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• Level: GCSE
• Subject: Maths
• Word count: 5017

# Maths Grids Totals

Extracts from this document...

Introduction

Module 4 Coursework

I am going to investigate squares of different sizes and on different grids.

I am going to draw a square around numbers on a grid, and find the product of the top-left and bottom-right numbers, and the top-right and bottom-left numbers. I will then calculate the difference.

10 x 10 grids

2 x 2 squares

 12 13 22 23

12 x 23 = 276

13 x 22 = 286

286 – 276 = 10.

 55 56 65 66

55 x 66 = 3630

56 x 65 = 3640

3640 – 3630 = 10.

 25 26 35 36

25 x 36 = 900

26 x 35 = 910

910 – 900 = 10

I have found that that there is a difference of 10 on any 2 x 2 square, on a 10 x 10 grid.

3 x 3 squares

 36 37 38 46 47 48 56 57 58

36 x 58 = 2088

38 x 56 = 2128

2128 – 2088 = 40.

 77 78 79 87 88 89 97 98 99

77 x 99 = 7623

79 x 97 = 7663

7663 – 7623 = 40.

4 x 4 squares

 25 26 27 28 35 36 37 38 45 46 47 48 55 56 57 58

25 x 58 = 1450

28 x 55 = 1540

1540 – 1450 = 90.

5 x 5 squares

 32 33 34 35 36 42 43 44 45 46 52 53 54 55 56 62 63 64 65 66 72 73 74 75 76

32 x 76 = 2432

36 x 72 = 3592

3592 – 2432 = 160.

I have now found out the differences for 2x2, 3x3, 4x4 and 5x5 squares on a 10x10, which will be shown in the following table:

 Size Difference 2 x 2 10 3 x 3 40 4 x 4 90 5 x 5 160

All the differences are square numbers of the previous number multiplied by 10 (e.g. 3 x 3 = 10 x 22 = 10 x 4 = 40). This gives me the formula 10(n-1)2.

Using this formula, I predict that the difference for an 8 x 8 square will be 10(8-1)2 = 10 x 72 = 10 x 49 = 490.

I will take a random 8 x 8 square from the 10 x 10 grid:

 12 13 14 15 16 17 18 19 22 23 24 25 26 27 28 29 32 33 34 35 36 37 38 39 42 43 44 45 46 47 48 49 52 53 54 55 56 57 58 59 62 63 64 65 66 67 68 69 72 73 74 75 76 77 78 79 82 83 84 85 86 87 88 89

12 x 89 = 1068

19 x 82 = 1558

1558 – 1068 = 490.

My prediction was correct; therefore this proves that the formula 10(n-1)2 is correct.

 n n x x+ (n-1) x+ 10(n-1) x+ 11(n-1)

I am going to prove 10(n-1)2 works, algebraically. First, I am going to show an n x n square on a 10 x 10 grid (algebraically).

“x” is the number in the top-left corner. The number to the right of it is “x+ (n-1)” because it is “x” added to the length of the square (take away one).

Middle

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32 x 112 = 3584

24 x 120 = 2880

3584 – 2880 = 704. The formula was correct.

I am now going to prove the formula algebraically.

“x” is the number in the top-left corner. The number to the right of it is “x+ (n-1)” because it is “x” added to the length of the square (take away one). Because the square is on an 11 x 11 grid, the formula for the number underneath the square is “x” added to (11 multiplied by “the squares length”-1). The bottom-right number is “x+ 12(n-1)” because it is “x+ 11(n-1) added to another (n-1), so therefore this becomes “x+ 10(n-1).

 n n x x+ (n-1) x+ 11(n-1) x + 12(n-1)

[x + (n-1)][x + 11(n-1)]= x2 + 11x(n-1) + x(n-1) + 11(n-1)2.

x[x+12(n-1)] = x2 + 12x(n-1).

This becomes: x2 + 11x(n-1) + x(n-1) + 11(n-1)2x2 – 12x(n-1).

The x2 and –x2 cancel each other out, while the 11x(n-1) and x(n-1) add together to cancel out the -12x(n-1). This leaves 11(n-1)2, which is the overall formula for an 11 x 11 grid.

A 10 x 10 grid’s formula is 10(n-1)2, a 9 x 9 grid is 9(n-1)2 and an 11 x 11 grid is

11(n-1)2. The size of the grid corresponds directly to the first digit of the formula.

