• 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
  11. 11
    11
  • Level: GCSE
  • Subject: Maths
  • Word count: 2057

Number grids

Extracts from this document...

Introduction

Number grids

My task is to find an algebraic rule for different sized squares in a set sized number grid.

To do this I will establish my algebraic rule by creating a 10×10 square and marking out 3 different sized squares inside this square. I will then work out the rules for these individual squares and combine them to create my overall rule.

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 have marked out my smaller squares inside the grid and will now work out an algebraic rule:

To find my algebraic rule I will times the opposite corners in the inset squares and take the numbers away from each other to find the difference.

2×2-

55×66=3630

56×65=3640

3640-3630 = 10

89×100=8900

90×99=8910

8910-8900=10

22×33=726

23×32=736

736-726=10

27×38=1026

28×37=1036

1036-1026=10

After multiplying the corners of the 2×2 squares I then took the lowest away from the highest. This number is always 10.

In this section: a represents the number in the top left hand corner of the inset square.

a

a+1

a+10

a+11

        (a+1)×(a+10)-a×(a+11)Here I have multiplied the opposite corners of the grid

        [a²+11a+10]-[a²+11a] Here I have multiplied out the brackets and simplified the rule

a²+11a+10    -  

a²+11a _        Here I have subtracted the two sections to prove my overall rule.

0 +  0 +10

            =10

...read more.

Middle

10×1²=10

The number in the 2×2 squares was 10 so this proves my formula correct.

To extend my work I will create a different sized number grid and calculate a formula that will work for the difference in any sized square in any sized 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

As in my original grid I have marked out individual squares and will now calculate the formula for them

2×2

1×9=9

2×8=16

16-9=7

6×14=84

7×13=91

91-84=7

25×33=825

26×32=832

832-825=7

36×44=1584

37×43=1591

1591-1584=7

41×49=2009

42×48=2016

2016-2009=7

a

a+1

a+7

a+8

        a×(a+8)-(a+1)×(a+7)

        [a²+8a+7]-[a²+8a]

a²+8a+7  -

a²+8a

0 +0 +7

           =7

3×3

1×17=17

3×15=45

45-17=28

10×26=260

12×24=288

288-260=28

29×45=1305

31×43=1333

1333-1305=28

33×49=1617

35×47=1645

1645-1617=28

5×21=105

7×19=133

133-105=28

a

a+1

a+2

a+7

a+8

a+9

a+14

a+15

a+16

        a×(a+16)-(a+2) ×(a+14)

        [a²+16a+28]-[a²+16a]

a²+16a+28   -

a²+16a

0 + 0+28

         =28

4×4

1×25=25

4×22=88

88-25=63

22×46=1012

25×43=1075

1075-1012=63

25×49=1225

28×46=1288

1288-1225=63

a

a+1

a+2

a+3

a+7

a+8

a+9

a+10

a+14

a+15

a+16

a+17

a+21

a+22

a+23

a+24

a×(a+24)-(a+3)×(a+21)

[a²+24a+63]-[a²+24a]

a2+24a+63 -

a²+24a

0+0+63

        =63

Before creating an overall formula I will create a formula for my 7×7 grid. This will allow me to create my overall formula:

a×(a+8(n-1)) – ((a+(n-1)×(a+7(n-1)))

This cancels down to

(a²+8a(n-1))-(a²+8a(n-1)+7(n-1)²)

To work out my overall formula I will subtract the first part of the Formula from the other:

...read more.

