T-Total.I was set the task of investigating the relationship between the T-number and T-total on different grid sizes and different positions on the grid.

EG/ 20 + 11 + 3 + 2 + 1 = 37

EG/ 100 – Total of differences (63) = 37. (Correct T-total)

Now, I will try the newly found way with other various T’s from within the grid.

21 – 12 = 9

21 – 4 = 17

21 – 3 = 18

21 – 2 = 19

Same differences as before, because each number goes up the same amount as any other number in the square wherever you move it so they are all going to stay the

same to each other.

If the T moved down one row it would look like the following:

 Each number has increased by 9 compared to its previous position.

2 = 11

3 = 12

4 = 13

12 = 21

21 = 30

It is inevitable that they would of increased by 9 when they moved down 1 row because each number anywhere on it does because the rows are made up of 9 numbers.

On this T the differences are the same yet again (-9, -17, -18, -19)

So, 21 x 5 = 105

                  - 63

                 = 42 (Correct T-total)

Again on the other T one row down:

T = 30 + 21 + 11 + 12 + 13 = 87

(30 x 5) –63 = 87

Beginnings of the formula:

5N - 63

Now I have to find what the 63 is related to apart from the differences, grid size is most likely as it is the only number that could be important that I haven’t tried yet, so:

63 divided by 9 = 7

This isn’t very helpful at the moment so I will now look at another sized grid to see if 7 appears again and to see if the formula works.

10 x 10

I will now see if the differences between the T-totals are still 5 like they were with the 9x9 T-totals.

T = 40

T-total = 45

T = 50

As predicted the T-totals are still 5 more each time.

The difference this time obviously between the T-no. and the other numbers in the T is 10 because the grid length is 10.

1 + 2 + 3 + 12 + 22 = 40        

Now I’ll try the formula:

5 x 22 = 110 – 63 = 47 (wrong T-total)

Now, what must be wrong is the 63 because it wouldn’t be the 5 because the number of numbers in the T hasn’t changed.

What was significant with 63 with the 9x9 grid was the fact that dividing 63 by 9 gave me 7. The 9 won’t be used this time because that only related to a 9 x 9 grid, so hopefully this idea should apply to a 10 x 10 by replacing the 9 with 10.

Therefore, 7 (I will have to find out where this 7 came from later) x 10 = 70.

So if I minus 70 from 110 that gives me 40! The correct T-total.

24 x 5 = 120

            - 70

           = 50.

Now for the moment of truth…

3 + 4 + 5 + 14 + 24 = 50!

And just to make sure, another T shape randomly picked from the 10x10 grid:

66 x 5 =330

            -70

          = 260

66 + 56 + 46 +45 + 47 = 260!

Now I can safely say that the equation looks like this:

Again, N = T-no.

            T = T-total  

            And X = grid size.

T = 5N – (7 x X)

Or simply T = 5N – 7X

If you don’t know the first T-number you could use 2X x 2 to figure it out.

>>>>>>>>>>>>>>>> 2 rows along (18) +2  

X, (in this case 9) x 2 = 18 + 2 = 20 (first T-no.)

This is because 20, the first possible T-no. is already on the third line of the grid, therefore there have already been 2 whole rows (18) and 1 number along on the next row (+2)  = 20. It’s very simple, but as normal hard to explain.

So this could be put into the formula as followed:

5(2X + 2) – (7 x X)

Then expanded to:

10X + 10 – 7X

 This is only for use when you don’t know the first T-number for some reason, which is unlikely because you’re bound to have a grid when doing this.

Now moving onto Translation and namely Vectors

What I aim to achieve is by using the information I gathered from the first part to use in investigating a new equation to work out the T-total when translating the T using vectors anywhere on the grid.

         1st T

The Vector of the second T from the first T is 4

                                                                         -4

So I can refer to them algebraically I will call them B

                                                                                  C

I can see from looking at the grid that the all the numbers have increased by 40 as I mentioned earlier.

If I am going to find the T-total of another T on the grid from the information I know from the first T then I’m sure I’ll use 37, the first T-total somewhere or another.

B refers to the amount of spaces the T moves horizontally, left or right.

C refers to the amount of space the T moves vertically, up or down.

This brings me to another point, if I was to move the T left of the first position one of the numbers in the T would be off the side of the grid.

It would look like this:

Now, would it work if I did this because:

1. There isn’t actually a 0 on the grid
2. The 9 and the 18 shouldn’t be there because they are meant to be on the other side of the grid.

I’ll try the equation with it:

5 x 19 – 63 = 32.

19 + 10 + 1 +2 = 32.

Strange, it should of deliberately not worked because I had put 5 x 19 but there were only 4 numbers (0 isn’t actually there, or is it?).

19 – 10 = 9

19 – 2 = 17

19 – 1 = 18

18 +17 +9 = 44

4 x 19 – 44 = 32

This is good, it shows that the equation will work with different amounts of numbers within the T, so it may work with enlargement and scale factors.

Back to Vectors:

1st T

The Vector of the second T from the first T is 4

                                                                         -4

So I can refer to them algebraically I will call them B

                                                                                  C    

I can see from looking at the grid that the all the numbers have increased by 40 as I mentioned earlier.

If I am going to find the T-total of another T on the grid from the information I know from the first T then I’m sure I’ll use 37, the first T-total somewhere or another.

B refers to the amount of spaces the T moves horizontally, left or right.

C refers to the amount of space the T moves vertically, up or down.

The first goes 4 across so it looks like this:

So, the T-no. is +4 obviously.

