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Investigation – T-Shapes

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Introduction

                        Kyla Middleditch

Investigation – T-Shapes

The task set is to investigate the way the totals within a 2D ‘T-shape’ can be calculated on a grid of various sizes in various positions.

1 . Investigate the relationship between the T-Total and the T-Number.

Diagram 1.

 The area defined by cyan shading on Diagram 1 is the “T-Shape”.image00.png

I will call the total of the numbers inside the T-Shape the T-Total.

I will also consistently use one number square of the T-Shape through out my investigation to relate my formulas to. I will call this number in the shape the T-Number, this number is indicated by the bold blue font on the T-Shape in Diagram 1.

For the purpose of consistency and methodical working all diagrams in Part 1 will be displayed on

    9x9 grids.

For the T-Shape shown in Diagram 1 the total is: 1 + 2 + 3 + 11 + 20 = 37

I continued to investigate the totals of the following T-Shapes, I created each new T-Shape by moving the shape one square to the right each time across the top row of the grid.

image01.png

...read more.

Middle

From my table of results (Diagram 7) I can predict that a T-Shape in the following location (Diagram 8) will have the following T-Total:

62 + 5 = 67.

However, I would like to investigate producing a consistent algebraic formula that could be used to find the T-Total anywhere on a 9x9 grid.

                                                Diagram 9.

image04.png

I will return to my predicted answer example, Diagram 8 and use the formula to work out its T-Total.

T-number of Diagram 8 = 26

For the purpose of developing algebraic formulae, the T-Number will be expressed as n.

T-total = (n) + (n – 9) + (n– 18) + (n – 19) + (n – 17)

We can simplify this to 5n – 63.

We get –63 because it is the total of the differences between the T-Number and the number in each of the other four squares. I can test the formula by using the numbers given in Diagram 8: (I subtracted each number in the T-shape from the T-Number except the actual T-Number square)

(26) + (26 – 9) + (26– 18) + (26 – 19) + (6 – 17)

This can be simplified as: 130 – 63 which equals a T-Total of 67.

To check the formula I can work out the T-Total by hand:

7 + 8 + 9 + +17 + 26 = 67.

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Conclusion

Diagram 11.

r = new number which must be investigated.image06.png

From Diagram 10 I can do the following calculations relating them to the framework of Diagram 11:

(8 – 5 = 3)  = N-r

(8 – 2 = 6 = 2 x 3) = N-2r

(8 – 6 + 1 = 3) = N-2r+1

(8 – 6 – 1 = 1) = N-2r-1

These calculations show that r=3.

So far the 3x3 formula looks like this:

(n) + (n-3) + (n-6) + (n-7) + (n-5) = T-Total                  Simplified as: 5n – 21.

Diagram 12.

image07.png

My previous formulas, for the 3x3 grid and the 9x9 grid, do not work on this 4x4 grid, they give the following answers:

5n-21 = 29

5n-63 = -13

        I know these formulas give me the wrong answer on this grid because I worked out the T-total by hand and it is: 22.

        I will use Diagram 11 and its explanation to try and work out a formula for the 4x4 grid.

N = (10)image06.png

N-r = (10 – r = 6)

N-2r = (10 – 2r = 2)

N-2r-1 = (10 – 2r - 1= 1)

N-2r+1 = (10 – 2r + 1 = 3)

10 – 6 = 4            10 = 1 = 11

10 – 8 = 2

10 – 1 = 9

        From looking at the equations above, the only number which would satisfy each one, is 4.

...read more.

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