+5 +5 +5 +5 +5
There is an addition of 5 again-(same as in the horizontal T-shapes of the 9 by 9 grid. Could this be a pattern?)
+40 +40 +40 +40 +40
These totals have an addition of 40 each time.
With this I can see that 40÷8= 5.
From looking at the previous vertical translations on the 9 by 9 grid and the ones done here on the 8 by 8 grid, I can see that the number that is added to get to the next total, when divided by the width of the grid (the grid number) gives you 5, which is the constant number added in the totals of the horizontal t-shapes.
Throughout both grids, the horizontal T-shape totals have an addition of 5. I predict that the next grid will show these same results.
I am now going to try the same thing using a 7 by 7 grid to see if my theory is correct:
Table for horizontal T-shapes:
+5 +5 +5 +5 +5
Once again, there is an addition of 5. The first part of my prediction is correct.
Table for vertical T-shapes:
+35 +35 +35 +35
Each time 35 is being added.
35÷grid number
=35÷7= 5
These results correspond directly to my theory, therefore it is correct.
Part 2:
Using grids of different sizes translate the T-shape to different positions. Investigate the relationship between the T-total, the T-numbers and the grid size.
Grid: 9 by 9
+5 +5 +5 +5 +5 +5
There is an addition of 5 each time.
So the first part of the formula is 5n.
To find the rest: 5n±X (where X= a linear bit)
(5×20)±X=37
100-X=37
X=63
Total formula: t=5n-63 _ This is equivalent to 7×9!
I will check to see that this formula is correct and whether it will work anywhere on the grid:
- 5n-63 2) 5n-63 3) 5n-63
(5×24) -63 (5×40)-63 (5×57)-63
120-63 200-63 285-63
57√ 137 √ 222 √
Another quicker way that this formula can be found out is using an algebraic method:
I noticed that the centre column of the T-Shape goes up in 9's.this is probably because of the grid size (9 by 9). Using this method a formula can be worked out to find any T-Total on this size grid.
n+n-9+n-18+n-19+n-17
= 5n-63
Grid: 8 by 8:
n+n-8+n-16+n-17+n-15
= 5n-56 This is equivalent to 7×8!
Check:
- 5n-56 2) 5n-56 3) 5n-56
(5×22)-56 (5×35)-56 (5×59)-56
110-56 175-56 295-56
54 √ 119 √ 239 √
From these previous grid sizes I can already see a pattern. The last parts of the formulas are all factors of 7. I will try and see if this pattern works out for the next grid size.
Grid: 7 by 7:
n+n-7+n-14+n-13+n-15
= 5n-49 This is equivalent to 7×7!
Once again this is a factor of 7.
Check:
- 5n-49 2) 5n-49 3) 5n-49
(5×18)-49 (5×30)-49 (5×48)-49
90-49 150-49 240-49
41 √ 101 √ 191 √
Since all the formulas contain numbers that are factors of 7, a formula can be calculated for finding out any T-total on any Grid.
For example, using the algebraic method, I will use a T-shape from the 9 by 9 grid:
n=T-number g=grid number
So: n + n-g + n-2g + n-2g+1 + n-2g-1 t=5n-7g
This can be checked using the grid numbers used in the pages above to see whether it is correct.
Grid: 9 by 9 Grid: 8 by 8: Grid: 7 by 7:
- 5n-7g 2) 5n-7g 3) 5n-7g
5n-(7×9) 5n-(7×8) 5n-(7×7)
=5n-63 √ =5n-56 √ =5n-49 √
So no matter what the grid size is, to find the T-total of a number the formula will always be 5n-7g—where g= the grid number.
I will now translate the T-shape to different positions on the grid to find out any more relationships.
For this next part, I have translated the T-shape upside down and tabulated the results going horizontally across the grids:
+5 +5 +5 +5 +5 +5
Using the algebraic method, I can find out the formula to find the T-total:
n + n+g + n+18 + n+2g-1 + n+2g+1
= 5n+7g
As you can see the formula has become positive as the T-shapes have been inverted, whereas before when they were upright, they were negative (5n-7g).
So 5n+(7×9)
= 5n+63
As you can see, this formula is exactly the same as the formula for the upright T-shapes, except now it has a positive sign in it as the T-shapes have been inverted.
Will this be the same for the 8 by 8 grid and the 7 by 7 grid?
8 by 8 Grid:
Again 5 is added each time.
By using the algebraic method we also find out that the formula is 5n+56. So far, my theory is correct.
7 by 7 grid:
Again 5 is being added each time. The formula comes out to be 5n+49.
By inverting the T-shapes, I find that the formulas all change to become positive instead of negative.
