• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  • Level: GCSE
  • Subject: Maths
  • Word count: 1834

Number Grids

Extracts from this document...

Introduction

Introduction

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

91

92

93

94

95

96

97

98

99

100

We were given this number grid and shown that if a two–by-two square was drawn anywhere on the grid, encompassing four numbers, the difference between the product of the top left number and the bottom right number and the top left number.

For example        (13x22) - (12x23) = 10

Also                (16x25) - (15x26) = 10

We were then told to investigate further.

To give us an idea of where we were headed, our teacher told us we should be aiming at least to find a formula that would allow a person to work out the difference between the products of the corners of any rectangle on such a grid of any length.

The Investigation

The fact that the difference came out to ten no matter where the square was drawn made sense, it could be easily demonstrated algebraically.

Let’s say the top left number was ‘n’. If followed along the grid, the other numbers would come to be: (bottom right) n+11, (top right) n+1 and (bottom left) n+10.

So if worked out, the sum would look like:

image16.png

After this, I tried squares of different sizes around the grid.

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

91

92

93

94

95

96

97

98

99

100

I found that the differences in all the 3x3 squares came out to 40

Eg.

(14x32) - (12x34) = 40

(18x36)

...read more.

Middle

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

91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

2x2:

(2x12) - (1x13) = 11

(5x15) – (4x16) = 11

3x3:

(36x56) – (34x58) = 44

(40x60) – (38x62) = 44

4x4:

(81x111) – (78x114) = 99

(86x116) – (83x119) = 99

The pattern came out as:

2x2

3x3

4x4

5x5

 11

44

99

176

So,

11        44        99        176image00.pngimage00.pngimage00.png

      33         55       77                1st Differenceimage00.pngimage00.png

        22        22                2nd Difference

image73.png, determining the coefficient to be used for n2.

n

2

3

4

5

pattern

11

44

99

176

11n2

44

99

176

275

pattern – 11n2

-33

-55

-77

-99

image18.png

Formula:                                

image19.pngimage01.pngimage01.pngimage01.png

                        -22        -22       -22         Formula for linear part:         

Put together:

image20.png

(Simplifying down to this)

Indeed the only part of the end formula that changed, was the part hypothesized to be directly linked to the length of the grid. With the grid length 10, the formula came out to beimage21.png, with the grid length as 11, it came out to beimage22.png. As this part of the formula appeared to be linked to the grid length, it could now be replaced with an expression representing the grid length so it would work for any grid. Thus the formula now becameimage23.png. If the assumption that one of the image24.pngs were in fact the length, and the other the width continued, the formula then becameimage25.png.

To prove this formula, I looked at the situation algebraically and came up with a list of expressions to represent the real life values.

...read more.

Conclusion

image49.png

Simplifying down to image51.png

This was my first 3D formula.

Now for the second set, the calculation was image31.pngimage45.png-image30.pngimage46.png. When the respective expressions were substituted for these values, it came out to:

image52.png

image53.png

Simplifying down to image54.png

This was my second 3D formula.

I tested the both of these to check that everything was ok with them.

The example I came up for the test was:

1

2

3

4

5

6

7

8

9

10

11image11.png

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27image12.png

28

29image13.png

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


91

92

93

94

95

96

97

98

99

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

image14.png

In the example, image55.png, image56.png, image57.png, image58.png, image59.png, image60.png, image61.png and image62.png. Other than that, image63.png, image65.png, image66.png, image67.png, image68.png.image15.png

For the first formula,

image69.png

If done with the formula instead,

image70.png

For the second formula,

image71.png


If done with the formula instead,

image72.png

In both cases, the answer came out the same, both formulas worked. One slight problem was that somewhere in the process I managed to mix up the order I subtracted in so that my first formula always gave a negative answer (as I subtracted a larger value from a smaller one) and my second formula always gave a positive answer (as I subtracted a smaller value from a larger one, like I was supposed to). This isn’t that much of a problem as what the actual number was would always came out the same, the sign in front of it solely depends on the order in which the numbers were subtracted from one another. As the investigation asks for a mere difference, the sign could be ignored altogether.

...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.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Number Stairs, Grids and Sequences essays

  1. Number Grids Investigation Coursework

    = w ((n - 1) (n - 1)) = w (n - 1)2 As the formula I worked out was D = w (n - 1)2 and the expression for the difference between

  2. Investigation of diagonal difference.

    N n + (X-1) n + G n + G + (X - 1) This cutout solution should give the correct diagonal difference of a horizontally aligned 2 x X cutout. I'll try it for a 2 x 4 cutout.

  1. Algebra Investigation - Grid Square and Cube Relationships

    = 10w-10 When finding the general formula for any number (n), both answers begin with the equation n2+nw+9n, which signifies that they can be manipulated easily. Because the second answer has +10w-10 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 10w-10 will always be present.

  2. Investigate the differences between products in a controlled sized grid.

    I am then going to represent the other numbers in relation to x. 1 2 11 12 21 22 31 32 x x+1 x+30 x+31 I am now going to multiply these together as I did with the numbers to form an algebraic equation.

  1. Number Grid Investigation

    the 2 x 3 to the 2 x 4 it has gone up 8 I think it will continue to go up 8 for the 2 x 5 21 22 29 30 37 38 45 46 53 54 21 x 54 = 1134 Difference= 32 22 x 53 = 1166

  2. Investigate Borders - a fencing problem.

    Diagram of Borders of square: 5x3 Table of results for Borders of square: 5x3 Formula to find the number of squares needed for each border (for square 5x3): Common difference = 4 First term = 16 Formula = Simplification = Experiment I will try to find the number of squares

  1. Investigate the difference between the products of the numbers in the opposite corners of ...

    - 3015 = 40 I will now try a random position on the grid to finalise my results so far. 77 78 79 87 88 89 97 98 99 77 x 99 = 7623 79 x 97 = 7663 7663 - 7623 = 40 I can now confirm that for

  2. Number Grids

    I will then place these results in a table before trying to work out a general rule between all the differences using an algebraic formula. Here is the investigation for a 3 x 3 grid... 1 2 3 11 12 13 21 22 23 12 13 14 22 23 24

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work