# Investigation &#150; 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”.

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.

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.

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.

Conclusion

Diagram 11.

r = new number which must be investigated.

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.

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)

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.

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

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