# T-Totals.For my investigational piece of coursework I will be investigating the T-Totals of number grids.

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Introduction

Mathematics Coursework 2003

T-Totals

For my investigational piece of coursework I will be investigating the ‘T-Totals’ of

number grids.

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 |

Look at the ‘T-Shape’ drawn on a 9 by 9 number grid:

The total of the numbers inside of 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 T-Shape is 20.

Part 1: Investigating the relationship between the T-Total and T-Number:

To begin with I drew some more T-Shapes (positions chosen at random) onto the grid to allow me to find a formula for the T-Totals a lot quicker. The results are listed 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 | 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 |

Red shape: T-Number = 20 and T-Total = 37

Blue shape: T-Number = 23 and T-Total = 52

Dark blue shape: T-Number = 43 and T-Total = 152

Gold shape: T-Number = 66 and T-Total = 267

I noticed that every time you translate the T-Shape one place to the right the T-Total increases by 5. This is because there are 5 numbers in the shape and moving the shape one space to the right every time increases each value in the shape by one. One multiplied by the five values in the shape equals 5.

I then began looking at the differences between the numbers in the red shape. The differences between the T-Number and the other numbers in the shape worked out to look like a shape like this:

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

N-9 | ||

N |

Where N is the T-Number.

I then tested the theory on the other shapes. All of them used the same formula regardless of their position on the grid. For example the T-Shapes

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

N-9 | ||

N | ||

4 | 5 | 6 |

14 | ||

23 |

And

Middle

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

6 by 6 grid: T-Number is 29, and following the formula worked out in part 1:

(N-13) + (N-12) + (N-11) + (N-6) +N

= 5N-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 |

7 by 7 grid: T-Number is 30, and following the formula worked out in part 1:

(N-15) + (N-14) + (N-13) + (N-7) +N

= 5N-49

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 |

8 by 8 grid: T-Number is 23, and following the formula worked out in part 1:

(N-17)+ (N-16) + (N-15) + (N-8) +N

= 5N-56

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 |

9 by 9 grid: T-Number is 38, and following the formula worked out in part 1:

(N-19)+ (N-18) + (N-17) + (N-9) +N

= 5N-63

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 |

10 by 10 grid: T-Number is 38, and following the formula worked out in part 1:

(N-21)+ (N-20) + (N-19) + (N-10) +N

= 5N-70

I noticed that every time the grid size is increased by 1 the variable at the end of the formula becomes another minus seven to what it already was before, e.g. 5N-21, 5N-28, 5N-35…… and so on. The constant value is always 5N, simply because there are always 5 numbers in the T-Shape. I also noticed a pattern between the grid size and the value to be subtracted at the end of the equation:

Grid size value (multiplied by itself) | Relative value to be subtracted in equation | Similarity |

3 | 21 | Grid size X 7 |

4 | 28 | Grid size X 7 |

5 | 35 | Grid size X 7 |

6 | 42 | Grid size X 7 |

7 | 49 | Grid size X 7 |

8 | 56 | Grid size X 7 |

9 | 63 | Grid size X 7 |

10 | 70 | Grid size X 7 |

From this table I can work out a simple formula to find the T-Total of any sized grid. Using another T-Shape will help me prove the formula:

1 | 2 | 3 |

5 | ||

8 | ||

N-2G-1 | N-2G | N-2G+1 |

N-G | ||

N |

(Using a 3 by

3 grid as an

Example)

Where G is the grid size (in this example G = 3)

When worked out:

T-Total = (N-2G-1) + (N-2G) + (N-2G+1) + (N-G) + (N

= 5N-7G

Therefore the equation for finding the T-Total for any grid size is in the following formula:

I can now deduce that for a 1 by 1 size grid (although this is not possible as a 3 by 3 sized grid is necessary for a T-Shape) the equation would be 5N-7, for a 2 by 2 (again not possible to do) the equation would be 5N-14, for a 3 by 3 it would be 5N-21, and so forth up to huge numbers that would be very hard to work out using arithmetic – for example 231 by 231 would have a T-Total equation of 5N- (231*7) which works out as 5N-1617. Similarly a 45 by 45 grid size would work out as 5N- 315 and a 74 by 74 grid size would have an equation of 5N-518.