This means the formulae can be re-written as g(n-1)2 (this is the overall rule for any square on any grid).

I am now going to predict differences for squares on a 14 x 14 grid, using the formula g(n-1)2. First, I am going to predict a 12 x 12 square on a 14 x 14 grid. The formula, with the numbers in place, is 14(12-1)2 = 14 X 112 = 14 X 121 = 1694.

I chose the following square:

 29 30 31 32 33 34 35 36 37 38 39 40 43 44 45 46 47 48 49 50 51 52 53 54 57 58 59 60 61 62 63 64 65 66 67 68 71 72 73 74 75 76 77 78 79 80 81 82 85 86 87 88 89 90 91 92 93 94 95 96 99 100 101 102 103 104 105 106 107 108 109 110 113 114 115 116 117 118 119 120 121 122 123 124 127 128 129 130 131 132 133 134 135 136 137 138 141 142 143 144 145 146 147 148 149 150 151 152 155 156 157 158 159 160 161 162 163 164 165 166 169 170 171 172 173 174 175 176 177 178 179 180 183 184 185 186 187 188 189 190 191 192 193 194

40 x 183 = 7320.

29 x 194 = 5626

7320 – 5626 = 1694.

The formula worked. Just to make sure it works universally, I will choose another square, however this time it will be 7 x 7.

A 7 x 7 square on the grid will have the formula 14(7-1)2 = 14 X 62 = 14 X 36 = 504. I chose the following square:

 64 65 66 67 68 69 70 78 79 80 81 82 83 84 92 93 94 95 96 97 98 106 107 108 109 110 111 112 120 121 122 123 124 125 126 134 135 136 137 138 139 140 148 149 150 151 152 153 154

70 X 148 = 10360

64 X 154 = 9856

10360 – 9856 = 504.

I have found that the formula g(n-1)2 works, and I will now prove it algebraically.

 n n x x+(n-1) x+ g(n-1) x+ g(n-1) + (n-1)

[x + (n-1)][x + g(n-1)] = x2 + xg(n-1) + x(n-1) + g(n-1)2

x[x + g(n-1) + (n-1)] = x2 + xg(n-1) + x(n-1).

This becomes x2 + xg(n-1) + x(n-1) + g(n-1)2 – x2 – xg(n-1) – x(n-1).

After cancelling out, only g(n-1)2 is left, which proves that it is the overall formula, not only for squares of different sizes, but also for different sized grids.

Now that I have found the rule for squares, I am going to try and find a rule for rectangles. To start off, I am going to use rectangles on a 10 x 10 gird.

I am going to use the smallest possible rectangles to start off with (2 x 3 rectangles):

 11 12 13 21 22 23

13 x 21 = 273

11 x 23 = 253

273 – 253 = 20.

 66 67 68 76 77 78

68 x 76 = 5168

66 x 78 = 5148

5168 – 5148 = 20.

 33 34 43 44 53 54

34 x 53 = 1802

33 x 54 = 1782

1802 – 1782 = 20.

 7 8 17 18 27 28

8 x 27 = 216

7 x 28 = 196

216 – 196 = 20.

All 2 x 3 rectangles have a difference of 20, regardless of whether they are bigger horizontally or vertically.

I will now get the differences for other sized rectangles:

3 x 4

 22 23 24 25 32 33 34 35 42 43 44 45

25 x 42 = 1050

22 x 45 = 990

1050 – 990 = 60.

3 x 4

 58 59 60 68 69 70 78 79 80 88 89 90

60 x 88 = 5280

58 x 90 = 5220

5280 – 5220 = 60.

4 x 5

 2 3 4 5 12 13 14 15 22 23 24 25 32 33 34 35 42 43 44 45

5 x 42 = 210

2 x 45 = 90

210 – 90 = 120.

4 x 5

 66 67 68 69 70 76 77 78 79 80 86 87 88 89 90 96 97 98 99 100

70 x 96 = 6720

66 x 100 = 6600

6720 – 6600 = 120.
5 x 6

 51 52 53 54 55 56 61 62 63 64 65 66 71 72 73 74 75 76 81 82 83 84 85 86 91 92 93 94 95 96

56 x 91 = 5096

51 x 96 = 4896

5096 – 4896 = 200.

