# Maths Coursework

Extracts from this document...

Introduction

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:

57 | 58 |

67 | 68 |

(58x67)-(57x68) = 10

3:

88 | 89 |

98 | 99 |

(89x98)

Middle

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 = 60

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 = 80

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 |

Conclusion

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 = 81

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 = 54

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 = 108

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)

=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

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