Formula = 5n + 63
= (5x16) + 63
= 143
Formula = 5n – 63
= (5x16) – 63
= 17
16 + 17 + 9 + 18 + 27 = 87
As neither formula worked I decided to use the difference method to again find a new formula.
T-number T-total Difference
10 57 5n
11 62 5n
12 67 5n
13 72 5n
14 77 5n
5n = 5x10
= 50 = 57
-50) = 0 = 7
= 5n + 7
Eg. Formula = 5n + 7
= (11x5)
= 55 + 7
= 62
11 + 12 + 13 + 4 + 22 = 62
Formula = 5n + 7
= (5x14) + 7
= 77
14 + 15 + 16 + 7 + 25 = 77
Formula = 5n + 7
= (5x57) + 7
= 285 + 7
= 292
57 + 58 + 59 + 50 + 68 = 292
I predict to work out the reverse of the sideways T in the opposite direction, you have to change the formula to 7 from –7.
Eg. Formula = 5n – 7
= (5x12) – 7
= 60 – 7
= 53
12 + 11 + 10 + 1 + 19 = 53
Formula = 5n –7
= (5x16) –7
= 80 – 7
= 73
16 + 15 + 14 + 5 + 23 = 73
Formula = 5n – 7
= (5x68) – 7
= 340 – 7
= 333
68 + 67 + 66 + 57 + 75 = 333
- Uses grids of different sizes. Translate the T– shape to different positions. Investigate relationships between the T-total, the T-numbers and the grid size.
Here is a grid 3 by 3:
I am going to concentrate on the original translation of the T. The T-total of 8 in this grid is 19. Adding another column (to make it 3 by 4) will give me two T’s to do the difference method with.
T-number T-total Difference
10 22 5n
11 27 5n
5n = (5x10)
= 50 –22 = 28
= 5n – 28
Eg. Formula = 5n – 28
= (5x11) – 28
= 55 – 28
= 27
T-number T-total Difference
12 25 5n
13 30 5n
14 35 5n
The difference between the T-totals is always 5n.
5n – x = (5x12) – x
= 60 – x
25 = 60 – x
x = 35
= 5n – 35
Eg. Formula = 5n –35
= (5x14) – 35
= 70 – 35
= 35
T-number T-total Difference
14 28 5n
15 33 5n
16 38 5n
17 43 5n
It is 5n again.
5n – X = (14x5) –X
= 70 – X
28 = 70 – X
X = 42
= 5n – 42
Eg. Formula = 5n – 42
= (5x17) – 42
= 85 – 42
= 43
I noticed a pattern in the formulas.
3x4 5n – 28 7
3x5 5n – 35 7
3x6 5n - 42 7
The difference between the number you take away increases by 7 every time. Therefore I predict that for a 3 by 3 grid the formula will be 5n – 21.
We already know that the T-total of this T is 19
Eg. Formula = 5n – 21
= (5x8) – 21
= 40 – 21
= 19
Therefore I predict that for a 3 by 8 grid (8 columns) the formula will be 5n – 56. This is because for a 3 by 6 grid the formula was 5n – 42. As the formula increases by 7 for each column, there are two more columns than by the 3 by 6 grid. Therefore there will be 14 more to take away.
Eg.
Formula = 5n –56
= (5x18) – 56
= 90 – 56
= 34
1 + 2 + 3 + 10 + 18 = 34
Formula = 5n – 56
= (5x22) – 56
= 110 – 56
= 54
5 + 6 + 7 + 14 + 22 = 54
So a column is added, the number taken away in the formula is increased by 7.
Eg.
From this a new formula can be formed, using 5n as a starting point.
5n – 7x
5n is 5 times the T-number.
7x is 7 times the number of columns.
Eg.
Formula = 5n – 7x
= (5x26) – (7x10)
= 130 – 70
= 60
5 + 6 + 7 + 16 + 26 = 60
Formula = 5n – 7x
= (5x29) – (7x10)
= 145 – 70
= 75
8 + 9 + 10 + 19 + 29 = 75
From varying the amount of columns is shows that the amount of columns does affect the formula. But does the amount of rows?
Formula = 5n – 7x
= (5x8) – (7x3)
= 40 – 21
= 21
8 + 5 + 1 + 2 + 3 = 21
Formula = 5n – 7x
= (5x11) – (7x3)
= 55 – 21
= 34
11 + 8 + 4 + 5 + 6 = 34
Formula = 5n – 7x
= (5x11) – (7x3)
= 55 - 21
= 34
11 + 8 + 4 + 5 + 6 = 34
Varying the number of rows of the 3 by 3 grid has not affected the 3 by 3 formula. Here is a 3 by 6 grid with 3 extra rows, so it is a 6 by 6 grid.
Formula = 5n – 7x
= (5x27) – (7x6)
= 135 – 42
= 93
14 + 15 + 16 + 21 + 27 = 93
Formula = 5n – 7x
= (5x35) – (7x6)
= 175 – 42
= 133
35 + 29 + 22 + 23 = 24 = 133
From this you can see that the number of rows does not affect the formula. Here is a list of T-totals with T-numbers and it’s grid size.
