‘5n-56’ where n = T-number
Therefore (5x59) – 56 = 295 – 56 = 239
As you can see whether we are translating to the right, left, up or
down the formula 5n-56 will find the T-total of the T-shape if it’s on an 8x8 grid.
Rotating 90°
I will rotate the T-shape in a 90° direction, and see if I can find a formula to find the T-total of rotated T-shapes.
I will be rotating T19 in a 90° direction; the T-number remains the same, ‘19’, but the T-total will change.
2+3+4+11+19=39
13+21+29+20+19=102
As you can see from the table above, the 90° rotated T19 has a T-total of 102, a ‘+63’ increment.
I will now try this on another T-shape to see if I will get the same ‘+63’ increment.
I will use the T-shape of T35
SUM method: 18+19+20+27+35=119
Algebraic Formula (5n-56): 5x35-56 = 119
29+37+45+36+35=182
We also have an increment of ‘+63’, therefore we know that, to find the T-total of a 90° rotated T-shape, we would be able to do so by simply adding ‘63’ to the current T-total.
x = current T-total + 63 (where ‘x’ is the new T-total to be found...)
Now I will find an algebraic formula for finding the T-total of any 90° rotated T-shape. To find this algebraic formula, I will find out a way to find the individual values in the T-shape:
Let’s refer to the T-number as ‘n’
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n+1+n+2+n-6+n+10
=n+n+n+n+n+1+2+10-6 (gather like terms) =‘5n+7’
Therefore my algebraic formula for finding the T-total of any 90° rotated t-shape is ‘5n+7’
I will now find the T-total of a T-shape rotated from a 90° angle starting position.
Rotating 90° ...2
I will rotate this T35 t-shape in another 90°’s making it a 180° rotation in total.
29+37+45+36+35=182
Algebraic formula (5n+7)
5x35+7= 182
50+51+52+43+35=231
x = current T-total + 49 (where ‘x’ is the new T-total to be found...)
As you can see rotating a further 90° adds ‘+49’ to the current T-total, this of course leads me to believe that if I rotated a T-shape from a 0° position, to a 180° position, I would be able to find its T-total by simply adding ‘(63+49)’ to its current T-total. This is my prediction I will now prove my prediction.
Rotating 180°
I will now rotate the T-shape, T23 in 180.
To find the T-total of this T-shape, I will use the algebraic formula 5n-56
Where ‘n’ is the T-number of the T-shape.
5x23-56=59
Just to prove that this is correct: 23+15+7+6+8=59
23+31+38+39+40=171
I will now check to see if my prediction is correct.
If I add (63+49) to my current T-total of ‘59’ will I get 171?
59+ (63+49)
59+112 =171
You can clearly see from the above table of results, the T-total increases by an increment of ‘+112’ every time. Just like I predicted it would.
I will now find an algebraic formula for finding the T-total of any 180° rotated T-shape.
To find this algebraic formula, I will find out a way to find the individual values in the T-shape:
Let’s refer to the T-number as ‘n’
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n+8+n+15+n+16+n+17
=n+n+n+n+n+8+15+16+17=5n+56
Check
5n+56
Substitute the T-number into ‘n’
5x23+56 = 171
As you can see this formula will find the T-total of any T-shape rotated by180°
So let’s have a recap of all the algebraic formulas we have uncovered so far, and what they solve:
Rotating 270°
I will now find out a formula for finding the T-total of any T-shape rotated by 270°
2+3+4+11+19=39
9+17+25+18+19=88
As you can see the increment was ‘+49’ this allows me to build a simple formula.
x = current T-total + 49 where ‘x’ is equal to the new T-total
Therefore: x = 39 +49
x = 88
New T-total = 88
Although a good formula an algebraic formula is better as it doesn’t require an established T-total to be worked out, all I need is the T-number of the T-shape and I will be able to work it out,
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n-1+n-2+n-10+n+6
=n+n+n+n+n+6-1-2-10=5n-7
My formula for finding the T-total of any 180 rotated T-shape is therefore ‘5n-7’
Let’s check to see if it works. I will use the T-shape, T19 to test if this works.
