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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.

...read more.

Middle

54

55

62

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65

72

73

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75

3 x 5 Rectangle

65

66

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69

75

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79

85

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89

4 x 5 Rectangle

11

12

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21

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31

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5 x 6 Rectangle

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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.

Algebraimage03.png

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.

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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

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76

82

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84

85

...read more.

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

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99

3 x 5 Rectangle

9

11

13

15

17

29

31

33

35

37

49

51

53

55

57

Algebra

image07.png

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

...read more.

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