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.

Part 3a: Rotations

I began this part by rethinking my initial formula. If I am to test the T-Shape in different rotations of 0˚, 90˚, 180˚ and 270˚ then the formula ‘5N-7G’ must be true for 0˚ as this is the angle the formula was based upon.

I next drew a T-Shape rotated 90˚ in the right on a 6 by 6 grid (I will test other grid sizes later) to give the following diagram:

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 |

The T-Number of this shape is 14 and the T-Total is 77.

From my previous findings on a 6 by 6 number grid the formula for finding a T-Total was 5N-42. If I attempt to use this formula I get 5(14)-42 which equals 28, so this is obviously not the formula I can use for 90˚ rotations. However, I simply used the same method I used to find out non-rotated T-Total formulas as I did with the rotated formulas. I drew another T-Shape and the differences between the T-Number and the rest of the numbers:

N-4 | ||

N | N+1 | N+2 |

N+8 | ||

10 | ||

14 | 15 | 16 |

22 |

=

I then put these figures into another equation:

T-Total for 90˚ rotations on a 6 by 6 grid = (N) + (N+1) + (N-4) + (N+2) + (N+8)

= 5N + 7

I then tested this formula on another shape in the 6 by 6 table, using the figures 8, 9, 10, 4 and 16.

4 | ||

8 | 9 | 10 |

16 |

The calculated T-Total for this shape is 32, with the T-Number being 8. When putting 8 into the formula I’d just calculated I got 5(8) + 7 which equals 32 also.

Before moving onto the next step I tested the formula on grids of different sizes. I chose a 4 by 4 grid to test the formula on:

1 | 2 | 3 | 4 |

5 | 6 | 7 | 8 |

9 | 10 | 11 | 12 |

13 | 14 | 15 | 16 |

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 |

Where

----- and

----- are original T-Shapes

And

----- and

----- are the same T-Shapes

translated 4 places down.

The T-Total for the RED T-Shape is 37 and its BLUE translation has a T-Total of 217

Likewise:

The T-Total for the GOLD T-Shape is 102 and its GREEN translation has a T-Total of 282.

Again I noticed a pattern immediately. Both translations were 180 values higher than their original T-Shapes. The reason for this is the difference between the numbers in the original T-Shapes and the ones in the translated T-Shapes are always 36 values higher respectively. For example ‘1’ in the original T-Shape becomes ‘37’ in the translated T-Shape. Multiply this 36 by the 5 numbers in the T-Shape and you end up with 180, which is the difference between the T-Totals of the untranslated and translated shapes. This is explained more clearly below:

1 | 2 | 3 |

11 | ||

20 |

(1+b) | (2+b) | (3+b) |

(11+b) | ||

(20+b) |

When translated 4 places actually becomes:

(Where ‘b’ is the value the T-Shape is being translated by).

My provisional translation formula for translating T-Shapes up and down a grid is:

T-Total of translated shape = (T-Total of original T-Shape) + (Difference __________________________between T-Numbers of the two T-Shapes x 5)

To confirm my theory I will try it on a different grid size:

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 |

Where

----- = Original T-Shape and

----- = Translated T-Shape 3 units across

The T-Total for the RED T-Shape is 36 and the T-Total for the BLUE is 141.

When put into the provisional formula this equals:

T-Total of translated shape = (36) + (21 x 5)

which equals 141, proving my formula correct.

I can now conclude that the formula for finding a T-Total of a number translated up or down is:

It is important to note that when translating down the units are negative and when translating upwards the units are positive.

This concludes my T-Totals investigation.

By James Muir

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

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