5x19 – 7 = 88
The formula works, just as expected.
I will now show ALL the formulas that can be used in an 8x8 grid to calculate the T-total:
Grid Change
I will move on to a 9x9 grid size
Translating Horizontally
I will be exploring the relationship between the T-number and the T-totals depending on the translation of the T-shape. In this 9x9 grid, I will be translating to the right, and working out the formula for doing so, I will also work out an algebraic formula for working out the T-total of any T-shape in a 9x9 grid.
To begin with, I will first find the T-totals of these 3 T-shapes.
1+2+3+11+20 = 37
2+3+4+12+21 = 42
3+4+5+13+22 = 47
Table of results
As you can see the T-totals of the 3 T-shapes are: 37, 42 and 47
As you can see they increase by the integer ‘+5’ each time, they are translated to the right. Thus if translated to the left we would ‘-5’
A formula for finding the new t-total could be, current T-total + 5, this would be right but it is not extensive enough a better formula would be: new T-total = T-total + (x*5), where x is the number of times you are moving to the right, so if I translate to the right once it would be new T-total = current T-total + (1*5), T-total = T-total + 5, if I translate 5 times to the right it would be new T-total = current T-total + (5*5), T-total +25.
Proof
At T-20, if I translated twice to the right,
Formula: ‘Current T-total + (x*5)’ x is the number of times you translate to a certain direction, in this case the right.
New T-total = 37 + (2*5) = 37 + 10 = 47
As you can see the T-total of T22 is ‘47’
Algebraic
I will now find the algebraic formula for this 9x9 grid, using the T-shape T20.
I will find this formula using an equation, to do so I will find the nth term
My prediction is ‘5*n - x = T-total’ where n = T-number
If I substitute my values it will get:
5*20 - x = 47
100 - x = 47
100 - 47 = x
63 = x
Therefore ‘5n – 63 = T-Total’ is my formula for finding the T-total of any t-shape.
I have used an equation method to find my formula; I could have used the algebraic difference method, to find it.
T20
Tn
As you can see from the above T-shape, we now know how to find all the individual values of the T-shape.
So the algebraic formula would be
n-19+n-18+n-17+n-9+n = ‘5n-63’
As you can see I have arrived at exactly the same formula of ‘5n-63’
Now I will test this formula, on T20
5x20-63 = 37
This formula will find the T-total of any T-shape on a 9x9 grid, as long as it is translated horizontally or vertically
Translating Vertically
Having found out an algebraic formula for finding the T-total of all horizontally/vertically translated T-shapes on a 9x9 grid, I will now find the simple formula for translating vertically, as I have already found the algebraic formula for a 9x9 grid.
To start of I will find the T-total of the following T-shapes:
T20 and T29
1+2+3+11+20 = 37
10+11+12+20+29 = 82
As you can see the T-total increased by ‘+45’, therefore an easy way to work out the T-total of a T-shape translated vertically downwards is:
New T-total = current T-total + 45
Although this is a perfectly valid formula, it is however not extensive enough, a better formula would be.
New T-total = current T-total + (x*45) where ‘x’ is the number of times the shape is translated downwards
For example if the T-shape T20 was translated once downwards to T29 it would be: new T-total = 37 + (1x45) = 37 + 45 = 82
If translated twice, it would be 37 + (2x45) = 37 + 90 = 127. This is the T-total of T38
I will now use the formula I found earlier to work out the T-total of this same T-shape (T29), just to prove that it works on all shapes as long as they are translated vertically or horizontally.
The formula is ‘5n-63’
So by substituting the values I get:
5x29 – 63 = 82
As you can see the formula works. I will now proceed unto rotation of the T-shapes in a 9x9 grid.
Rotating 90°
I will rotate the T-shape in a 90° direction, and see if I can find a formula to find the T-total of rotated T-shapes.
I will be rotating T20 in a 90° direction; the T-number remains the same, ‘20’, but the T-total will change.
1+2+3+11+20 = 37
13+22+31+21+20=107
As you can see from the table above, the 90° rotated T19 has a T-total of 102, a ‘+70’ increment.
