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Math Investigation Essay

4261 words - 18 pages

Math Investigation of 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 ...view middle of the document...

I looked closely for patterns throughout the numbers. I found that to find the number of cubes with three faces painted, it was always a constant number (8).

I then looked for a pattern in my results to find how many cubes had two faces painted. I wrote the numbers down, to see if they had a pattern between them.

0 - 12 - 24 - 36 - 48
12 12 12 12

As shown above, I did find a pattern in the numbers. I found the numbers increased by 12. I assumed the formula was '12n'. I tried out the formula by using 'n = 2' and multiplied it by 12, summing up to 24. This was incorrect as I needed the answer 0. This made me realize I would have to minus a number in order to get my formula right.

I tried subtracting 2 from 'n' to find it was the correct method. I tested it on 2 x 2 x 2.

12(n – 2)= 0
n = 2
12(2 – 2) = 0
12 x 0 = 0

This was correct. I now went on to see how many cubes had one face painted. Again, I looked for patterns between the numbers.

0 - 6 - 24 - 54 - 96
6 - 18 - 30 - 42
12 12 12

This time I found that there was a second difference, which was constant. This told me that my formula contained 'n²'.
Using my previous knowledge of patterns, I knew I had to half the 12 to make it work with the n². Therefore this became '6n²'.

Taking 2 as my 'n', I tested this formula to see if it was correct.

n = 2
6 x (2²) = 24
The answer was supposed to be 0, which meant my formula was yet again wrong. As this was the same problem I had before, (the final answer was too big), I knew I had to subtract a number. I looked at the formula I was previously working on, and noticed I used (n - 2). I decided to use it in my present formula to see if it would help or not.

6(n - 2)² = 0
6(2 - 2)² = 0
6(0)² = 0

Now my formula was complete and correct. Finally I moved onto cubes which had no faces painted.

0 - 1 - 8 - 27 - 64
1 - 7 - 19 - 37
6 - 12 - 18
6 6

This time I found that the third difference was constant. This meant the formula contained 'n³'. I noticed the original numbers were all cube numbers, so I tried the formula n³ by using 'n = 2'.

n = 2
2 x 2 x 2 = 8

This was wrong. The answer I was looking for was 0. As I had used it in my previous formulas (n - 2), I decided to see if it would work again.

n = 2
(2 - 2) x (2 - 2) x (2 - 2) = 0

This formula worked which meant I had now obtained all my formulas. Just to check them, I tested them on a cube sized 10 x 10 x 10. I drew the diagram, and started counted the cubes as I was now confident enough to use the drawing rather than building the cube. Then I predicted the number of cubes with the faces.

For a cube sized 10 x 10 x 10

The number of cubes with three faces painted will be 8.

The number...

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