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• Level: GCSE
• Subject: Maths
• Word count: 4928

# T-Total the investigation relies on investigating relationships between our three variables; t-number, t-total, and grid size

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Introduction

T-Total

Introduction

The T-Total investigation involves looking at the relationship between what are known as the t-number and the t-total in an n grid. The t-number can be thought of as the number at the base of the t shape, and the t-total the sum of all the numbers in the T. Therefore taking the following grid:

 n 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 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

The t-number is 31, grid size is 9 and t-total is 31+22+12+13+14=92

Therefore the investigation relies on investigating relationships between our three variables; t-number, t-total, and grid size. Further variable factors may be added along the way to increase the scope of the investigation.

## Grid size

n

T-total

t

T-number

x

Individual numbers in ‘T’

a,b,c,d (see below)

 b c d a x

Individual variables within the ‘T’

We shall first look at how we can derive t through looking at n and x.

Contents

 Investigating t, n, and x 3 Verification 4 Alternate formula 5 Transformations 6 Vector Translations 8 Variable dimensions 9 Conclusion and Evaluation 11

Investigating t, n, and x

The first logical step to finding t is to look at the individual values x, a, b, c, and d. If we know these values, we can simply add them up to find t. As we already know x, it simply becomes a matter of finding the other values. There are specific rules governing the placement of each value in relation to the others- we can see that a is always directly above x, and that it is directly below c. These may seem obvious, but they are key to deriving solutions to the problem. Simplifying the problem even further, we can see that there are only two values we need to know to calculate the values of the other squares; the row and column of x.

Middle

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that t = 59 + 4(9[59/9] + 59 mod 9) – 7*9

thus t = 59 + 4(9*6 + 5) – 63

thus t = 59 + 236 – 63

therefore t = 232

Looking at the grid we see that t = 59+50+40+41+42=232

1. Take grid size 13 and the starting number 72.
 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 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169

T = 72 + 4(13*[72/13] + 72 mod 13) – 7*13

Thus T = 72 + 4(65 + 7) – 91

Thus T = 72 + 288 – 91

Therefore T = 269

Looking at the grid we see that t = 72+59+45+46+47 = 269

In both these cases, the value calculated by the formula is the same as the actual total, proving to all extensive purposes that the formula works.

Alternate formula

The above formula is perfectly capable of calculating the T-total for any values of x and n. However, it does not take long to notice that it has some drawbacks. Foremost, but not necessarily prohibitive, it is rather complicated to use and explain. The major problem, though, is that it is almost impossible to rewrite the formula to calculate x when given n and t, or n given x and t. The use of modulo and step functions means that when reversing the formula, certain numbers cannot be established or distinguished between (i.e., in mod 3 arithmetic, there is no difference between 11 and 17). I do not believe it is impossible to achieve, but it is convoluted enough that I will not attempt to do so in this piece of work. Instead, we shall take a step sideways to look at an alternate formula to solve this problem.

My original idea when first looking at the problem was that, by calculating the row and column of the t-number, I could work out similar values for the other numbers, and thus calculate their values. However, after I had worked out the above formulas, it became obvious that this step is in fact, unnecessary. To look at this next step in a ‘real world’ scenario, it is rather like trying to calculate your latitude and longitude in order to describe where the person standing next to you is. It is a lot more logical to say ‘half a meter to the left of me’. After all, if we are to believe Einstein, then everything only has a position in relation to something else. To return and apply this scenario to our problem, we can see that it is unnecessary to calculate the position of x in order to get the locations of the others when we can just say ‘one row above me’. Because rows loop every n (ie, each row is n long, and thus the distance between two points in the same column is n), we can just say that a (returning to our above notation) is equal to x-n. Doing this for all the numbers in the T, we get:

 b c d a x

Conclusion

I maintain, as was my original belief, that this investigation was poorly expandable, with little extra to look at beyond the obvious that doesn’t just recover the same topics. The fact that we have had to stray into n-dimensions, something that is clearly not totally applicable to this, is testimony to the fact. This investigation was, to be blunt, far too simple to provide any truly interesting results, or to be extended.

While I say this, there are many ways we could have extended this investigation, but few of them moved further from the original problem. For example, we could have looked at every single rotation of the T in 3 dimensions, but this would have not moved the investigation any further beyond the one rotation we looked at. We would simply have been faced with pages of tedious, repetitive algebra; something which is of no help to anyone. Looking at the original formula in 3 dimensions would have been interesting, but would have been completely anomalous to the investigation; it would not have shown us anything further than the alternate formula, and would have been unnecessarily complicated. There is no point to such an expansion. Had I not discovered it first, the inclusion of the original formula in this investigation would have been purely superfluous.

To surmise, I believe this investigation was covered as thoroughly as it could be, and that the results produced are valid and justified. I do believe that a lot of the material is tedious and repetitive; I also feel this was unavoidable due to the nature of the problem set. I further believe that there is nothing further I could do to extend this investigation without either moving completely out of the purview of the original problem, or covering the same territory already covered.

Nicholas Clarke        11AH

Clevedon Community School

Mathematics Investigation 2

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