I will now try this on another T-shape to see if I will get the same ‘+70’ increment.
I will use the T-shape of T34
SUM method: 15+16+17+25+34=107
Algebraic Formula (5n-63): 5x34-63 = 107
27+36+45+35+34=177
We also have an increment of ‘+70’, therefore we know that, to find the T-total of a 90° rotated T-shape, we would be able to do so by simply adding ‘70’ to the current T-total.
x = current T-total + 70 (where ‘x’ is the new T-total to be found...)
Now I will find an algebraic formula for finding the T-total of any 90° rotated T-shape in a 9x9 grid. To find this algebraic formula, I will find out a way to find the individual values in the T-shape:
Let’s refer to the T-number as ‘n’
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n+1+n+2+n-7+n+11
=n+n+n+n+n+1+2+11-7 (gather like terms) =‘5n+7’
Therefore my algebraic formula for finding the T-total of any 90° rotated t-shape is ‘5n+7’
I will now test this formula to see if it works, i will test it using the T34 T-shape
Substituting the T-number in place of ‘n’ I get:
5x34 + 7 = 177
The formula works!
Rotating 180°
I will now rotate the T-shape, T23 in 180.
To find the T-total of this T-shape, I will use the algebraic formula 5n-63
Where ‘n’ is the T-number of the T-shape.
5x23-63=52
Just to prove that this is correct: 23+14+5+4+6=52
23+32+40+41+42=178
You can clearly see from the above table of results, the T-total increases by an increment of ‘+126’ every time.
I will now find an algebraic formula for finding the T-total of any 180° rotated T-shape.
To find this algebraic formula, I will find out a way to find the individual values in the T-shape:
Let’s refer to the T-number as ‘n’
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n+9+n+17+n+18+n+19
=n+n+n+n+n+9+17+18+19=5n+63
Check
5n+63
Substitute the T-number into ‘n’
5x23+63= 178
As you can see this formula will find the T-total of any T-shape rotated by180°
So let’s have a recap of all the algebraic formulas we have uncovered so far, and what they solve:
Rotating 270°
I will now find out a formula for finding the T-total of any T-shape rotated by 270°
2+3+4+12+21=42
10+19+28+20+21=98
As you can see the increment was ‘+56’ this allows me to build a simple formula.
x = current T-total + 56 where ‘x’ is equal to the new T-total
Therefore: x = 42 +56
x = 98
New T-total = 98
Although a good formula an algebraic formula is better as it doesn’t require an established T-total to be worked out, all I need is the T-number of the T-shape and I will be able to work it out.
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n-1+n-2+n-11+n+7
=n+n+n+n+n+7-1-2-11=5n-7
My formula for finding the T-total of any 180 rotated T-shape is therefore ‘5n-7’
Let’s check to see if it works. I will use the T-shape, T19 to test if this works.
5x21 – 7 = 98
The formula works, just as expected.
I will now show ALL the formulas that can be used in a 9x9 grid to calculate the T-total:
Grid Change
I will move on to a 10x10 grid size
Translating Horizontally
I will be exploring the relationship between the T-number and the T-totals depending on the translation of the T-shape. In this 10x10 grid, I will be translating to the right, and working out the formula for doing so, I will also work out an algebraic formula for working out the T-total of any T-shape in a 10x10 grid.
To begin with, I will first find the T-totals of these 3 T-shapes.
1+2+3+12+22 = 40
2+3+4+13+23 = 45
3+4+5+14+24 = 50
Table of results
As you can see the T-totals of the 3 T-shapes are: 40, 45 and 50
As you can see they increase by the integer ‘+5’ each time, they are translated to the right. Thus if translated to the left we would ‘-5’
A formula for finding the new t-total could be, current T-total + 5, this would be right but it is not extensive enough a better formula would be: new T-total = T-total + (x*5), where x is the number of times you are moving to the right, so if I translate to the right once it would be new T-total = current T-total + (1*5), T-total = T-total + 5, if I translate 5 times to the right it would be new T-total = current T-total + (5*5), T-total +25.