5 x 6

 16 17 18 19 20 26 27 28 29 30 36 37 38 39 40 46 47 48 49 50 56 57 58 59 60 66 67 68 69 70

20x 66 = 1320

16 x 70 = 1120

1320 – 1120 = 200.

6 x 7

 34 35 36 37 38 39 44 45 46 47 48 49 54 55 56 57 58 59 64 65 66 67 68 69 74 75 76 77 78 79 84 85 86 87 88 89 94 95 96 97 98 99

39 x 94 = 3666

34 x 99 = 3366

3666 – 3366 = 300.

After trying out rectangles of 2 x 3, 3 x 4, 4 x 5, 5 x 6 and 6 x 7 size, I assembled the differences and put it into a table:

 Size Difference 2 x 3 20 3 x 4 60 4 x 5 120 5 x 6 200 6 x 7 300

The formula for squares was g(n-1)2 which is equal to g(n-1)(n-1). The g represented the grid size, and the n represented the width and height of the square. It could be re-written as g(h-1)(w-1), with h representing the height and w representing the width. This means the overall rule could be g(h-1)(w-1). I am going to test it in several ways. First, I am going to try it with rectangles I have already investigated:

For a 2 x 3 rectangle on a 10 x 10 grid, the rule would be 10(2-1)(3-1) which is 10 x 1 x 2 = 10 x 2 = 20. This is correct: I will now try the rule with a bigger rectangle.

A 5 x 6 rectangle is 10(5-1)(6-1), which become 10 x 4 x 5, which is 40 x 5 = 200. This is also correct. However, there is a possibility that this rule is not universal and that it only applies to rectangles with 1 square difference between the height and width.

I am now going to investigate the differences for other rectangles (i.e. rectangles with more than 1 square difference between the height and width (e.g. 2 x 4, 3 x 6 etc.)).

 23 24 25 26 33 34 35 36

26 x 33 = 858

23 x 36 = 828

858 – 828 = 30.

 1 2 11 12 21 22 31 32

2 x 31 = 62

1 x 32 = 32

62 – 32 = 30

 36 37 38 39 40 46 47 48 49 50 56 57 58 59 60

Conclusion

This leave g(h-1)(w-1), which is the universal rule for any square or rectangle on any size grid.

Matrices

A Matrix is a rectangular order of numbers (it can also be a square), i.e. in this case, it is the squares that are found on the grids (that are used to find the differences).
There are different things that can be done with matrices of different sizes. These include addition, subtraction, and multiplication.

Addition of a matrix is adding up the corresponding values of the matrices.

e.g. 2 matrixes on a 10 x 10 grid are:

 12 13 22 23 54 55 64 65

and

 54 + 12 55 + 13 22 + 64 23 + 65

The addition of these matrices would be:

…which in turn would become:

 66 68 86 88

This is basic addition for matrices.

Matrix addition can be applied to any square (or rectangle for that matter) of any size, as I am going to show with a 5 x 8 rectangle on an 11 x 11 grid.

 13 14 15 16 17 18 19 20 24 25 26 27 28 29 30 31 35 36 37 38 39 40 41 42 46 47 48 49 50 51 52 53 57 58 59 60 61 62 63 64 68 69 70 71 72 73 74 75 79 80 81 82 83 84 85 86 90 91 92 93 94 95 96 97 101 102 103 104 105 106 107 108 112 113 114 115 116 117 118 119

And

This becomes

 81 83 85 87 89 91 93 95 103 105 107 109 111 113 115 117 125 127 129 131 133 135 137 139 147 149 151 153 155 157 159 161 169 171 173 175 177 179 181 183

I am going to show how algebraically, Matrix addition works:

Let a = top-left number of first matrix.

Written algebraically, the first matrix can be written as

 a a+1 a+2 a+3 a+4 a+5 a+6 a+7 a+g a+g+1 a+g+2 a+g+3 a+g+4 a+g+5 a+g+6 a+g+7 a+2g a+2g+1 a+2g+2 a+2g+3 a+2g+4 a+2g+5 a+2g+6 a+2g+7 a+3g a+3g+1 a+3g+2 a+3g+3 a+3g+4 a+3g+5 a+3g+6 a+3g+7 a+4g a+4g+1 a+4g+2 a+4g+3 a+4g+4 a+4g+5 a+4g+6 a+4g+7

There is a “g” in the cells: “g” stands for the size of the grid on which the squares are.

The second matrix would be exactly the same as this one, but with “b” replacing every “a”.

--

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