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Investigating 'Painted cubes'.

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Introduction

INTRODUCTION

I am investigating about ‘Painted cubes’. I have been given the following task. ‘Imagine that there is a very large cube which measures 20 by 20 by 20 (20 x 20 x 20) small cubes. The outer surface of the cube is painted red. When it is cut up into smaller cubes there are 8000 small cubes altogether.’

The ultimate aim is to find how many small cubes have 0 faces, 1 face, 2 faces, 3 faces, 4 faces, 5 faces and 6 faces that can be seen. I have to also work out formulas for the nth term, so I can work out how many cubes have 0 faces, 1 face, 2 faces, etc for any size cube.

TABLE OF RESULTS

I counted the number of

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Middle

216

7 x 7 x 7

125

150

60

8

343

8 x 8 x 8

216

216

72

8

516

9 x 9 x 9

343

294

84

8

729

10 x 10 x 10

512

384

96

8

1000

PATTERNS

After filling in the table with the data I started looking for patterns so that I could work out formulas. I had to investigate and find formulas that would work out how many cubes had different amount of faces, e.g. 0 faces, 1 face, 2 faces, 3 faces.

I noticed that the columns for 3 faces had a pattern except 1 x 1 x 1. All of the cubes had 8 cubes with 3 different faces painted. All of these 8 are the vertices of the cube and so had three faces painted.

For 2 faces I noticed that each one got higher by 12, so it was + 12. This told me that somewhere in the formula that there would be +12.

For 1 face I knew that there would be a 6 in the formula, because all the numbers were multiples of 6. Whilst for 0 faces I could not see a pattern straight away, because there was a big leap for the numbers in the column for 0 faces.

The column for 4 and 5 faces was empty because on the large cube there can only be cubes with 0, 1, 2, 3 faces showing. However the cube with dimensions 1 x 1 x 1 had 6 faces showing which is an exception.  

PREDICTION

3 faces

2 faces

1 face

0 faces

Total

8

108

486

729

1331

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Conclusion

GENERAL FORMULA

  • I then checked to see if the formulas worked because if you add them up then it should be equal to the total number of small cubes in the large cube. Which has the formula x=n3, (x representing the total number of cubes).
  • With this it would also work out the general formula, so I added the formulas for 0 faces, 1 face, 2 faces and 3 faces.

Here are the formulas when added together:

= (n – 2)3 + 6(n – 2)2 + (12n +24) + (8)

= (n3 – 6n2 + 12n – 8) + (6n2 – 24n + 24) + (12n – 24) + (8)

= n3 – 6n2+ 12n – 8 + 6n2– 24n + 24 + 12n – 24 + 8

= n3 + 12n – 8 -24n + 24 + 12n – 24 + 8

= n3 + 12n – 8 –24n + 24 + 12n – 24 + 8

= n3 – 8 +24 –24 + 8

= n3 – 8 +24 –24 + 8

= n3 +24 –24

= n3 +24– 24

= n3

CONCLUSION

I found that the shape of the cube had a part in the formulas, like the number of cubes with painted faces was 8, because there are 8 vertices. Also on 2 faces 12 was to be multiplied by something because there are 12 edges.

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