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Maths T-totals coursework
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Introduction
GCSE T-totals Coursework
Introduction
In this project I will be investigating the formula, patterns and relationships between the t-numbers, t-shape and t-totals in different sized grids 10x10, 9x9, 8x8, 7x7.
I have a grid nine by nine starting with the numbers 1-54. There is a shape in the grid called the t-shape which is highlighted in red shown in the table below.
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 |
28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 |
37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 |
The t-number is the number at the bottom of the t-shape which is 20
The t-total is all the numbers in the t added up together which is 1+2+3+11+20=37
T-number | T-total |
20 | 37 |
21 | 42 |
22 | 47 |
23 | 52 |
24 | 57 |
25 | 62 |
26 | 67 |
As you can see:
The t-number increases by 1 each time.
The t-total increases by 5 each time is there a link?
20x5=100
100-63=37 the t-total
The link between 63 and 9 is 7 because 7x9=63
So the formula is T-number x 5 (7x9)
5n –number-7x9
How did I work out this and what can we do with this formula?
The formula starts with 5 as there is a rise between the t-total of 5 each time. We then -63. I got this number by working out the difference between the t-number and the other numbers in the t-shape. E.g.
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 |
28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 |
37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 |
Middle
51
...read more.
52
53
54
5x48-63=t-total
5x48-63=177
Check to prove formula
T-total=29+30+31+39+48=177
This proves my formula works.
I will now try using grids of different sizes and translating the t-shape into different positions. I will then investigate the relationship between the t-number, t-total and grid size.
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
11 | 12 | 13 | 14 | 15 | 16 | 171 | 181 | 19 | 20 |
21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |
41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |
51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |
T-number=22
T-total=40
The t-number and t-total have both risen even though the t-shape is in the same place as it was before. The t-number has risen by 2 and the t-total by 3. This is because I am now using a ten by ten grid. I will now try and find some more t-numbers and t-totals for some more t-shapes on this grid.
T-number=55
T-total=34+35+36+45+55=205
Now I will try and find a formula for the ten by ten grid. I will first find the difference between the t-total and other numbers in the t-shape.
22-1=21
22-2=20
22-3=19
22-12=10
Total=70
Here is another way to figure out the formula
n-21 | n-20 | n-19 |
n-10 | ||
n | ||
n+n-10+n-20+n-21+n-19
So this would mean that my formula will be 5n-70
I will try and test my new formula
5n-70=t-total
5x22-70=40
The same formula can be used for each grid with only changing the last number in the formula. I will now try it on a different sized grid to prove this.
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 |
25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 |
33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |
41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 |
49 | 50 | 51 | 52 | 53 | 54 | 55 | 56 |
57 | 58 | 59 | 60 | 61 | 62 | 63 | 64 |
T-number=18
T-total=1+2+3+10+18=34
Difference
18-1=17
18-2=16
18-3=15
18-10=8
Total=56
n-17 | n-16 | n-15 |
n-8 | ||
n | ||
n+n-8+n-17+n-16+n-15= 5n-56
5n-56=t-total
5x18-56=34
As you can see by changing the grid size we have to change the end of the formula but we still keep the rule of how you get the number to minus in the formula.
So I can find formulas for all grid sizes, which are shown below in the table
Grid width | Formula |
4x4 | 5n-28 |
5x5 | 5n-35 |
6x6 | 5n-42 |
7x7 | 5n-49 |
8x8 | 5n-56 |
9x9 | 5n-63 |
10x10 | 5n-70 |
Conclusion
The difference between each t-total is 5 so again the formula will start with 5, I will then multiply 5 with 12 the t-number. 5x12=60 we then minus 7 to get 53 the t-total.
So therefore the formula will be:
5n-7
To check and make sure the formula works we do:
n-11 | ||
n-2 | n-1 | n |
n+7 | ||
n+n-1+n-2+n-11+n+7=5n-7
I will now rotate the t-shape 180 degrees from its last position. I predict that the same will happen as before and the equation will stay the same but the minus sign will change to a plus sign. I will prove this below.
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 |
28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 |
37 | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
46 | 47 | 48 | 49 | 50 | 51 | 52 | 53 | 54 |
T-number=10
T-total= 10+11+12+3+21=57
To find the formula 5n+7 that I predicted we do:
5x10(t-number) =50+7=57
To check the formula we do:
n-7 | ||
n | n+1 | n+2 |
n+11 | ||
n+n+1+n+2+n+11+n-7=5n+7
The formula is correct.
Conclusion
In this coursework I found many ways to solve the problem with the t-shape being in various different positions and different grid sizes. I have made my calculations less difficult by creating a formula which changes according to the different circumstances.
Below I have shown the different formulas I have come up with these only apply to the 9x9 grid.
5n-63
After 180 degrees rotation
5n+63
270 degrees rotation
5n-7
90 degrees rotation
5n+7
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