I predict that if you rotate the T 180 degrees that the formula will be, 5n + 7x.
Formula = 5n + 7x
= (5x2) + (7x3)
= 10 + 21
= 31
2 + 5 + 7 + 8 + 9 = 31
Formula = 5n + 7x
= (5x4) + (7x6)
= 20 + 42
= 62
4 + 10 + 15 + 16 + 17 = 62
Formula = 5n + 7x
= (5x20) + (7x8)
= 100 + 56
= 156
20 + 28 + 35 + 36 = 37 = 156
I then went to find a formula for varying grid size, with the T on its side.
Eg.
T-number = 4
T-total = 27
T-number T-total Difference
5 32 5
6 37 5
Formula = 5n + 7
T-number T-total Difference
6 37 5
7 42 5
8 47 5
Formula = 5n + 7
This formula did not seem to change when I added more columns, so the columns might not affect the formula, but the rows could. This formula also works for the 3 by 3 grid.
Formula = 5n + 7
= (5x4) + 7
= 20 + 7
= 27
4 + 5 + 6 + 3 + 9 = 27
So I decided to vary the number of rows in the grid. Here is a 3 by 3 grid.
T-number T-total Difference
4 27 15
7 42 15
Formula = 15n – X
27 = (15x4) – X
27 = 60 – X
X = 33
= 15n – 33
= (15x7) –33
= 105 – 33
= 73
Using the difference method does not give a correct formula for this grid. I tried a grid with more rows, so more T-numbers and T-totals.
T-number T-total Difference
4 27 15
7 42 15
10 57 15
13 72 15
16 87 15
19 102 15
22 117 15
Formula = 15n
= (15x19)
= 285 – 102
= 183
= 15n –183
This formula only works for one T-number, therefore each T-number has an individual formula.
The number taken away from 15n increases by 30 for every row.
So each row is worth an extra 30 taken away from 15n. This was not particularly useful though for finding a formula. So, I decided to see if I could see any patterns in the data. This is what I found.
22 – 4 = 18x10 = 180
13 – 4 = 9x10 = 90
What I was trying to do was to develop a formula from the first formula with 3 rows and T-number 4. For example the number 90, + 33 = 123, which is the number taken away for the formula for T-number 13. 180 added to 33 is 213, which is the number taken away from 15n, for the T-number 22. This has to be expressed into a formula, so here is.
15n – (10(n - 4) + 33)
Formula = 15n – (10(n - 4) + 33)
= (15x7) – (10(7 – 4) + 33)
= 105 – (10x3 + 33)
= 105 (30 + 33)
= 105 – 63
= 42
7 + 8 + 9 + 6 _+ 12 = 42
Formula = 15n – (10(n - 4) + 33)
= (15x13) – (10(13-4) + 33)
= 195 – (10x9 + 33)
= 195 – 123
= 72
13 + 14 + 15 + 12 + 18 = 72
Formula = 15n – (10(n - 4) + 33)
= (15x22) – (10(22-4) + 33)
= 330 – (10x18 + 33)
= 330 – (180 + 33)
= 330 – 213
= 117
22 + 23 + 24 + 21 + 17 = 117
As the formula doesn’t change when a column is added, this is the general formula for this sideways T.
Eg.
Formula = 15n – (10(n – 4) + 33)
= (15x7) – (10(7 - 4) + 33)
= 105 – (10x3 + 33)
= 105 – (30 + 33)
= 105 – 63
= 42
7 + 8 + 9 + 3 + 15 = 42
Formula = 15n – (10(n – 4) + 33)
= (15x16) – (10(16 - 4) + 33)
= 240 – (10x12 + 33)
= 240 –153
= 87
16 + 17 + 18 + 12 + 24 = 87
Formula = 15n – (10(n – 4) + 33)
= (15x27) – (10(27 - 4) + 33)
= 405 – (10x23 + 33)
= 405 – ( 230 + 33)
= 405 – 263
= 142
27 + 28 + 29 + 23 + 35 = 142
I therefore predict if you reverse the T round the other way that the formula will be, 15n + (n + 4)-10 + 33).
Formula = 15n + (n + 4)-10 + 33)
= (15x9) + (9 + 4)-10 + 33)
= 135 + (13x-10 + 33)
= 135 + (-130 + 33)
= 135 – 97
= 38
9 + 8 + 7 + 1 + 13 = 38
Formula = 15n + (n + 4)-10 + 33)
= (15x30) + (30 + 4)-10 + 33)
= 450 + (34x-10 + 33)
= 450 + (-340 + 33)
= 450 – 373
= 72
30 + 29 + 28 + 22 + 34 = 143
Formula = 15n + (n + 4)-10 + 33)
= (15x22) + (22 + 4)-10 + 33)
= 330 + (26x-10 + 33)
= 330 + (-260 + 33)
= 330 – 227
= 103
22 + 21 +14 + 20 + 26 = 103
