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• Level: GCSE
• Subject: Maths
• Word count: 4480

# Maths Investigation -Painted Cubes

Extracts from this document...

Introduction

Maths Investigation –

Painted Cubes

## Introduction

I was given a brief to investigate the number of faces on a cube, which measured 20 small cubes by 20 small cubes by 20 small cubes (20 x 20 x 20)

To do this, I had to imagine that there was a very large cube, which had had its outer surface painted red.  When it was dry, the large cube was cut up into the smaller cubes, all 8000 of them.  From there, I had to answer the question, ‘How many of the small cubes will have no red faces, one red face, two red faces, and three faces?’

From this, I hope to find a formula to work out the number of different faces on a cube sized ‘n x n x n’.

## Solving the Problem

To solve this problem, I built different sized cubes (2 x 2 x 2,  3 x 3 x 3,  4 x 4 x 4,  5 x 5 x 5,  6 x 6 x 6,  7 x 7 x 7,  8 x 8 x 8,  9 x 9 x 9) using multi-links.

I started by building a cube sized ‘2 x 2 x 2’.  As I looked at the cube, I noticed that all of them had three faces.  I then went onto a ‘3 x 3 x 3’ cube.  As I observed the cube, I saw that the corners all had three faces, the edges had two, and the faces had one.  I looked into this matter to see if this was true…

As I went further into the investigation, I found this was true.  This made it much easier for me to count the cubes, and be more systematic.  Now I could carry on building the cubes, and be more confident about not missing any out.

Whilst building the cubes, I also drew them and decided to colour code the different faces (Red = Three faces,  Green = Two faces,  Blue = One face).

Middle

For cubes with two faces, I used the number 12 because the numbers went up in multiples of twelve.

For cubes with three faces, I used the number 6 because the formula contained ‘n²’ and so I had to divide the second difference (12) by 2.

For cubes with 0 faces painted, I used the term ‘n³’ because the numbers were the cube numbers.

## EXTENSION 1

For my first extension, I investigated further into the number of faces painted, but this time in cuboids.

For these cuboids I am going to keep two of the lengths constant, and change one (2 x 2 x 2,  2 x 2 x 3,  2 x 2 x 4,  2 x 2 x 5,  2 x 2 x 6,   2 x 2 x 7,  2 x 2 x 8,  2 x 2 x 9).

Again I drew diagrams, and began counting the faces of the cubes…

## Results Table

 Cube Size2 x 2 x n No. of cubes with 3 faces painted No. of cubes with 2 faces painted No. of cubes with 1 face painted No. of cubes with 0 faces painted Total No. of cubes 2 x 2 x 3 8 4 0 0 12 2 x 2 x 4 8 8 0 0 16 2 x 2 x 5 8 12 0 0 20 2 x 2 x 6 8 16 0 0 24 2 x 2 x 7 8 20 0 0 28 2 x 2 x 8 8 24 0 0 32 2 x 2 x 9 8 28 0 0 36

## Patterns and Formulas

As I looked at my results, I realised I only had to find formulas for cubes with three face and two faces painted, as the others had a constant zero.

I saw that again there was no formula for three faces painted as it was the same constant 8.

Now, I looked for patterns within the results of cubes with two faces painted.

0        4        8        12        16

4         4         4         4

I found the first difference was a constant 4, and so I thought the formula may be ‘4n’.  I again tried it out on 2 x 2 x 3…

4n

4 x 3 = 12

This was wrong, as it was too big.  The answer I needed was four, and so I tried subtracting 8.  This was right, but just in-case I tested it on 2 x 2 x 4.

(4 x 4) – 8

16 – 8 = 8

This was correct.

Conclusion

To do this, I would have to take values of a cube (all three lengths constant), and substitute them into my general formulas that I found for my cuboids.  Firstly, I would test the cube with sides 3 x 3 x 3…

Three faces - 8

Two faces - 4(a – 2) + 4(b – 2) + 4(c – 2)

4a + 4b + 4c – 24

(4 x 3) + (4 x 3) + (4 x 3) – 24

12 + 12 + 12 – 24 = 12

One Face - 2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)

2ab + 2ac + 2bc – 8a – 8b – 8c + 24

(2 x 3 x 3) + (2 x 3 x 3) + (2 x 3 x 3) – (8 x 3) – (8 x 3) – (8 x 3) + 24

18 + 18 + 18 – 24 – 24 – 24 + 24 = 6

Zero Faces - (a – 2)(b – 2)(c – 2)

abc – 2ab – 2ac – 2bc + 4a + 4b + 4c – 8

(3³) – (2 x 3 x 3) – (2 x 3 x 3) – (2 x 3 x 3) + (4 x 3) + (4 x 3) + (4 x 3) – 8

27 – 18 – 18 – 18 + 12 + 12 + 12 – 8 = 1

The values I got for this cube were all correct.  Just to be sure, I checked once again, with the cube sized 7 x 7 x 7

Three faces - 8

Two faces - 4(a – 2) + 4(b – 2) + 4(c – 2)

4a + 4b + 4c – 24

28 + 28 + 28 – 24 = 60

One face - 2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)

2ab + 2ac + 2bc – 8a – 8b – 8c + 24

98 + 98 + 98 – 56 – 56 – 56 + 24 = 150

Zero faces - (a – 2)(b – 2)(c – 2)

abc – 2ab – 2ac – 2bc + 4a + 4b + 4c – 8

343 – 98 – 98 – 98 + 28 + 28 + 28 – 8 = 125

This was again correct, which meant that my theory was also right.  A cube was a special form of a cuboid, and this was demonstrated by my general formulas working for my cubes.

## Conclusion

Consequently, I did find a general formula that would work for any sized cube or cuboid.  In the process, I also found formulas for keeping all the lengths constant, two same numbers as constants, and two different numbers for constants…

n x n x n  (All lengths constant)

Three faces –  8

Two faces –  12(n – 2)

One face –  6(n – 2)²

Zero faces –  (n – 2)³

TOTAL FORMULA –  8 + 12(n – 2) + 6(n – 2)² + (n – 2)³

2 x 2 x n  (Two same numbers constant)

Three faces –  8

Two faces –  4n – 8

One face –  0

Zero faces –  0

TOTAL FORMULA –  8 + (4n – 8)

2 x 3 x n  (Two different numbers constant)

Three faces –  8

Two faces –  4n – 4

One face –  2n –  4

Zero faces –  0

TOTAL FORMULA –  8 + (4n – 4) + (2n – 4)

a x b x c  (All lengths different)

Three faces –  8

Two faces –  4(a – 2) + 4(b – 2) + 4(c – 2)

One face –  2(a – 2)(b – 2) + 2(a – 2)(c – 2) + 2(b – 2)(c – 2)

Zero faces –  (a – 2)(b – 2)(c – 2)

=        +                         +                          +        +

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