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• Level: GCSE
• Subject: Maths
• Word count: 1657

# T-Totals Investigation.

Extracts from this document...

Introduction

Ayesha butt 10G         Maths Coursework        28/04/2004

T-Totals Investigation:

In this investigation I will be exploring the different relationships between the letter ‘T’ and its numbers and totals using a variety of different grid sizes and translations to effectively find any correlations between them.

 1 2 3 11 20

This is the ‘T’ shape that I will be starting with.

The total of the numbers inside the T-shape is:

1+2+3+11+20= 37

This is called the T-total.

The number found at the bottom of the whole T-shape is called the T-number. In this case, the T-number for this T-shape is 20.

Part 1:

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

Starting with a 9 by 9 grid, a table for T-shapes going horizontally across the grid can be drawn:

 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-number 20 21 22 23 24 25 26 T-total 37 42 47 52 57 62 67

+5     +5     +5     +5     +5    +5

A T-shape with the T-number 27 cannot be drawn as this would give an incomplete ‘T’.

As can be seen above, there is always an addition of 5.

Middle

32

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46

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 T-number 18 19 20 21 22 23 T-total 34 39 44 49 55 59

+5     +5       +5     +5      +5

There is an addition of 5 again-(same as in the horizontal T-shapes of the 9 by 9 grid. Could this be a pattern?)

 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

 T-number 18 26 34 42 50 58 T-total 34 74 114 154 194 234

+40    +40   +40     +40    +40

These totals have an addition of 40 each time.

With this I can see that 40÷8= 5.

From looking at the previous vertical translations on the 9 by 9 grid and the ones done here on the 8 by 8 grid, I can see that the number that is added to get to the next total, when divided by the width of the grid (the grid number) gives you 5, which is the constant number added in the totals of the horizontal t-shapes.

Throughout both grids, the horizontal T-shape totals have an addition of 5. I predict that the next grid will show these same results.

I am now going to try the same thing using a 7 by 7 grid to see if my theory is correct:

Table for horizontal T-shapes:

 T-number 16 17 18 19 20 58 T-total 31 36 41 46 51 56

+5     +5     +5     +5     +5

Once again, there is an addition of 5. The first part of my prediction is correct.

Table for vertical T-shapes:

 T-number 16 23 30 37 44 T-total 31 66 101 136 171

+35    +35      +35    +35

Each time 35 is being added.

35÷grid number

=35÷7= 5

These results correspond directly to my theory, therefore it is correct.

Part 2:

Using grids of different sizes translate the T-shape to different positions. Investigate the relationship between the T-total, the T-numbers and the grid size.

Grid: 9 by 9

 T-number 20 21 22 23 24 25 26 T-total 37 42 47 52 57 62 67

+5       +5     +5     +5    +5    +5

There is an addition of 5 each time.

So the first part of the formula is 5n.

To find the rest: 5n±X (where X= a linear bit)

(5×20)±X=37

100-X=37

X=63

Total formula: t=5n-63 _                       This is equivalent to 7×9!

I will check to see that this formula is correct and whether it will work anywhere on the grid:

1. 5n-63                2) 5n-63                3) 5n-63

(5×24) -63            (5×40)-63            (5×57)-63

120-63                   200-63                    285-63

57√                     137 √                    222

Another quicker way that this formula can be found out is using an algebraic method:

 1 2 3 11 20 n-19 n-18 n-17 n-9 n

Conclusion

I will now translate the T-shape to different positions on the grid to find out any more relationships.

 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

For this next part, I have translated the T-shape upside down and tabulated the results going horizontally across the grids:

 T-number 2 3 4 5 6 7 8 T-total 73 78 83 88 93 98 103

+5    +5     +5      +5     +5     +5

Using the algebraic method, I can find out the formula to find the T-total:

 2 11 19 20 21 n n+9 n+2g-1 n+18 n+2g+1

n + n+g + n+18 + n+2g-1 + n+2g+1

= 5n+7g

As you can see the formula has become positive as the T-shapes have been inverted, whereas before when they were upright, they were negative (5n-7g).

So 5n+(7×9)

= 5n+63

As you can see, this formula is exactly the same as the formula for the upright T-shapes, except now it has a positive sign in it as the T-shapes have been inverted.

Will this be the same for the 8 by 8 grid and the 7 by 7 grid?

8 by 8 Grid:

 T-number 2 3 4 5 6 7 T-total 66 71 76 81 86 91

Again 5 is added each time.

By using the algebraic method we also find out that the formula is 5n+56. So far, my theory is correct.

7 by 7 grid:

 T-number 2 3 4 5 6 T-total 59 64 69 74 79

Again 5 is being added each time. The formula comes out to be 5n+49.

By inverting the T-shapes, I find that the formulas all change to become positive instead of negative.

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

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