# Investigating a T shape which will be on a 9x9 grid and have an area of 5 squares.

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Introduction

T-Total

Maths Coursework February 2004

Introduction

In this piece of coursework I will be investigating a T shape which will be on a 9x9 grid and have an area of 5 squares. The T-total is all the numbers in the T shape added up together and the T-number is the bottom number in the T highlighted in green below.

2 | 3 | 4 |

12 | ||

21 |

In this T there are five numbers 2, 3, 4, 12 and 21 if you add these numbers up it makes the T-total

2+3+4+12+21=42

The highest number in this T is 21 so 21 would be the T-number.

Once I have investigated that I will investigate whether or not using grids of different sizes would make a difference to the formulae and any relationship found. I will try and find a relationship between the T-Total, the T-Number and the grid size.

In the third part of this coursework I will use different transformations and combinations of transformations. I will investigate the relationship between the T-Total, the T-Number, the grid size and the transformations.

After that I drew a table of the first few T-Numbers and T-Totals for the first few T’s.

T20 | T21 | T22 | T23 | |

T-Total | 37 | 42 | 47 | 52 |

T-Number | 20 | 21 | 22 | 23 |

The first T is the T shown in the introduction, the second T is

2 | 3 | 4 |

12 | ||

21 |

And so on.

In the table there is a pattern, when the T-Number goes up one the T-Total goes up five, this pattern also works for T’s anywhere on the grid

This pattern occurs because each number inside the T goes up one as you move it across the grid and there are five numbers in the T. This pattern also works for every T on the grid.

T50 | T51 | T52 | T53 | |

T-Total | 187 | 192 | 197 | 202 |

T-Number | 50 | 51 | 52 | 53 |

The next thing I did was name each of the squares in relation to the T-Number

N-19 | N-18 | N-17 |

N-9 | ||

N |

The N stands for the T-Number and the other numbers are in relation to it and it works for every T. This helps because it makes it easier for me to work out a piece of algebra to solve the problem for every T.

In this T I have noticed that the first difference from N is 9 which is also the width of the grid.

I’ll put that idea into another T. Note W= Width Number (9)

N- (2W-1) | N-2W | N-(2w+1) |

N-W | ||

N |

Middle

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

I have highlighted the first T that can be made on this grid. The T-Number is 22. If the formula works for this T then the T-Total should be 40. The T-Total is 1+2+3+12+21= 40.

This formula works on that T but does it work on any otherT’s in the 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 | 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 |

This T’s T-Number is 69. If the formula works for this T the T-Total should be 275. The T-Total is 48+49+50+59+69= 275.

The formula seems to work for this size grid.

As there are 10 in each row it’s obvious that the row above will be 10 less than the row below. So 59 is 10 less than the T-Number 69. If you calculate the whole T you realise that row 2 is 10 less than row 1 and row 3 is 20 less than row 1, but there are three relevant numbers in row 3 which are 19 less, 20 less and 21 less that the T-Number. The 19 and the 21 can both be changed into 20’s because if you take the 1 from 21 and put it on the 19 they both make 20.

N-21 | N-20 | N-19 |

N-10 | ||

N |

Conclusion

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 | 101 | 102 | 103 | 104 | 105 | 106 | 107 | 108 |

109 | 110 | 111 | 112 | 113 | 114 | 115 | 116 | 117 | 118 | 119 | 120 |

121 | 122 | 123 | 124 | 125 | 126 | 127 | 128 | 129 | 130 | 131 | 132 |

133 | 134 | 135 | 136 | 137 | 138 | 139 | 140 | 141 | 142 | 143 | 144 |

This T will be called ST 64 as it starts at 64, it is a stretch of 2 in both directions.

26 | 27 | 28 | 29 | 30 |

40 | ||||

52 | ||||

64 |

ST 64=26+27+28+29+30+40+52+64= 296

I think I can work out the formula using my previous method so:

12+24+ (36x5) =216

Now if I see how many times 12 goes into 216

216/12 = 18

This means the formula is:

8N-18W= T-Total

8N= 8=number of integers in the T shape N= T-Number

18W= 18= Number calculated W=Width

Just to make sure this formula works here is another T.

ST 142 = 104+105+106+107+108+118+130+142= 920

Using the formula the T-Total would be (8x142) + (18x12) = 920

You would be able to use this way of finding out the formula for any T.

Conclusion

In conclusion I have found that the number of integers in the T calculates the number before the N in the formula and the width of the grid calculates the value of W. The number before the W is found by looking at the numbers in the rows and relating them to the T-Number. When the T is regular the W-Number is negative e.g.

-18W |

8N

Whereas if the T is flipped upside down the W-Number is positive.

This student written piece of work is one of many that can be found in our GCSE T-Total section.

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