# 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

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:

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.

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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