T-total= 37
+5
T-total=42
+5
T-total=47
Every time the T moves right the T-total is +5 so this means all the t-totals have a relationship with the 5* table.
So if I * the T-number by 5 I get 100 and then if we go back to diagram that showed T algebraically and use 63 that I worked out from it. Then if I -63 from 100 I get 37 and that is the t-number.
So if I put that in a formula it looks something like this N=T+63
5
The relationship between the T-number, T-total and the grid size
The examples I used in the first section only worked for a 9*9 size grid and to find so to find the relationship between the grid size I have to try the same thing with different grid sizes and then notice a pattern.
This is the T is written algebraically for a 9*9 size grid add all the numbers in the T up I get 63
1st 3 rows of a 8*8 grid
If I change the grid size we change this part of the formula. Because every time you move up or down you + or – the grid size. I have chosen W to represent this in the formula.
This is the t written algebraically for a 8*8 size grid. Add all the numbers in the T up and you get 56
This is the T written algebraically for a 10*10 size grid, add all the numbers up in that T and you get 70
So this means to find ? in N=T+?
5
I need to find a pattern in the numbers 56,63,70. Because I found that for an 8,9 and 102 size grids those were the numbers. If I added all the numbers up in the T up.
So there is an obvious pattern in those numbers and that is +7 and they are in the 7 times tables. And the ? is the grid size*7 because:
and these all link with what I found out on the bottom of the last page. So to put this into the formula it looks like this N=T+(7w)
5
w=the size of the grid and I need to put it in brackets because it need to be worked out before the rest of the formula.
Now I have found this out I can now put the formula onto practise to see if it is correct. I will practise on this 5*5 grid.
So to work out if my formula is correct I should be able to find the T-number from the T-total so 25+(7*5)=12
5
25+35 =12
5
60 =12
5
and that works out correct so now I know that my formula is correct.
The relationship between the T-total, T-number, grid size and the transformation
So get from N to N we have translated it so I need to include the translation in. the translation is (4,3) and as I have explained at the very beginning every time you move across you +1 and if you move down you +9 on this grid. So instead of writing (4,3) i can write (+4,+27) because I moved across the X axis 4 time and down the Y axis 3 times. So to give it a general term and use X to represent times moved across the X axis, Y to represent times moved across the Y axis and w to represent the grid size it looks like this (+X,(+wY) so it will work if I put it in any grid.
That is only to work out how to get from the first N(N1) to the second N(N2) what I need to find out is how to get from the N1 to the second T-total(T2). So if we put the translation into a formula to find T2 from N1 this is what it looks like. T2=5[(n+x)+(wY)]-(7w)
Replacing the letters for numbers in the example I should be able to work it out like this T2=5[(20+4)+(3*9)]-(7*9)
=5[24+27]-63
=5[51]-63
=255-63
=192
so now to find out if my formula is correct I will work out T2 by adding the numbers in the T, 32+33+34+42+51=192
That concludes my maths coursework…
