T-Total Investigation

GCSE Maths

Firstly I looked at how the T-shapes would overlap each other -

Then I experimented with a 9X9 grid and realized

the following points;

Sum of T2=T+5
Difference between T-total and

T-number goes up in 4

Difference between each T-total is 5

I thought of an equation that would be able to tell a person the T-total of

the second T just by using the first T.

T is made up of 5 squares and moves across 1 to make T2

Squares in T X Movement = 5 ∆ Second T = T + 5

ST X M = 5 ∆T2 = T + 5

This proved to be unnecessary in my investigation

I thought it would be useful to create a table of results based

on point 3 that may be easier to work from:

Values 20 and 37 are used first because they are the first possible values of a 9X9 grid

This table shows that when 20 is taken from any T-number and then the answer is multiplied by 5 and 37 is added, you will get the T-total. I had to test this method with a few other numbers to check that it worked.

I could now make an equation to summarize my work on task 1

5(n-20)+37

This can be simplified to: 5n-63

Now I had completed task1 I felt I had more of a ...

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Values 20 and 37 are used first because they are the first possible values of a 9X9 grid

I could now make an equation to summarize my work on task 1

5(n-20)+37

This can be simplified to: 5n-63

Now I had completed task1 I felt I had more of a feel to the investigation and was confident in successful completion. I would now begin to experiment with different grid sizes and investigate the relation of this alteration to the T-total and T-number. As my first grid size was 9X9 I thought it would be both sensible and interesting to venture 2 above and 2 below these measurements, therefore I chose to investigate on a 7X7 grid, and an 11X11 grid.

On the 7X7 grid it was possible to take the full sum of the entire top row (28) and then take away the central T-number (9) and then take away the amount of cells highlighted in the top row (3) and be left with the T-number. r-c-h=n (top row - central T-number – highlighted cells of top row= T-number)

This didn’t work on a 5X5 grid or an 11X11 grid so I began to investigate again.

If you take the grids length or height (7) and then multiply it by 2, and then add 2 you will get the T-number (16). This worked for a 5X5 grid and an 11X11 grid. 2g+2 (grid sizeX2+2)

I also noticed that there was a reason why this formula worked - the T-number is on the third row, you can get to 14 by traveling the grids full width twice (7x2=14) from there you need only to travel two more across to reach the T-number, which is always 3 down, and 2 from the left.

Finding this formula helped me in looking at the problem in a different way and would help more later on.

I drew another table based on the results gained from using a 7X7 grid

I now needed an equation to find the first possible T-total. When looking at the grid a few things become apparent about the relationships between numbers inside the T and factors such as the grid size and T-total. The lower stem of the T will always have numbers 2 digits greater than the grids length and double the grids length plus 2, this is because of their positioning on the grid. Also notice that the top bar of the T will always have the digits 1,2 and 3 in them while in that position whatever the grid size may be. So working upwards the equation would be:

2n+2+(n+2)+6 (you add six because the top bar of the T will always be 1,2 and 3 which equal six when added together)

Because there are so many digits scattered everywhere you can simplify this equation to:

3n+10 (where n equals grid size)

I had already found a formula using the 7X7 grid so I began using the 11X11 grid; I decided to draw a table as I had done for the other grid sizes to make my investigation easier.

In every table I had looked at, the difference between T-totals was always 5.