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• Level: GCSE
• Subject: Maths
• Word count: 1729

# Math Grid work

Extracts from this document...

Introduction

Number Grids Coursework – Maths – Mr Danes

## Introduction

We are looking at number grids and are using the numbers in the edges of any square or rectangle. I am multiplying the opposite edged numbers and subtracting the smaller number from the bigger one.

## This is a 10 x 10 square

 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 pick any square from this grid and do the multiplying and subtracting and see what I get. I multiply the top right hand number to the bottom left and number and then subtract the number from the other two numbers multiplied from it.

2 x 2 square

 24 25 34 35
 46 47 56 57

 89 90 99 100

I notice that I get the same number for any same size square so to prove this I will use X and prove why the answer is always 10. I can use a square in which the boxes are the edges of the square on the grid.

 X X + 1 X + 10 X + 11

If I multiply this out if I were doing what I was doing with the numbers in the grid I would,

(X+1)(X+10)…which is = X2+11X+10…and…

X(X+11)…which is = X2 +11X

[X2+11X+10]- [X2 +11X] = 10

X can be any number and will always equal 10!

3 x 3 square

 36 37 38 46 47 48 56 57 58

X

## X + 2

X + 20

X + 22

If I multiply this out if I were doing what I was doing with the numbers in the grid I would,

(X+2)(X+20)…which is = X2+22X+40…and…

X(X+22)…which is = X2 +22X

[X2+22X+40]-[X2 +22X] = 40

X can be any number and will always equal 40!

4 x 4 square

 4 5 6 7 14 15 16 17 24 25 26 27 34 35 36 37

 X X + 3 X + 30 X + 33

Middle

54

55

61

62

63

64

65

71

72

73

74

75

81

82

83

84

85

91

92

93

94

95

 X X + 4 X + 40 X + 44

If I multiply this out if I were doing what I was doing with the numbers in the grid I would,

(X+4)(X+40)…which is = X2+44X+160…and…

X(X+44)…which is = X2 +44X

[X2+44X+160]-[X2 +44X] = 160

X can be any number and will always equal 160!

### Working out a formula for any square size on a 10 by 10 number grid

The sequence I get is 10, 40, 90 and then 160.

So I can do the use a square to find out a formula that works out N for any square size on a 10 by 10 number grid.

I do what I do with these figures if they were numbers:

1. [X+(S-1)][X+10(S-1)

I get X2+11X(S-1)+10(S-1)2

And…

2. X[X+11(S-1)]

I get X2+ 11X(S-1)

I then subtract 2 from 1…

[X2+11X(S-1)+10(S-1) 2] -  [X2+ 11X(S-1)]

And whatever that is left over must be the formula, so therefore the formula to work out N for any square size in a 10 by 10 grid square is…

N= 10(S-1)2

##### Test

When the square size is 2…

 89 90 99 100

Using the formula, (2-1) 2 = 1.. x 10 = 10

###### When the square size is 5…
 2 6 42 46

Using the formula, (5-1) 2 = 16… x 10 = 160

The formula works!

### What happens if I change the grid size?

I think that I have come to the stage where I can use algebra to work out what the number each time will be. So, I don’t have to show examples with numbers all the time.

## A 6 by 6 grid square on a 2x2 square

 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

If I multiply this out if I were doing what I was doing with the numbers in the grid I would,

(X+1)(X+6)…which is = X2+7X+6…and…

X(X+7)…which is = X2 +7X

## An 8 by 8 grid square on a 2x2 square

 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 51 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

Conclusion

N= g(S-1)2

##### Test

When the square size is 2 and the grid size is 10

 89 90 99 100

Using the formula, (2-1) 2 = 1.. x 10 = 10

###### When the square size is 3 and the grid size is 6
 2 4 14 16

Using the formula, (3-1) 2 = 4… x 6 = 24

The formula works!

#### Final formula: for any size square/rectangle in any grid square

Common sense tells me that, this is basically replacing the S by the length and width of any square/rectangle.

I do what I do with these figures if they were numbers:

1. [X+(w-1)][X+g(L-1)]

I get X2+gX(L-1)+X(w-1)+g(w-1)(L-1)

And…

2. X[X+g(L-1)+(w-1)]

I get X2+ gX(L-1)+X(w-1)

I then subtract 2 from 1…

[X2+gX(L-1)+X(w-1)+g(w-1)(L-1)]  - [X2+ gX(L-1)+X(w-1)]

The only thing left over is g(w-1)(L-1)

And whatever that is left over must be the formula, so therefore the formula to work out N for any square/rectangle in a g by g grid square is…

N= g(w-1)(L-1)

##### Test

When the grid size is 10 and the length and width are 2

 89 90 99 100

Using the formula, 10(2-1)(2-1) = 10x1x1= 10

###### When the grid size is 5, length is 2 and width is 4
 2 5 7 10

Using the formula, 5(4-1)(2-1) = 5x3x1 = 15

The formula works!

***

By: Haroon Motara

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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