Proof
At T-22, if I translated twice to the right,
Formula: ‘Current T-total + (x*5)’ x is the number of times you translate to a certain direction, in this case the right.
New T-total = 40 + (2*5) = 40 + 10 = 50
As you can see the T-total of T24 is ‘50’
I will now find the algebraic formula for this 10x10 grid, using the T-shape T22.
I will find this formula using an equation, to do so I will find the nth term
My prediction is ‘5*n - x = T-total’ where n = T-number
If I substitute my values it will get:
5*22 - x = 40
110 - x = 40
110 - 40 = x
70 = x
Therefore ‘5n – 70 = T-Total’ is my formula for finding the T-total of any t-shape.
I have used an equation method to find my formula; I could have used the algebraic difference method, to find it.
T22
Tn
As you can see from the above T-shape, we now know how to find all the individual values of the T-shape.
So the algebraic formula would be
n-21+n-20+n-19+n-10+n = ‘5n-70’
As you can see I have arrived at exactly the same formula of ‘5n-70’
Now I will test this formula, on T22
5x22-70=40
This formula will find the T-total of any T-shape on a 9x9 grid, as long as it is translated horizontally or vertically
Translating Vertically
Having found out an algebraic formula for finding the T-total of all horizontally/vertically translated T-shapes on a 10x10 grid, I will now find the simple formula for translating vertically, as I have already found the algebraic formula for a 10x10 grid.
To start off I will find the T-total of the following T-shapes:
T22 and T32
1+2+3+12+22 = 40
11+12+13+22+32 = 90
As you can see the T-total increased by ‘+50’, therefore an easy way to work out the T-total of a T-shape translated vertically downwards is:
New T-total = current T-total + 50
Although this is a perfectly valid formula, it is however not extensive enough, a better formula would be.
New T-total = current T-total + (x*50) where ‘x’ is the number of times the shape is translated downwards
For example if the T-shape T22 was translated once downwards to T32 it would be: new T-total = 40 + (1x50) = 40 + 50 = 90
If translated twice, it would be 40 + (2x50) = 40+ 100 = 140. This is the T-total of T42
I will now use the formula I found earlier to work out the T-total of this same T-shape (T32), just to prove that it works on all shapes as long as they are translated vertically or horizontally.
The formula is ‘5n-70’
So by substituting the values I get:
5x32 – 70 = 90
As you can see the formula works. I will now proceed unto rotation of the T-shapes in a 10x10 grid.
Rotating 90°
I will rotate the T-shape in a 90° direction, and see if I can find a formula to find the T-total of rotated T-shapes.
I will be rotating T23 in a 90° direction; the T-number remains the same, ‘23’, but the T-total will change.
2+3+4+13+23 = 45
15+25+35+24+23=122
As you can see from the table above, the 90° rotated T23 has a T-total of 122, a ‘+77’ increment.
I will now try this on another T-shape to see if I will get the same ‘+77’ increment.
I will use the T-shape of T37
SUM method: 16+17+18+27+37=115
Algebraic Formula (5n-70): 5x37-70 = 115
29+39+49+38+37=192
We also have an increment of ‘+77’, therefore we know that, to find the T-total of a 90° rotated T-shape, we would be able to do so by simply adding ‘77’ to the current T-total.
x = current T-total + 77 (where ‘x’ is the new T-total to be found...)
Now I will find an algebraic formula for finding the T-total of any 90° rotated T-shape in a 10x10 grid. To find this algebraic formula, I will find out a way to find the individual values in the T-shape:
Let’s refer to the T-number as ‘n’
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n+1+n+2+n-8+n+12
=n+n+n+n+n+1+2+12-8 (gather like terms) =‘5n+7’
Therefore my algebraic formula for finding the T-total of any 90° rotated t-shape is ‘5n+7’
I will now test this formula to see if it works, I will test it using the T37 T-shape
Substituting the T-number in place of ‘n’ I get:
5x37 + 7 = 192
The formula works!
Rotating 180°
I will now rotate the T-shape, T25 in 180.
