# Investigate the relationship between the T-total and the T-number

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Introduction

## GCSE Maths Investigation – T shapes

Part 1 – Investigate the relationship between the T-total and the T-number

When the “T-total” equals 37, the T-number is 20. I will investigate the other T-totals to see if there is a relationship.

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 |

T-total | T-number |

1 + 2 + 3 + 11 +20 = 37 | 20 |

2 + 3 + 4 + 12 + 21 = 42 | 21 |

3 + 4 + 5 + 13 + 22 = 47 | 22 |

4 + 5 + 6 + 14 + 23 = 52 | 23 |

5 + 6 + 7 + 15 + 24 = 57 | 24 |

6 + 7 + 8 + 16 + 25 = 62 | 25 |

7 + 8 + 9 + 17 + 26 = 67 | 26 |

The difference between the T-totals is:

37 42 47 52 57 62 67

+5 +5 +5 +5 +5 +5

I will now see if there is any relationship between the T-total and the T-number. In order to see this I will subtract the T-number from the T-total.

T-number – T-total = Difference

42 – 21 = 21

47 – 22 = 25

52 – 23 = 29

57 – 24 = 33

62 – 25 = 37

I will put this information into a formula.

Here; n = the T number

T = T-total

So if n = 20

20 + (20 – 19) + (20 – 18) + (20 – 17) + (20 – 9) = T

1 2 3 11

Therefore:

n + (n – 19) + (n – 18) + (n – 17) + (n – 9) = T

To simplify: 5n – 63 = T

### T = 5n - 63

I will use one of the T-shapes on the 9 by 9 grid to see if this formula works.

Eg1: 20 x 5 – 63 = 37

This is correct

Middle

94

95

96

97

98

99

100

T-total | T-number |

1 + 2 + 3 + 12 + 22 = 40 | 22 |

2 + 3 + 4 + 13 + 23 = 45 | 23 |

3 + 4 + 5 + 14 + 24 = 50 | 24 |

4 + 5 + 6 + 15 + 25 = 55 | 25 |

5 + 6 + 7 + 16 + 26 = 60 | 26 |

6 + 7 + 8 + 17 + 27 = 65 | 27 |

7 + 8 + 9 + 18 + 28 = 70 | 28 |

The difference between the T-totals is:

40 45 50 55 60 65 70

+5 +5 +5 +5 +5 +5

I am going to see if the formula that I calculated earlier works for a 10 by 10 grid.

Therefore: T = n + (n-21) + (n – 20) + (n – 19) + (n – 10)

## To simplify: T = 5n – 70

This formula is similar except the number that is being subtracted. However I will see if this works with a T-shape in the 10 by 10 grid.

T = 5 x 26 – 70

T = 60

This is true because 5 + 6 + 7 + 16 +26 = 60

The equation is correct.

However, in order to find the relationship between the grid size and the T-total and T-number it must be included in the equation.

So, when n = T-number

T = T-total

W = Grid size

## This simplifies to: 5n – 7w

If the grid decreases to 6 by 6, the T-numbers will be closer together. I noticed that on the previous 2 grids, the numbers being subtracted can be divided by the grid size and both equal –7.

I think that a 6 by 6 grid should be:

T = 5n – 42 (-7 x 6 = 42)

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 |

T-total | T-number |

1 + 2 + 3 + 8 + 14 = 28 | 14 |

2 + 3 + 4 + 9 + 15 = 33 | 15 |

3 + 4 + 5 + 10 + 16 = 38 | 16 |

4 + 5 + 6 + 11 + 17 = 43 | 17 |

Part 3: Use grids of different sizes again. Try other transformations and combinations of transformations. Investigate relationships between the T-total, the T-numbers, the grid size and the transformations.

REFLECTION

Horizontal reflection

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 |

38 , 43 39 , 44 40 , 45

+5 +5 + 5

Conclusion

210 + -20 - -90 – 63 = T

T = 210 – 20 + 90 – 63

T = 217

This is true because the T-total is 217 for this T-shape. I deliberately used a translation with negative numbers to check that this did not alter the equation.

COMBINING REFLECTION AND TRANSLATION

n = T-number

W = Grid size

Z = across translation

U – Up or down translation

N = new T-number

a = horizontal distance from reflection line

b = vertical distance from the reflection line

T = T-total

HORIZONTAL REFLECTION TO TRANSLATION:

5n – 7w + 10a + 5 + 5N + 5z – 5wu – 7w = T

The new T-number needed to be included in this equation otherwise it would not work, because the translation would occur from the original T-number.

This formula works because when tested on a 9 by 9 grid with the T-number being 21 and a being 2, the new t-number is 26 and when translated by 2 and -2 the T-total equals 147. This agrees with the equation.

TRANSLATION TO VERTICLE REFLECTION:

5n + 5z – 5wu – 7w + 5N + 10bw + 12w = T

The new T-number needed to be included in this equation otherwise it would not work, because the reflection would occur from the original T-number.

This formula works because when tested on a 9 by 9 grid with the T-number being 76 and the translation being +2 by +6, the new T-number is 24. When “b” is 1, the T-total equals 318. This agrees with the equation.

T-shape investigation.doc 5/4/2007 11:53 AM

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

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