# Task Two T-Totals.

Extracts from this document...

Introduction

Task Description

Look at this shape drawn on a 9 x 9 number grid. The total of the numbers in the T-Shape is 1+2+3+11+20 =37. This is called the T-Total.

The number at the bottom of the T-shape is called the T-Number.

The T-number for this shape is 20.

Investigate

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 | 42 | 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 task that I have been set is to investigate the relationships between the T-Totals, as outlined in the task on page one.

In order to achieve the task I will begin with a 9 x 9 square and move on to different number grids to see if there is a relationship between any of the grids.

To demonstrate how this works, I will draw numerous grids showing the T-Totals and how they have been solved.

My predictions that I aim to prove are that the T-Totals have an Algebraic Rule to link the two amounts together.

An example of which is below;

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 | 42 | 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 |

As you can see, on this gird the T-Total has been highlighted in orange. In order to find the T-Total, you must add all of the numbers inside the T- shape together, which will look like this; 1+2+3+11+20 =37.

The T-number is the number at the bottom of the T-shape, so for the shape it is 20.

Middle

49

50

51

42

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-Number is 21 & the T-Total is 42.

As you can see from the first two grids, the T-Number moves a square along, the T-Total goes up by five. This means that the ratio between the T-Number & the T-Total is 1:5.

This is useful for when you are changing the position of the T-Shape. For example when you move it up 5 places to 6, it will look like this;

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 | 42 | 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-Number is 25 & the T-Total is 54. Another way to work out the numbers in green, besides adding them together is to work out the difference between the T-Numbers, which for this example is 54.

If you then multiply the 54 by 5, as the number rises by 5 every time the T-Number goes up. Then add the T-Total from the original shape to get the new T-Total number for the green shape.

This relationship between the T-Totals & T-Numbers can be formed together to make an algebraic formula.

5tn – 63 = T-Total. (tn = T-Number)

As the formula starts with 5 x the T-Number, because I have found that there is a rise of each time in the grid for every T-Number. Then I subtracted the 63, which I got by working out the difference between the T-Number & another number within the T-Shape.

This has to be done to all four numbers in the T-Shape in order for the formula to work correctly. An example of which is below in yellow.

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 | 42 | 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 yellow T-Shape has a T-Number of 32. My workings out for the T-Number & the rest of the numbers in this T-Shape are as follows;

32-13=19, 32-14=18, 32-15=17, 32-23=9 Total=63

To prove that my findings are correct, I have drawn out another T-Shape in lilac on the above grid and followed the same rule as above. My workings for this second theory are below;

70-51=19, 70-52=18, 70-53=17, 70-61=9 Total=63

As you can see from both sets of results the final number is 63. This is also where the 63 in the equation came from. Another place where the 63 would originate from is the size of the grid, 7x9 = 63. However, if the grid size was 6 x 6, then it would be 6 x 9.

When all of the formulae are put together, the numbers that I used to plus or minus by are actually divisible by 7.

Therefore when all of the formulae are added together you get the following formula; 5tn-63=T-Total.

To prove that this formula has worked, I have provided an example below;

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 | 42 | 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 |

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 | 42 | 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 |

5tn + 63 = T-Total 5 x 2 +63 = 73

Check that the formula has worked by changing the signs;

T-Number = 2

T-Total= 2 + 11 + 19 +20 + 21 = 73

This example shows that I was correct to change the formulas around slightly.

By moving the T-Shape 180 degrees will not entirely prove that the theory is correct, so I will have to move it again to another position.

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 | 42 | 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 |

Here you can see that I have rotated the T- Shape around on its side. The formula for this shape will be similar to the one above by using the minus to calculate the differences in the T-Number to the other numbers in the T-Shape.

Calculation:

12 – 1 = 11, 12 – 10 = 2, 12- 19 = 7, 12 – 11 = 1 Total = 7

Formula:

5tn – 7 = T-Total 5 x 12 – 7 = 53

Check to see the formula works:

T- Number = 12

T-Total = 1 + 10 + 19 + 11 + 12 = 53

From this investigation of T-Totals, I conclude that there is a relationship between the T-Totals & the T-Numbers.

However, it all depends on where the shapes are placed & at what degree as the formulas have to be changed accordingly.

To investigate the relationships even further I could have perhaps tried them on a diagonal line or even increased the size of the T-Shape to see if the same rules applied here too.

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

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month