To find the T-total of this T-shape, I will use the algebraic formula 5n-70
Where ‘n’ is the T-number of the T-shape.
5x25-70=55
Just to prove that this is correct: 25+15+5+4+6=52
25+35+45+44+46=195
You can clearly see from the above table of results, the T-total increases by an increment of ‘+140’ every time.
I will now find an algebraic formula for finding the T-total of any 180° rotated T-shape.
To find this algebraic formula, I will find out a way to find the individual values in the T-shape:
Let’s refer to the T-number as ‘n’
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n+10+n+19+n+20+n+21
=n+n+n+n+n+10+19+20+21=5n+70
Check
5n+70
Substitute the T-number into ‘n’
5x25+70= 195
As you can see this formula will find the T-total of any T-shape rotated by180°
So let’s have a recap of all the algebraic formulas we have uncovered so far, and what they solve:
Rotating 270°
I will now find out a formula for finding the T-total of any T-shape rotated by 270°
2+3+4+13+23=45
11+21+31+22+23=108
As you can see the increment was ‘+63’ this allows me to build a simple formula.
x = current T-total + 63 where ‘x’ is equal to the new T-total
Therefore: x = 45 +63
x = 108
New T-total = 108
Although a good formula an algebraic formula is better as it doesn’t require an established T-total to be worked out, all I need is the T-number of the T-shape and I will be able to work it out.
The image above shows exactly what needs to be done to n, the T-number to find the individual numbers in the T-shape.
n+n-1+n-2+n-12+n+8
=n+n+n+n+n+8-1-2-12=5n-7
My formula for finding the T-total of any 180 rotated T-shape is therefore ‘5n-7’
Let’s check to see if it works. I will use the T-shape, T23 to test if this works.
5x23 – 7 = 108
The formula works, just as expected.
I will now show ALL the formulas that can be used in a 10x10 grid to calculate the T-total:
CONCLUSION
To conclude I will now show all the formulas, or the collection of formulas that can be used in an 8x8, 9x9 and 10x10 grid. And highlight their relationships.
What pattern can you notice?
Looking at the table above, I can see that the formula for calculating the horizontal and vertical translations increase by an increment of ‘-7’ every time…which shows that it is a multiple of ‘7’
5n-56(8x8), 5n-63(9x9)
5n-56 + - 7
5n-56 – 7 = 5n-63
Looking at the table above, I can see that the formula for calculating the 90° rotations stays the same… ‘5n+7’
This means if you want to find the 90° rotation of any T-shape on any grid, simply using this formula: ‘5n+7’ will work out its T-total.
I can also see that the formula for calculating the 180° rotations increase by an increment of ‘+7’ as we increased the grid size…which shows yet again that the number is always a multiple of ‘7’
5n+56(8x8), 5n+63(9x9)
5n+56 + 7 = 5n+63
Looking at the table above, I can see that the formula for calculating the 270° rotations stays the same… ‘5n-7’
This means if you want to find the 270° rotation of any T-shape on any grid, simply using this formula: ‘5n-7’ will work out its T-total.
I have noticed all the numbers used in the formulas are multiple of ‘7s’
7, 56, 63, 70…
I have also noticed that all the formulas begin with ‘5n’; this is because they are always 5 numbers in a T-shape.
I have also noticed a clear relationship between the formula and the Grid number.
Tn where ‘n’ = the T-number
And where‘d’ = the grid number (8x8, dxd)
This is an algebraic formula showing the relationship between the T-total formula and the grid size number.
n+n-d+n-2d+n-2d-1+n-2d+1 = n+n+n+n+n+-d+-2d+-2d-1+-2d+1
= 5n-7d
The relationship between the T-total and the grid size is 5 * the T-number of the T-shape – 7 * the grid size.
Proof
Let’s use T23 in the 10x10 grid to prove this relationship
2+3+4+13+23=45
5n-7d where ‘n’ = T-number and‘d’ = grid size number
By substituting the values I get:
5x23 – 7x10
115 – 70 = 45 as you can see there is a relationship.
This concludes my T-total project.
JJ Emelle
