# GCSE module 5 AQA Mathematics

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Introduction

Ginny Faulkner maths coursework 2008 year 11

I was asked to investigate drawing boxes on a 10 x 10 grid and multiply the top left number with the bottom right and vice versa and find the difference. I will try to obtain a formula to make any sized box from one number. I will then explore what happens when the size of the grid is increased or enlarged.

Size of box | Result |

2 x 2 | 10 |

2 x 3 | 20 |

2 x 4 | 30 |

2 x 5 | 40 |

2 x 6 | 50 |

2 x 7 | 60 |

The pattern is that for every extra row that is added, another 10 is added to the result. I can prove this by drawing boxes from the grid.

I decided that I would work out by trying to make an equation to find out the nth term.

I will at first work out the equation for a 2 x 2

Middle

28(28+11)

28 x 39 = 1092

(28+10)(28+1)

38 x 29 = 1102

1102-1092= 10

Simplify.

(n +10)(n +1)= y

I have switched the two separate

Parts of the equation so that y will come

first in the final calculation which will

make the subtraction easier.

n (n+11)= x

y – X = 10

y= n + 10(n+1) – x= n (n+11)

test: n = 13

y = 13+10 (13+1) – x= 13(13+11)

y = 23 x 14 – x = 13 x 24

y = 322 – x = 312

322 – 312 = 10

The formula to work out the product of a 2 x 2 box is:

y= n + 10(n+1) – x= n (n+11)

3 x 2 boxes.

I will now use a similar method to prove that a 2 x 3 box will always produce 20.

14 | 15 |

24 | 25 |

34 | 35 |

The numbers highlighted in yellow are the numbers that I need to use.

14 x 35 = 490

34 x 15 = 510

510 – 490 =20

(This box is a different way round so I will expand my boxes across instead of down to make an equation although both = 20)

45 | 46 | 47 |

55 | 56 | 57 |

45 x 57 = 2565

55 x 47 = 2585

2585 – 2565 = 20

This proves that any randomly selected number is likely to produce 20.

Conclusion

I will call the size of the box B. for example 2 x B will be the number that one wishes to use to multiply with 2 to make the box size.

The part of the equation which needs to change is:

n+3/2/1 and n+13/12/11

I have noticed that this number will always be one less than the number that is ‘xing’ by 2 to create the box.

So, the part of the formula could be:

n+(B-1) and the other part could be n+ (B+9)

therefore the formula would be:

y= n+10(n+(B-1)) – x = n (n+(B+9))

So assuming n was 7 and B was 3 the equation would be:

y= 7+10(7+ (3-1)) – x = 7 (7 + (3+9))

y= 17 (7+2) – x = 7 (7+12)

y= 17 x 9 – x =7 x 19

y= 153 – x= 133

153- 133 = 20

This proves that my formula correctly shows that one can work out the total of a box of whichever size from just one equation.

7 | 8 | 9 |

17 | 18 | 19 |

7 x 19 = 133

17 x 9 = 153

153 – 133= 20.

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