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• Level: GCSE
• Subject: Maths
• Word count: 2556

# Number Grid Investigation

Extracts from this document...

Introduction

Mathematics Coursework

Higher Tier Task – Number Grid

Introduction

I am going to investigate for my coursework something that I feel will require a lot of thorough investigation and could prove to have some valuable and interesting answers. I am going to investigate the differences in product of the diagonally opposite corners of randomly selected squares within a 10 X 10 grid, numbered 1- 100. Finding the differences of squares that are aligned differently and that are of different lengths and widths is my main objective and I hope to do this so that there is a more thorough explanation for my results.

These squares and rectangles will be randomly selected and aligned in the 10 x 10 grid should and back up my theories with explanations and also with the of use algebra to prove them and to improve my investigation. I will make several predictions and theories that vary greatly so that they can relate to the patterns and record my results in a table.

Firstly I will start with the squares because this was the investigation I was originally set and then I will move onto investigating the number gird with randomly selected squares and see whether using the different sized shapes will affect the difference that we are trying to find.

Middle

54

55

62

63

64

65

72

73

74

75

3 x 5 Rectangle

 65 66 67 68 69 75 76 77 78 79 85 86 87 88 89

4 x 5 Rectangle

 11 12 13 14 15 21 22 23 24 25 31 32 33 34 35 41 42 43 44 45

5 x 6 Rectangle

 2 3 4 5 6 7 12 13 14 15 16 17 22 23 24 25 26 27 32 33 34 35 36 37 42 43 44 45 46 47

Results

 n m n + m n x m (n x m) – (n + m) D 3 2 5 6 1 20 4 2 6 8 2 30 5 2 7 10 3 40 4 3 7 12 5 60 5 3 8 15 7 80 5 4 9 20 11 120 6 5 11 30 19 200

From these results, I can clearly see that there is a correlation between:

The Overall Difference we have been trying to calculate from the start of the experiment.

=D

And the difference between the product of the length (m) and width (n)

and the sum of the length (m) and width (n)

= (n x m) – (n + m)

I can expand upon this to devise a formula that:

= 10 x [1 +(n x m) – (n + m)]

= 10(n x m)(n – 1)

D = 10(n x m)(n – 1)

This formula has only been devised by testing many different sized rectangles and can only be proved with the use of algebra.

Algebra

D = (B x C) – (A x D)

D =   x + (n-1)    X       x + 10(m-1)

-        x           X       x + 10(m-1) + (n-1)

D = (x + n - 1) (x +10m -10)

-  x (x +10m -10 + n -1)

D = (x² + 10mx - 10x + nx - 10mn – 10n - x -10m + 10)

- (x² + 10mx - 10x + nx - x)

D = 10mn - 10n – 10m +10

D = 10(mn – m – n + 1)

D = 10 (n – 1) (m - 1)

The difference is therefore 10 (n – 1) (m – 1)

We cannot be 100% sure that this formula is correct without testing it. So, to test this formula, we have previously established that the difference in a 5 x 4 Rectangle is 120 so we can therefore test this with our equation for squares.

So when:

n = 5

m = 4

10 (n – 1) (m – 1) = 120

10 (5 - 1)(4 – 1)

= 10(4 x 3)

= 10 x 12

= 120

The formula worked.

Grid Size

By changing the size of the overall grid used, the numbers should all be shifted in a different direction than before, depending on the size of the grid I use, previously I had used a 10 x 10 square.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90

In this case, the Grid size (G) is 9 as it is an alternative number to the previous recurring grid size used, 10, which was used in the previous equations to calculate the various factors of my investigation. Hopefully this change should have an effect on the difference produced in the squares I choose.

Squares

2 x 2 Square

 4 5 13 14

3 x 3 Square

 48 49 50 57 58 59 66 67 68

4 x 4 Square

 55 56 57 58 64 65 66 67 73 74 75 76 82 83 84 85

Conclusion

-        x              X       x + 11L(n - 1)

D = (x² + x10L(n - 1) + xL(n – 1) + 10L²(n -1)² )

- (x² +  x11L(n - 1) )

D = (x² + x11L(n - 1) + 10L²(n -1)² )

- (x² +  x11L(n - 1) )

D = 10L²(n -1)²

The difference is therefore 10L²(n -1)²

Now we can test the formula that I appear to find to see if it is, in fact, correct.

So when:

L = 2

n = 4

10L²(n -1)² = 360

D  = 10 x 2² (4 - 1)²

= 10 x 4  x 3²

= 40 x 9

= 360

The formula worked again. So I will therefore conclude that it is correct for this investigation and can be used as a valid result of my research.

Rectangles

2 x 3 Rectangle

 1 3 5 21 23 25

3 x 4 Rectangle

 25 27 29 31 45 47 49 51 65 67 69 71

4 x 5 Rectangle

 31 33 35 37 39 51 53 55 57 59 71 73 75 77 79 91 93 95 97 99

3 x 5 Rectangle

 9 11 13 15 17 29 31 33 35 37 49 51 53 55 57

Algebra

D = (B x C) – (A x D)

D =   x + L(n-1)     X      x + 10L(m - 1)

-        x             X       x + 10L(m - 1) + L(n – 1)

D = (x² + x10L(m - 1) + xL(n – 1) + 10L²(m -1)(n  - 1) )

- (x² +  x10L(m - 1) + xL(n - 1) )

D = 10L²(m -1)(n - 1)

The difference is therefore 10L²(m -1)(n - 1)

I can test this now, using the results that I have previously gathered when investigating the affect of changing the Number Gap (L). I know that in a 3 x 5 Rectangle, the difference is 320.

So, in this case:

D = 10L²(m -1)(n - 1)

=  10 x 2² x (5 - 1)(3 - 1)

=     10 x 4 x (4 x 2)

=     40 x 8

=     320

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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