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  • Level: GCSE
  • Subject: Maths
  • Word count: 3623

Maths Coursework

Extracts from this document...

Introduction

image00.png

Introduction:

The task we have been set for this piece of coursework is to investigate numbers presented in a number grid. A box is placed around 4 of the number and the question asked was to find the product of the top left number and the bottom right number. This solution was then taken away from the top right number and the bottom left number like this:

12

13

22

23

(13 x 22)-(12 x23) = 10

Aim:

My aim in this coursework is to experiment using 3 different variables in boxes.

These will be: size, shape and size of grid.

I will investigate boxes that are the same size, bigger, smaller, in different proportions and those that are different in shape.

What I mean by this is that I will take boxes of the following proportions:

2x2, 3x3, 4x4, 5x5, 2x3, 2x4, 2x5, 3x4, 3x5, 4x5.

I will take 4 boxes of each of type of the above that are squares. I will take these results by using the random function on my calculator.  The figure given will be any number in the box.

Investigation:

2x2 boxes:

My first boxes that I will be investigating are the 2x2 boxes of which I am taking 4.

Theses are the boxes I have taken:

1:

35

36

 45

46

(36x45)-(35x46) = 10
2:
image01.png

57

58

67

68

(58x67)-(57x68) = 10
3:
image07.pngimage09.pngimage05.pngimage06.png

88

89

98

99

(89x98)

...read more.

Middle

4

5

6

7

14

15

16

17

24

25

26

27

(7x24)-(4x27) = 60

33

34

35

36

43

44

45

46

53

54

55

56

(36x53)-(33x56) = 60

Lorenzo Brusini

Proving algebraically the 3x4 boxes:

The answers for 3x4 boxes all answer to 60. I shall try to use the algebraic box once again where n is the smallest number in the box:

N

N+1

N+2

N+3

N+10

N+11

N+12

N+13

N+20

N+21

N+22

N+23

I shall try to use the formula:

(N+3)x(N+20)-N(N+23) = 60

(N2+20N+3N+60)-(N2+23N) = 60

N2+23N+60-N2-23N = 60image03.pngimage03.pngimage03.pngimage03.png

Therefore 60 = 60

3x5 boxes:

54

55

56

57

58

64

65

66

67

68

74

75

76

77

78

(58x74)-(54x78) = 80

13

14

15

16

17

23

24

25

26

27

33

34

35

36

37

 (17x33)-(13x37) = 80

2

3

4

5

6

12

13

14

15

16

22

23

24

25

26

 (6x22)-(2x26) = 80

36

37

38

39

40

46

47

48

49

50

56

67

68

69

60

 (40x56)-(36x60) = 80

Proving algebraically the 3x5 boxes:

The answers for 3x5 boxes all answer to 80. I shall try to use the algebraic box once again where n is the smallest number in the box:

N

N+1

N+2

N+3

N+4

N+10

N+11

N+12

N+13

N+14

N+20

N+21

N+22

N+23

N+24

Lorenzo Brusini

I shall try to use the formula:

(N+4)x(N+20)-N(N+24) = 80

(N2+20N+4N+80)-(N2+24N) = 80

N2+24N+80-N2-24N = 80image03.pngimage03.pngimage03.pngimage03.png

Therefore 80 = 80

4x5 boxes:

43

44

45

46

47

53

54

55

56

57

63

64

65

66

67

73

74

75

76

77

(47x73)-(43x77) = 120

22

23

24

25

26

33

34

35

36

37

43

44

45

46

47

53

54

55

56

57

 (26x53)-(22x57) = 120

66

67

68

69

70

76

77

78

79

80

86

87

88

89

90

96

97

98

99

100

 (70x96)-(66x100) = 120

35

36

37

38

39

45

46

47

48

49

55

56

57

58

59

65

66

67

68

69

 (39x65)-(35x69) = 120

Proving algebraically the 4x5 boxes:

The answers for 4x5 boxes all answer to 120. I shall try to use the algebraic box once again where n is the smallest number in the box:

N

N+1

N+2

N+3

N+4

N+10

...read more.

Conclusion

2+10N+9-N2-10N = 9image03.pngimage03.pngimage03.pngimage03.png

                Therefore 9 = 9

N

N+3

N+27

N+30

(N+3)x(N+27)-N(N+30) = 81

(N2+3N+27N+81)-(N2+30N) = 81

N2+30N+81-N2-30N = 81image03.pngimage03.pngimage03.pngimage03.png

Therefore 81 = 81

N

N+3

N+18

N+21

(N+3)x(N+18)-N(N+21) = 54

(N2+3N+18N+54)-(N2+21N) = 54

N2+21N+54-N2-21N = 54image03.pngimage03.pngimage03.pngimage03.png

Therefore 54 = 54

N

N+4

N+27

N+31

(N+4)x(N+27)-N(N+31) = 108

(N2+4N+27N+108)-(N2+31N) = 108

N2+31N+108-N2-31N = 108image03.pngimage03.pngimage03.pngimage03.png

Therefore 108 = 108

AxB grid:

N

N+A-1

N+G(B-1)

N+G(B-1)+(A-1)

Where: A = The horizontal side of the box,

         B = The vertical side of the box,

         N = The smallest number in the box,

                                                         G = The size of the grid.

I shall put this box into the formula I used for all the other boxes:

(N+A-1)(N+G(B-1))-N(N+1G(B-1)+A-1)

=(N+A-1)(N+GB-G)-N(N+GB-G+A-1)image04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.pngimage04.png

=N2+GBN-GN+AN+GAB-GA-N-GB+G-N2-GBN+GN-AN+N

= GAB-GA-GB+G

This can be factorised to make:

G(A-1)(B-1)

Lorenzo Brusini

I shall show that this formula works by using it on every rectangular box:

Size of box

Formula

Answer

2x2

6(1x1)

6

2x2

7(1x1)

7

2x2

8(1x1)

8

2x2

9(1x1)

9

4x4

6(3x3)

54

4x4

7(3x3)

63

4x4

8(3x3)

72

4x4

9(3x3)

81

3x4

6(2x3)

36

3x4

7(2x3)

42

3x4

8(2x3)

48

3x4

9(2x3)

54

4x5

6(3x4)

72

4x5

7(3x4)

84

4x5

8(3x4)

4x5

9(3x4)

Looking at the results of the formula it shows that it works and can be used for any square or rectangle in any size grid.

Lorenzo Brusini

...read more.

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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