# T-Total.I was set the task of investigating the relationship between the T-number and T-total on different grid sizes and different positions on the grid.

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Introduction

## Mathematics GCSE Coursework

## T-Total

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Say you had a 9 by 9-number grid with a T-shape drawn on it. The total of the numbers inside the T is 1 + 2 + 3 + 11 + 20 =37.

This is called the T-total.

The number at the bottom of the T-shape (20 in this case) is called the T-number.

I was set the task of investigating the relationship between the T-number and T-total on different grid sizes and different positions on the grid.

1 | 2 | 3 |

10 | 11 | 12 |

19 | 20 | 21 |

What I will aim to find out is a formula that will link the T-number and T-total in some way or another. First I will find out the difference between the different T-numbers and T-totals in the grid (sticking to only a 9x9 for now)

So it is easier and quicker to write I will use the following letters to represent certain numbers: -

- X = grid length

- N = T-number
- T = T-total

The difference between the first T-number (20) and T-total (37) is 17.

2 | 3 | 4 |

11 | 12 | 13 |

20 | 21 | 22 |

This is the next T (not T-total but T-shape) along.

The T-number here is 21.

The T-total = 2 + 3 + 4 + 12 + 21 = 42

42 – 21 = 21

3 | 4 | 5 |

12 | 13 | 14 |

21 | 22 | 23 |

N = 22

T = 47

Difference = 25

4 | 5 | 6 |

13 | 14 | 15 |

22 | 23 | 24 |

N = 23

T = 52

Difference = 29

So, the difference so far starting from the first T look as followed:

- 17
- 21
- 25
- 29

The differences are going up by 4 each time i.e. the 2nd difference is constant therefore it is a quadratic equation. (Ax² + Bx + C)

n | T-number | T-Total | 1st Difference (between T-no. and it’s T-total) | 2nd Difference |

1. | 20 | 37 | 17 | 4 |

2. | 21 | 42 | 21 | 4 |

3. | 22 | 47 | 25 | 4 |

4. | 23 | 52 | 29 | 4 |

It appears that using these results will not work in a quadratics table or equation because the differences have to be in the same category EG/ just T-total.

The T-total appears as followed:

- 37
- 42
- 47
- 52

The T-total is going up by 5 each time.

Middle

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X, (in this case 9) x 2 = 18 + 2 = 20 (first T-no.)

This is because 20, the first possible T-no. is already on the third line of the grid, therefore there have already been 2 whole rows (18) and 1 number along on the next row (+2) = 20. It’s very simple, but as normal hard to explain.

So this could be put into the formula as followed:

5(2X + 2) – (7 x X)

Then expanded to:

10X + 10 – 7X

This is only for use when you don’t know the first T-number for some reason, which is unlikely because you’re bound to have a grid when doing this.

Now moving onto Translation and namely Vectors

What I aim to achieve is by using the information I gathered from the first part to use in investigating a new equation to work out the T-total when translating the T using vectors anywhere on the grid.

1st T

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 Vector of the second T from the first T is 4

-4

So I can refer to them algebraically I will call them B

C

I can see from looking at the grid that the all the numbers have increased by 40 as I mentioned earlier.

If I am going to find the T-total of another T on the grid from the information I know from the first T then I’m sure I’ll use 37, the first T-total somewhere or another.

B refers to the amount of spaces the T moves horizontally, left or right.

C refers to the amount of space the T moves vertically, up or down.

This brings me to another point, if I was to move the T left of the first position one of the numbers in the T would be off the side of the grid.

It would look like this:

0? | 1 | 2 |

9? | 10 | 11 |

18? | 19 | 20 |

Now, would it work if I did this because:

- There isn’t actually a 0 on the grid
- The 9 and the 18 shouldn’t be there because they are meant to be on the other side of the grid.

I’ll try the equation with it:

5 x 19 – 63 = 32.

19 + 10 + 1 +2 = 32.

Strange, it should of deliberately not worked because I had put 5 x 19 but there were only 4 numbers (0 isn’t actually there, or is it?).