The T-total now = 57

(20 more than 37)

B (4) x 5 = 20

I’ve used 5 because:

1. It makes 4 into 20 and
2. I’ve already used in the other equations so it’s bound to fit in hear.

So in the first move T = 37 + 5B

T = 37 + 20 = 57

37 = 5(2X + 2) – 7X

Therefore 37 = 10X + 2 - 7X, as I showed before for working out the first T-total if you didn’t have the first T-number.

I could call 37 by the letter F (for first t-total) so that it can represent the first T-total of any sized grid as opposed to just 37, which only applies to a 9x9.

The second part of the vector, the vertical one, -4 will be a bit trickier as you are not just moving it simply sideways as the numbers increase by one but by 9 instead.

Therefore this is going to involve the grid length, 9.

Because it is moving down, nine at a time it’s going to involve multiplying as opposed to adding.

Goes to >

All the numbers have increased by 36, which isn’t that significant.

If for each row it’s going up at nine at a time then something’s going to be multiplied by 9, namely the 5 x the vector (-4), hopefully.

C (-4) x 5 = -20 x 9

                 = -180

Ok, it’s a negative number so if a minus the negative from the 57 we got with the first part of the equation I should hopefully get the T-total.

57 - - 180

= 57 + 180

= 237

           

Now to see if it’s correct:

5 x 60 – 63 = 237!

The equation then looks like this:

T = (F + 5b) – (X x 5c)

Now to see if it works with other vectors:

The vector is 2

                     -4

Using T = (F + 5b) – (X x 5c)

T = (37 + 10) – (9 x –20)

T = 47 - -180

T = 227

Check answer:

5 x 58 – 63 = 227!

On to Rotation:

The T-no. for this sideways T is 10, the T-total is 57.

Using what information I have already I’ll follow the original equation:

5 x 10 = 50

Obviously if I now minus 63 it’s going to be wrong. What it seems is that all I need to do is add 7.

50 + 7 = 57

Now try in another position on the grid:

T-no. = 12

T-total = 72

13 x 5 + 7 = 72

It appears that the equation quite simply for rotating a T by 90degrees is:

5N + 7

Rotating the T by 270degrees.

T-no. = 12

T-total = 53

12 x 5 = 60

 

If I minus 7 it will give me the correct T-total of 53.

So 5N – 7 seems to be the equation for rotating 270degrees.

Test out on another area to make sure:

 

41 x 5 – 7 = 198

41 + 40 + 39 + 30 + 48 = 198

Now for a 180degree rotation:

I know that a 0degree turn is 5N – 7X so like the similarity between 90 and 270 rotations the equation for a 180degree might be 5N + 7X as opposed to – 7X.

T-no. = 2

T-total = 73

5 x 2 = 10

         + 63 = 73!

Equations so far:

All equations = T

 Standard upright T:

5N – 7X (where N is the T-no. and X is the grid length)

90degree rotation:

5N + 7 (no X)

180degree rotation:

5N + 7X  

270degree rotation:

5N – 7 (no X)

Translating the T using Vectors:

 (F + 5B) – (X x 5C) (where F= first T-total of the first T on the grid EG/ 37 for a 9x9 grid. B = top vector, i.e. horizontal. C = bottom vector, vertical.

To work out F only with the grid length:

 

F = 10X + 10 – 7X

All these equations work on different sizes of grids.

What puzzles me now is where this 7 comes from?

It’s in all the equations for no obvious reason totally unlike everything else that has been reasonably obvious.

It appears to have no link to the grid size as it stays the same in all equations with whatever grid you have.

Investigating the mysterious number 7:

All the numbers in the shaded area shown above add up to 99.

If you minus the ones that aren’t normally in the T you get the T-total.

The numbers which you minus, shaded slightly darker than the rest add up to 62. (99-37). This appears to have no significance with anything so far.

Just a thought:

As I said before when doing the vertical vectors you multiply the vertical vector by 9 because every time you move down one row you are going up 9 more. The reason it multiplies instead of just adding nine is because adding nine would be if you were working out the T-no. but for working the t-total you have to multiply. Hence 7 x X in all the equations.

So if I move down from 20 to 11 I’m going down 9 (so it’s dividing by nine once), then you have to divide by nine twice to get to the first row, which has three numbers on, 1, 2, 3.

So for each number you divide by 9 twice to get from 20, not literally.

In total then, you have divided by 9, 7 times. Once to 11, and then twice to each of the three bottom numbers. This is why you then multiply by 9 in the equation because I’m just dividing to try and find the origins of the number 7.

I know this sounds very strange but it’s nearly impossible to explain it.

I suppose 7 is just one of those numbers that are just there for no obvious reason that is just part of the whole T-total work, like other numbers in maths that are just part of a certain equation for no particular reason.

Reflection:

Reflection is merely a combination of two things I have already found out:

Vectors and Rotation

By translating the T by vectors and then rotating on the spot or just doing one of them you can get the T into any position a reflection would. If anything, getting a separate single equation for reflection would be near impossible because there is no way of putting the reflection line into the equation (I.e. what numbers it passes through, what angle is it going at)

Enlargement:

Enlarging T by scale factors would cause problems:

1. A scale factor of 2 would mean that the T wouldn’t have a T-no. because it would be 2 squares wide.

2. What side would it go of the original T because you can’t centre if it were only 2 squares wide?

All T’s would need to be enlarged with a odd numbered scale factor, so that that they had a number in the middle which could be used as a T-no.

Ok, here’s a long shot.

Try using the equation for this one:

There are 45 numbers in the T

So, 45x77 – 63 = 3402, which I can guess is wrong anyway.

The 7 probably will have to change somehow.

Maybe it would work it you enlarged the T by all dimensions apart from width, which seems to cause a problem.

There are 17 numbers in the T:

17 x 77 – 63 =1246, which again I can guess is wrong.