# T-Totals. Aim: To find the relationship between the T-Total and the T-number. To find a formula that works for every grid size. To find how rotating the T-shape effects the T-Total To find a formula that works f

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Introduction

T-Totals

Aim:

- To find the relationship between the T-Total and the T-number.
- To find a formula that works for every grid size.
- To find how rotating the T-shape effects the T-Total
- To find a formula that works for every translation.

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 1=1+2+3+11+20

=37

T-Number=20

T-Total=2+3+4+12+21

=42

T-Number=21

## T-Total=3+4+5+13+22

=47

## T-Number=22

T-Total=4+5+6+14+23

=52

### T-Number=23

#### I have noticed that the difference off the T-totals is 5:

37 42 47 52 57

5 5 5 5

(This is because there are 5 numbers in the T-totals and they are going up 1 every time)

The formula must equal T=5n+C (T=T-Totals n=T-number C=Constant)

We know the T-totals and T-numbers so we need to find what the constant is, there is too ways of doing this by going back to the 0th formula or just imputing the other totals into the formula. I will use the second method to succeed in finding the Constant.

##### T-total 1=37

37=(5*20)+C C=37-5*20 C=37-100 C= -63

##### T-number=20

###### Therefore T=5n-63

I will try this formula on T-Number 50:

Using my formula, T-number=69 so

T=(5*69)-63

=345-63

=282

Now I need to show it is correct:

T=50+51+52+60+69

=282

My prediction was correct so my formula is right.

Middle

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I am going to use T-total 56 (which I have chosen by using the random button on my calculator) to show that the formula for grid size 10 is T=5n+70.

T-number=77

T=(77*5)-70

=315

By adding all the numbers in the T-total together we get:

T=56+57+58+67+77

=315

The formula has now been proven correct.

We now know that the formula for all grids is T=5n-7G where G is grid size.

We can show this by using Grid size 7:

n-2G-1 | n-2G | n-2G+1 | 4 | 5 | 6 | 7 |

8 | n-G | 10 | 11 | 12 | 13 | 14 |

15 | n | 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 |

We can right the other numbers in forms of n and G:

T=n+(n-G)+(n-2G)

+(n-2G-1)

+(n-2G+1)

T=5n-7G

This is an alternate proof that T=5n-7G. I have succeeded in my second aim.

Rotation

Leaving out Enlargement, rotation is one of the three transformations I am going to look at and how it affects the T-Total.

Grid size 9

n-7 | ||||

n | n+1 | n+2 | ||

n+11 | ||||

n | ||||

n+G | ||||

n+2G-1 | n+2G | n+2G+1 | ||

n-11 | ||||

n-2 | n-1 | n | ||

n+7 | ||||

2 | 3 | 4 | ||

10 | 12 | 14 | ||

19 | 20 | 21 | 22 | 23 |

28 | 30 | 32 | ||

38 | 39 | 40 |

I have noticed that when you rotate it 180 degrees the –7G turns into a +7G, which is totally relevant, however when u rotate it 90 degrees, there is no connection, the formula is T=5n+7, but this is relevant when u rotate it 270 degrees where T=5n-7. I believe this will be similar for all grid sizes; I will try it out with grid size 7.

Grid Size 7

n-5 | ||

n | n+1 | n+2 |

n+9 | ||

n | ||

n+G | ||

n+2G-1 | n+2G | n+2G+1 |

n-9 | ||

n-2 | n-1 | n |

n+5 |

Conclusion

Grid size 7

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 |

I have translated the shape (4,0). So x=4

T-number is 16.

T=(5*16)-7G+(5*4)

=51

T=5+6+7+13+20

=51

My formula works, but what if the translation is not y=0. This is what I will find out next.

Grid size 6

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 |

There is a pattern when moving the T-shape down one, you add on 30. I am pretty sure that this changes with the grid size, so I need to input G into the equation 30=5G. So T=5n-7G-5Gy.

I need to try this out with another Grid size.

Grid Size 7

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 |

I have translated the T-shape (0, -4).

T-Number=18

T=(5*18)-(7*7)-(5*7*-4)

=90-49+140

=181

T=31+32+33+39+46

=181

Now I have a formula for translating parallel to the y and x-axis, I will try to merge them and see if the formula works.

T=5n-7G-5Gy+5x

=5n-G(7+5y)+5x

So when you translate a shape by say, (5, -6), you would input this into the formula and come out with T=5n-G(7+(5*-6))+(5*5).

Let me prove this formula is correct:

Grid size 8

1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

9 | 10 | 11 | 12 | 23 | 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 |

I have translated the T-shape (4, -4)

T-number=18

T=5*18-8(7+5*-4)+5*4

=90+104+20

=214

T=37+38+39+46+54

=214

My formula was correct.

I have completed my last aim.

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

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