Conclusion

a2+aB(A-1)+a(Z-1)+(Z-1)B(A-1)

a2+aB(A-1)+a(Z-1) –

0 +     0    +    0   +(Z-1)B(A-1)

=B(Z-1)(A-1)

=15(Z-1)(A-1)

Looking at this formula and my previous work with squared I can see that the general formula for the difference could be:

B(Z-1)(A-1)

I will now prove this by creating a 7×5 grid and using the same process before, make up a general rule for any sized rectangle in any sized 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

2×3

1×16=16

2×15=30

30-16=14

18×33=594

19×32=608

608-594=14

a

a+1

a+2

a+7

a+8

a+9

a×(a+9) – (a+2)×(a+7)

a2+9a – a2+9a+14

a2+9a+14

a2+9a –

0+0+14

=14

6×2

15×27=405

22×20=440

440-405=35

22×34=748

29×27=783

783-748=35

a

a+1

a+2

a+3

a+4

a+5

a+7

a+8

a+9

a+10

a+11

a+12

a×(a+12) – (a+5)×(a+7)

a2+12a – a2+12a+35

a2+12a+35

a2+12a –

0+0+35

=14

I will now create my overall formula for any sized rectangle in any sized grid.

The letters represent the same as previously.

a×(a+B(A-1)+(Z-1)) – a+(Z-1)) × (a+B(A-1))

a2+aB(A-1)×a(Z-1)+(Z-1)B(A-1)

a2+aB(A-1)×+a(Z-1) –

0 +     0    +    0   +(Z-1)B(A-1)

=B(Z-1)(A-1)

Therefore I conclude that my overall formula for the difference is B(Z-1)(A-1). I will check this by inserting numbers into my formula.

B=7

Z=6

A=2

B(Z-1)(A-1)

7×(6-1)(2-1)

=7×(5×1)

=7×5=35

This proves my formula for the difference correct as in the 6×2 grids the difference was always 35.

...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. Investigation of diagonal difference.

    A vertically aligned cutout containing only n expressions reads - n n + 1 n + 30 n + 31 And from analysing this cutout we can see a clear relation between g and the number added to n in the bottom two corners.

  2. Algebra Investigation - Grid Square and Cube Relationships

    This will remain constant. w: The width of the cube in question. h: The height of the cube in question. d: The depth of the cube in question. g: The overall gridsize contained in the cube. s: The increment (step)

  1. Maths - number grid

    My aim still remaining the same, to identify a formula. By drawing up the table above formed from my results in Chapter Four I was able to see a trend of multiples of 12 and was successful in establishing a formula, which was: 12(s - 1)(r - 1)

  2. Number Stairs

    am going to the 4 step stair investigation on the 9x9 grid. 28 19 20 10 11 12 1 2 3 4 Stair number (N) =1, stair total=1+2+3+4+10+11+12+19+20+28= 110 Stair number (n) = 2 Stair Total = 2+3+4+5+11+12+13+20+21+29= 120. Stair number = 3 Stair total = 3+4+5+6+12+13+14+21+22+30= 130 Stair number=4

  1. Investigate The Answer When The Products Of Opposite Corners on Number Grids Are Subtracted.

    If I take a number grid, 3 x 3 for example, and replace the numbers in the grid with algebraic expressions, then I may be able to work out a formula. I will assume the starting number is always one, for this case.

  2. Maths Grids Totals

    A 3 x 3 square would be 9(n-1)2 = 9(3-1)2 = 9 x 4 = 36. A 4 x 4 square would be 9(n-1)2 = 9(4-1)2 = 9 x 9 = 81. A 5 x 5 square would be 9(n-1)2 = 9(5-1)2 = 9 x 16 = 144.

  1. Mathematical Coursework: 3-step stairs

    89 90 91 92 93 94 95 96 97 98 99 78 79 80 81 82 83 84 85 86 87 88 67 68 69 70 71 72 73 74 75 76 77 56 57 58 59 60 61 62 63 64 65 66 45 46 47 48 49 50

  2. My coursework task is to investigate why, in a number grid square of 1-100, ...

    We can once again show why the result is always 90 by using algebra as formulas. (P+30) (P+3) - P (P+33) Now we can put the formula to the test by using it with numbers: (5x32)- (2x35) =90 = As we can see the formula has proven to work well

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