# T-Total Investigation

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Introduction

Daniel Hughes 4C T-Total Investigation 28/04/07

T-Total Investigation

I am going to look at and investigate the relationship between the T-total, R-number and grid size of a T-shape on a numbered grid.

- Investigate the relationship between the T-total and the T-number drawn on a 9 by 9 number 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 |

There is a possibility of 196 different translations of T-shape on this grid. I am going to investigate on this grid the T-shape shown above.

Here is a table with T-totals from the grid above:

T-number | T-total |

20 | (20+11+1+2+3) 37 |

21 | (21+12+2+3+4) 42 |

22 | (22+13+3+4+5) 47 |

23 | (23+14+4+5+6) 52 |

24 | (24+15+5+6+7) 57 |

To find a relationship between the T-total and T-number I used the difference method.

## T-TotalT-numberDifference

20 37 5

21 42 5

22 47 5

23 52 5

24 57

T = T-total

The reason why the difference increases by 5 each time, is because there is 5 numbers in the T. As the T moves across the each number increases by one. Five times each one equals 5.

T = 5n – X

= (5 x 20) - X = 100 – X = 37

= 63

So, 5n – 63 is the formula to work out the t-total for this grid.

Eg. Formula = 5n – 63

= (5x24) - 63

= 120 – 63

= 57

24 + 15 + 5 + 6 + 7 = 57

Formula = 5n – 63

= (5x50) – 63

= 250 – 63

= 187

50 + 41 + 31 + 32 + 33 = 187

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 |

The T though can be translated to different positions.

Eg.

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 |

In different positions though the formula does not work.

Eg. = 5n – 63

= (5x4) – 63

= 20 – 63

= -43

Middle

= 285 + 7

= 292

57 + 58 + 59 + 50 + 68 = 292

I predict to work out the reverse of the sideways T in the opposite direction, you have to change the formula to 7 from –7.

Eg. Formula = 5n – 7

= (5x12) – 7

= 60 – 7

= 53

12 + 11 + 10 + 1 + 19 = 53

Formula = 5n –7

= (5x16) –7

= 80 – 7

= 73

16 + 15 + 14 + 5 + 23 = 73

Formula = 5n – 7

= (5x68) – 7

= 340 – 7

= 333

68 + 67 + 66 + 57 + 75 = 333

- Uses grids of different sizes. Translate the T– shape to different positions. Investigate relationships between the T-total, the T-numbers and the grid size.

Here is a grid 3 by 3:

1 | 2 | 3 |

4 | 5 | 6 |

7 | 8 | 9 |

I am going to concentrate on the original translation of the T. The T-total of 8 in this grid is 19. Adding another column (to make it 3 by 4) will give me two T’s to do the difference method with.

1 | 2 | 3 | 4 |

5 | 6 | 7 | 8 |

9 | 10 | 11 | 12 |

T-numberT-total Difference

10 22 5n

11 27 5n

5n = (5x10)

= 50 –22 = 28

= 5n – 28

Eg. Formula = 5n – 28

= (5x11) – 28

= 55 – 28

= 27

1 | 2 | 3 | 4 | 5 |

6 | 7 | 8 | 9 | 10 |

11 | 12 | 13 | 14 | 15 |

T-numberT-total Difference

12 25 5n

13 30 5n

14 35 5n

The difference between the T-totals is always 5n.

5n – x = (5x12) – x

= 60 – x

25 = 60 – x

x = 35

= 5n – 35

Eg. Formula = 5n –35

= (5x14) – 35

= 70 – 35

= 35

1 | 2 | 3 | 4 | 5 | 6 |

7 | 8 | 9 | 10 | 11 | 12 |

13 | 14 | 15 | 16 | 17 | 18 |

T-numberT-totalDifference

14 28 5n

15 33 5n

16 38 5n

17 43 5n

It is 5n again.

5n – X = (14x5) –X

= 70 – X

28 = 70 – X

X = 42

= 5n – 42

Eg. Formula = 5n – 42

= (5x17) – 42

= 85 – 42

= 43

I noticed a pattern in the formulas.

3x4 5n – 28 7

3x5 5n – 35 7

3x6 5n - 42 7

The difference between the number you take away increases by 7 every time.

Conclusion

22 – 4 = 18x10 = 180

13 – 4 = 9x10 = 90

What I was trying to do was to develop a formula from the first formula with 3 rows and T-number 4. For example the number 90, + 33 = 123, which is the number taken away for the formula for T-number 13. 180 added to 33 is 213, which is the number taken away from 15n, for the T-number 22. This has to be expressed into a formula, so here is.

15n – (10(n - 4) + 33)

Formula = 15n – (10(n - 4) + 33)

= (15x7) – (10(7 – 4) + 33)

= 105 – (10x3 + 33)

= 105 (30 + 33)

= 105 – 63

= 42

7 + 8 + 9 + 6 _+ 12 = 42

Formula = 15n – (10(n - 4) + 33)

= (15x13) – (10(13-4) + 33)

= 195 – (10x9 + 33)

= 195 – 123

= 72

13 + 14 + 15 + 12 + 18 = 72

Formula = 15n – (10(n - 4) + 33)

= (15x22) – (10(22-4) + 33)

= 330 – (10x18 + 33)

= 330 – (180 + 33)

= 330 – 213

= 117

22 + 23 + 24 + 21 + 17 = 117

As the formula doesn’t change when a column is added, this is the general formula for this sideways T.

Eg.

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 |

Formula = 15n – (10(n – 4) + 33)

= (15x7) – (10(7 - 4) + 33)

= 105 – (10x3 + 33)

= 105 – (30 + 33)

= 105 – 63

= 42

7 + 8 + 9 + 3 + 15 = 42

Formula = 15n – (10(n – 4) + 33)

= (15x16) – (10(16 - 4) + 33)

= 240 – (10x12 + 33)

= 240 –153

= 87

16 + 17 + 18 + 12 + 24 = 87

Formula = 15n – (10(n – 4) + 33)

= (15x27) – (10(27 - 4) + 33)

= 405 – (10x23 + 33)

= 405 – ( 230 + 33)

= 405 – 263

= 142

27 + 28 + 29 + 23 + 35 = 142

I therefore predict if you reverse the T round the other way that the formula will be, 15n + (n + 4)-10 + 33).

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 |

Formula = 15n + (n + 4)-10 + 33)

= (15x9) + (9 + 4)-10 + 33)

= 135 + (13x-10 + 33)

= 135 + (-130 + 33)

= 135 – 97

= 38

9 + 8 + 7 + 1 + 13 = 38

Formula = 15n + (n + 4)-10 + 33)

= (15x30) + (30 + 4)-10 + 33)

= 450 + (34x-10 + 33)

= 450 + (-340 + 33)

= 450 – 373

= 72

30 + 29 + 28 + 22 + 34 = 143

Formula = 15n + (n + 4)-10 + 33)

= (15x22) + (22 + 4)-10 + 33)

= 330 + (26x-10 + 33)

= 330 + (-260 + 33)

= 330 – 227

= 103

22 + 21 +14 + 20 + 26 = 103

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

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