0? | 1 | 2 |

9? | 10 | 11 |

18? | 19 | 20 |

19 – 10 = 9

19 – 2 = 17

19 – 1 = 18

18 +17 +9 = 44

4 x 19 – 44 = 32

This is good, it shows that the equation will work with different amounts of numbers within the T, so it may work with enlargement and scale factors.

Back to Vectors:

1st T

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 Vector of the second T from the first T is 4

-4

So I can refer to them algebraically I will call them B

C

I can see from looking at the grid that the all the numbers have increased by 40 as I mentioned earlier.

If I am going to find the T-total of another T on the grid from the information I know from the first T then I’m sure I’ll use 37, the first T-total somewhere or another.

B refers to the amount of spaces the T moves horizontally, left or right.

C refers to the amount of space the T moves vertically, up or down.

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 first goes 4 across so it looks like this:

5 | 6 | 7 |

14 | 15 | 16 |

23 | 24 | 25 |

So, the T-no. is +4 obviously.

The T-total now = 57

(20 more than 37)

B (4) x 5 = 20

I’ve used 5 because:

- It makes 4 into 20 and
- I’ve already used in the other equations so it’s bound to fit in hear.

So in the first move T = 37 + 5B

T = 37 + 20 = 57

37 = 5(2X + 2) – 7X

Therefore 37 = 10X + 2 - 7X, as I showed before for working out the first T-total if you didn’t have the first T-number.

I could call 37 by the letter F (for first t-total) so that it can represent the first T-total of any sized grid as opposed to just 37, which only applies to a 9x9.

The second part of the vector, the vertical one, -4 will be a bit trickier as you are not just moving it simply sideways as the numbers increase by one but by 9 instead.

Therefore this is going to involve the grid length, 9.

Because it is moving down, nine at a time it’s going to involve multiplying as opposed to adding.

5 | 6 | 7 |

14 | 15 | 16 |

23 | 24 | 25 |

Goes to >

41 | 42 | 43 |

50 | 51 | 52 |

59 | 60 | 61 |

All the numbers have increased by 36, which isn’t that significant.

If for each row it’s going up at nine at a time then something’s going to be multiplied by 9, namely the 5 x the vector (-4), hopefully.

C (-4) x 5 = -20 x 9

= -180

Ok, it’s a negative number so if a minus the negative from the 57 we got with the first part of the equation I should hopefully get the T-total.

57 - - 180

= 57 + 180

= 237

Now to see if it’s correct:

5 x 60 – 63 = 237!

The equation then looks like this:

T = (F + 5b) – (X x 5c)

Now to see if it works with other vectors:

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 |

Conclusion

I suppose 7 is just one of those numbers that are just there for no obvious reason that is just part of the whole T-total work, like other numbers in maths that are just part of a certain equation for no particular reason.

Reflection:

Reflection is merely a combination of two things I have already found out:

##### Vectors and Rotation

By translating the T by vectors and then rotating on the spot or just doing one of them you can get the T into any position a reflection would. If anything, getting a separate single equation for reflection would be near impossible because there is no way of putting the reflection line into the equation (I.e. what numbers it passes through, what angle is it going at)

Enlargement:

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 |

Enlarging T by scale factors would cause problems:

- A scale factor of 2 would mean that the T wouldn’t have a T-no. because it would be 2 squares wide.

- What side would it go of the original T because you can’t centre if it were only 2 squares wide?

All T’s would need to be enlarged with a odd numbered scale factor, so that that they had a number in the middle which could be used as a T-no.

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 |

Ok, here’s a long shot.

Try using the equation for this one:

There are 45 numbers in the T

So, 45x77 – 63 = 3402, which I can guess is wrong anyway.

The 7 probably will have to change somehow.

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 |

Maybe it would work it you enlarged the T by all dimensions apart from width, which seems to cause a problem.

There are 17 numbers in the T:

17 x 77 – 63 =1246, which again I can guess is wrong.

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

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