the overlaps along shared edges. The formulae above assume all six faces painted in one colour.

A second variant asks for cubes with at least one painted face, which is simply

$n^3 - (n-2)^3$

For $n = 4$ that gives $64 - 8 = 56$.

Part 2: Dice

The standard die

On a conventional die the numbers on opposite faces sum to seven:

$1 \leftrightarrow 6, \qquad 2 \leftrightarrow 5, \qquad 3 \leftrightarrow 4$

If a question states or implies a standard die, this rule answers it immediately. Note that a die has 6 faces, 8 vertices and 12 edges, and that any face is adjacent to four faces and opposite to exactly one.

Reading two or more views

The general questions give two or more pictures of the same die and ask which number lies opposite a given one. Two rules settle nearly all of them:

Rule 1 — Adjacency rule. If two numbers appear together in any single view, they are on adjacent faces and therefore cannot be opposite each other. Collect all adjacencies across all views; the number that never appears alongside a given number is its opposite.

Rule 2 — Common face rule. If two views share exactly one common visible number, the die has been rotated about that face. The remaining numbers in the two views can be matched by rotating in the same sense.

Worked example

Two views of a die show:

• View 1: top 1, front 2, right 3
• View 2: top 1, front 3, right 5

Both views show 1 on top, so the die was rotated about the vertical axis. From View 1, the faces adjacent to 1 include 2 and 3; from View 2 they also include 5. So 2, 3 and 5 are all adjacent to 1, and by extension so is the fourth side face. The only number that never appears with 1 is 6, so 6 is opposite 1 — consistent with the standard rule.

For 2: it appears with 1 and 3 in View 1. Rotating from View 1 to View 2 moves 2 out of sight and brings 5 into view on the right, which places 5 where 2 is not. Combined with the adjacency of 2 and 3, the face opposite 2 is 5.

Dice from a net

When a flattened net is given, use the rule that in a straight line of squares, faces one apart are adjacent and faces two apart are opposite. Squares separated by one intervening square in a row lie on opposite faces once folded.

Part 3: Figure Counting

Counting triangles or rectangles in a composite figure is systematic, not intuitive. The method is to count by size class.

Counting triangles

1. Count the smallest, indivisible triangles first.
2. Count triangles made of two smallest triangles, then three, and so on.
3. Finally count the whole figure if it is itself a triangle.
4. Add the classes.

Label every vertex before starting so that no triangle is counted twice.

Counting rectangles in a grid

For a grid of $m$ vertical lines and $n$ horizontal lines, any rectangle is fixed by choosing two vertical and two horizontal lines:

$\text{Number of rectangles} = \binom{m}{2}\binom{n}{2}$

For a chessboard-style grid of $p$ columns and $q$ rows of cells, there are $p+1$ and $q+1$ lines, so

$\text{Rectangles} = \binom{p+1}{2}\binom{q+1}{2}$

Worked example. In a 3 by 3 grid of cells there are 4 vertical and 4 horizontal lines, so the number of rectangles is

$\binom{4}{2}\binom{4}{2} = 6 \times 6 = 36$

Of these, the number of squares is $3^2 + 2^2 + 1^2 = 14$.

In general, an $n$ by $n$ grid contains $\sum k^2$ squares for $k = 1$ to $n$, and $\left[\binom{n+1}{2}\right]^2$ rectangles.