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• Level: GCSE
• Subject: Maths
• Word count: 2735

# T-Total Investigation

Extracts from this document...

Introduction

Daniel Hughes 4C        T-Total Investigation        28/04/07

T-Total Investigation

I am going to look at and investigate the relationship between the T-total, R-number and grid size of a T-shape on a numbered grid.

1. Investigate the relationship between the T-total and the T-number drawn on a 9 by 9 number grid.
 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

There is a possibility of 196 different translations of T-shape on this grid.  I am going to investigate on this grid the T-shape shown above.

Here is a table with T-totals from the grid above:

 T-number T-total 20 (20+11+1+2+3) 37 21 (21+12+2+3+4) 42 22 (22+13+3+4+5) 47 23 (23+14+4+5+6) 52 24 (24+15+5+6+7) 57

To find a relationship between the T-total and T-number I used the difference method.

## T-TotalT-numberDifference

20                37                5

21                42                5

22                47                5

23                52                5

24                57

T = T-total

The reason why the difference increases by 5 each time, is because there is 5 numbers in the T.  As the T moves across the each number increases by one.  Five times each one equals 5.

T = 5n – X

= (5 x 20) - X = 100 – X = 37

= 63

So, 5n – 63 is the formula to work out the t-total for this grid.

Eg.        Formula = 5n – 63

= (5x24) - 63

= 120 – 63

= 57

24 + 15 + 5 + 6 + 7 = 57

Formula = 5n – 63

= (5x50) – 63

= 250 – 63

= 187

50 + 41 + 31 + 32 + 33 = 187

 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 T though can be translated to different positions.

Eg.

 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

In different positions though the formula does not work.

Eg.                = 5n – 63

= (5x4) – 63

= 20 – 63

= -43

Middle

= 285 + 7

= 292

57 + 58 + 59 + 50 + 68 = 292

I predict to work out the reverse of the sideways T in the opposite direction, you have to change the formula to 7 from –7.

Eg.        Formula = 5n – 7

= (5x12) – 7

= 60 – 7

= 53

12 + 11 + 10 + 1 + 19 = 53

Formula = 5n –7

= (5x16) –7

= 80 – 7

= 73

16 + 15 + 14 + 5 + 23 = 73

Formula = 5n – 7

= (5x68) – 7

= 340 – 7

= 333

68 + 67 + 66 + 57 + 75 = 333

1. Uses grids of different sizes.  Translate the T– shape to different positions.  Investigate relationships between the T-total, the T-numbers and the grid size.

Here is a grid 3 by 3:

 1 2 3 4 5 6 7 8 9

I am going to concentrate on the original translation of the T.  The T-total of 8 in this grid is 19.  Adding another column (to make it 3 by 4) will give me two T’s to do the difference method with.

 1 2 3 4 5 6 7 8 9 10 11 12

T-numberT-total        Difference

10                22                5n

11                27                5n

5n = (5x10)

= 50 –22 = 28

= 5n – 28

Eg.        Formula = 5n – 28

= (5x11) – 28

= 55 – 28

= 27

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

T-numberT-total        Difference

12                25                5n

13                30                5n

14                35                5n

The difference between the T-totals is always 5n.

5n – x = (5x12) – x

= 60 – x

25 = 60 – x

x = 35

= 5n – 35

Eg.        Formula = 5n –35

= (5x14) – 35

= 70 – 35

= 35

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

T-numberT-totalDifference

14                28                5n

15                33                5n

16                38                5n

17                43                5n

It is 5n again.

5n – X = (14x5) –X

= 70 – X

28 = 70 – X

X = 42

= 5n – 42

Eg. Formula = 5n – 42

= (5x17) – 42

= 85 – 42

= 43

I noticed a pattern in the formulas.

3x4        5n – 28        7

3x5        5n – 35        7

3x6        5n - 42         7

The difference between the number you take away increases by 7 every time.

Conclusion

22 – 4 = 18x10 = 180

13 – 4 = 9x10 = 90

What I was trying to do was to develop a formula from the first formula with 3 rows and T-number 4.  For example the number 90, + 33 = 123, which is the number taken away for the formula for T-number 13.  180 added to 33 is 213, which is the number taken away from 15n, for the T-number 22.  This has to be expressed into a formula, so here is.

15n – (10(n - 4) + 33)

Formula = 15n – (10(n - 4) + 33)

= (15x7) – (10(7 – 4) + 33)

= 105 – (10x3 + 33)

= 105 (30 + 33)

= 105 – 63

= 42

7 + 8 + 9 + 6 _+ 12 = 42

Formula = 15n – (10(n - 4) + 33)

= (15x13) – (10(13-4) + 33)

= 195 – (10x9 + 33)

= 195 – 123

= 72

13 + 14 + 15 + 12 + 18 = 72

Formula = 15n – (10(n - 4) + 33)

= (15x22) – (10(22-4) + 33)

= 330 – (10x18 + 33)

= 330 – (180 + 33)

= 330 – 213

= 117

22 + 23 + 24 + 21 + 17 = 117

As the formula doesn’t change when a column is added, this is the general formula for this sideways T.

Eg.

 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

Formula = 15n – (10(n – 4) + 33)

= (15x7) – (10(7 - 4) + 33)

= 105 – (10x3 + 33)

= 105 – (30 + 33)

= 105 – 63

= 42

7 + 8 + 9 + 3 + 15 = 42

Formula = 15n – (10(n – 4) + 33)

= (15x16) – (10(16 - 4) + 33)

= 240 – (10x12 + 33)

= 240 –153

= 87

16 + 17 + 18 + 12 + 24 = 87

Formula = 15n – (10(n – 4) + 33)

= (15x27) – (10(27 - 4) + 33)

= 405 – (10x23 + 33)

= 405 – ( 230 + 33)

= 405 – 263

= 142

27 + 28 + 29 + 23 + 35 = 142

I therefore predict if you reverse the T round the other way that the formula will be, 15n + (n + 4)-10 + 33).

 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

Formula = 15n + (n + 4)-10 + 33)

= (15x9) + (9 + 4)-10 + 33)

= 135 + (13x-10 + 33)

= 135 + (-130 + 33)

= 135 – 97

= 38

9 + 8 + 7 + 1 + 13 = 38

Formula = 15n + (n + 4)-10 + 33)

= (15x30) + (30 + 4)-10 + 33)

= 450 + (34x-10 + 33)

= 450 + (-340 + 33)

= 450 – 373

= 72

30 + 29 + 28 + 22 + 34 = 143

Formula = 15n + (n + 4)-10 + 33)

= (15x22) + (22 + 4)-10 + 33)

= 330 + (26x-10 + 33)

= 330 + (-260 + 33)

= 330 – 227

= 103

22 + 21 +14 + 20 + 26 = 103

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

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