To Find the Diagonal Difference Taken From Small Nxn Grids, Like 3x3.
Maths Course Work
TASK B: Diagonal Difference
Aim:
The aim of this investigation is to find the diagonal difference taken from small nxn grids, like 3x3. Then to experiment with larger grid sizes to find a pattern or formula to deduce further results
Introduction:
I will be compiling grids taken from an 8x8 grid like the one below; the grids will start at 2x2 then work up to the full 8x8. (Note it is impossible to find the diagonal difference in a 1x1 grid because their needs to be four or more numbers).
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9
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63
64
This is the full 8x8 grid. From this grid I will be taking smaller grids. The shaded numbers is just an illustration of the 3x3 grid.
0
1
2
8
9
20
26
27
28
This is the 3x3 grid; I will be multiplying the opposite corners then finding the difference. For example 10 x 28 = 280
12 x 26 = 312
312 - 280 = 32
Note take away the smaller number from the bigger number, this is to ensure you get a consistent result.
The investigation
I will start my investigation with the 2x2 grids. As there are many values for the smaller grids like 2x2 I will take my values from the larger grid in a distinct fashion. That is starting from the lowest value, top left, then moving down and right towards the bottom right. With grids like the 7x7 once I have taken the values from one angle I will then take them from the opposite angle. Like illustrated below.
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64
2
9
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9
0
27
28
2 x 9 - 1 x 10 = 8
20 x 27 - 19 x 28 = 8
0
1
28
29
8
9
36
37
1 x 18 - 10 x 19 = 8
29 x 36 - 28 x 37 = 8
The 2x2 Grids
The diagonal difference from the 2x2 grids = 8
Then 3x3 Grids
2
3
46
47
48
9
0
1
54
55
56
7
8
9
62
63
64
3 x 17 - 1 x 19 = 32
48 x 62 - 46 x 64 = 32
0
1
2
6
7
8
8
9
20
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5
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26
27
28
22
23
24
2 x 26 - 10 x 28 = 32
8 x 22 - 6 x 24 = 32
The Diagonal Difference from the 3x3 grids = 32
Then 4x4 Grids
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9
20
21
22
9
0
1
2
27
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30
7
8
9
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35
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28
43
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46
4 x 25 - 1 x 28 = 72
22 x ...
This is a preview of the whole essay
8 x 22 - 6 x 24 = 32
The Diagonal Difference from the 3x3 grids = 32
Then 4x4 Grids
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9
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9
0
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27
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30
7
8
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35
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4 x 25 - 1 x 28 = 72
22 x 43 - 19 x 46 = 72
0
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28
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31
8
9
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36
37
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39
26
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34
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37
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55
3 x 34 - 10 x 37 = 72
31 x 52 - 28 x 55 = 72
The Diagonal Difference from the 4x4 grids = 72
Then 5x5 Grids
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9
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9
0
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27
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31
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5 x 33 - 1 x 37 = 128
23 x 51 - 19 x 55 = 128
0
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32
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9
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36
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46
60
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64
4 x 42 - 10 x 46 = 128
32 x 60 - 28 x 64 = 128
The Diagonal Difference from the 5x5 Grids = 128
Then 6x6 Grids
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9
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9
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32
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6 x 41 - 1 x 46 = 200
24 x 59 - 19 x 64 = 200
0
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48
5 x 50 - 10 x 55 = 200
8 x 43 - 3 x 48 = 200
The Diagonal Difference from the 6x6 Grids = 200
Then 7x7 Grids
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7 x 49 - 1 x 55 = 288
8 x 50 - 2 x 56 = 288
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6 x 58 - 10 x 64 = 288
5 x 57 - 9 x 63 = 288
The Diagonal Difference from the 7x7 grids = 288
I've decided to leave out the 8x8 grid until I find a formula, I will then use the formula to find out the diagonal difference of an 8x8 grid then check using the above process (long hand).
Now I will try to find a formula to predict the diagonal difference in an 8x8 square.
8
2x2 Grid
208
6
216
24
224
32
3x3 Grid
232
40
240
48
248
56
256
64
264
72
4x4 Grid
272
80
280
88
288
7x7 Grid
96
296
04
304
12
312
20
320
28
5x5 Grid
328
36
336
44
344
52
352
60
360
68
368
76
376
84
384
92
392
?8x 8 Grids?
200
6x6 Grid
400
I noticed when I compared the 8 times table to the diagonal differences taken from the above grid sizes, (2x2, 3x3, 4x4, 5x5, 6x6, 7x7) that they all fitted into the 8 times table, and that there might be pattern concerning how many PRIME numbers in between each square grid.
In between 2x2 and 3x3 = 2
3x3 and 4x4 = 4
4x4 and 5x5 = 6
5x5 and 6x6 = 8
6x6 and 7x7 = 10
There is a pattern, the number of prime numbers in between each square grid goes up in the two times table.
This means that the difference between 7x7 and 8x8 would be 12; therefore the 8x8 grid may have the diagonal difference of 392.
It would seem that I have found something, though at the moment it is not all that useful unless I can turn it into a formula. The advantages of this being in the form of a formula is that, (when I find the right one) I could instantly work out the diagonal difference of a 8x8 grid without even knowing the diagonal difference of the 7x7grid.
FORMULAS
The first formula I came up with:
3x3 = 8 + 8 x 3 =32
4x4 = 32 + 8 x 5 =72
5x5 = 72 + 8 x 7 =128
6x6 = 128 + 8 x 9 =200
7x7 = 200 + 8 x 11 = 288
Therefore
8x8 = 288 + 8 x 13 =392 (again)
The reason why I put 13 down is because this formula use's a similar pattern to above (2times table) for the next formula you have to +2 to the multiplier. This formula does work, though you need to know the values of previous grid sizes for it two work; which is why I can not work out the 2 x 2 grid because I do not know the values of the grid size 1x1. (I need at a minimum of four numbers to get the diagonal difference values, also note that I cannot work out any grid sizes above 8x8; because this is the biggest and whole number square).
Second attempt:
(n-1)² x y
n = the size grid that you want work out. I.e. 8x8.
Y = the size of the main grid, or the grid that the smaller grids were compiled from. So in this scenario Y = 8. Therefore;
(n-1)² x 8
And as it's the 8x8 grid I' am investigating
(8-1)² x 8 = 392
This formula could be the one I set out to find, it does not need any values from previous grids.
(2-1)² x 8 = 8
(3-1)² x 8 = 32
(4-1)² x 8 = 72
(5-1)² x 8 = 128
(6-1)² x 8 = 200
(7-1)² x 8 = 288
(8-1)² x 8 = 392
Out of curiosity I wonder what 1x1 grid equals.
(1-1)² x 8 = 0
Formula Explained (technical)
The main part to the formula is the (n-1) part, this solves not needing smaller grid values because when you take the grid size your researching (8-1) then you take away the 1 you get (8-1) = (7).
The next part is to square the number in this case (7)=(7)² = 49.
Then you need to times the 49 by the size of the main grid (y=8) therefore 49 x 8 = 392.
The diagonal Difference of 8x8 Grid found. Now let's see if I was right.
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8 x 57 = 456
x 64 = 64
456 - 64 = 392
I was correct, the 8x8 grid has a diagonal difference of 392, and this means that all of my methods proved accurate. The most efficient of my methods is the main formula, (n-1)² x y, this is purely because you do not need any values from previous grids to complete the formula.
I have tested the formula on various size grids from 2x2 through to 8x8, and my formula returns the same results as those I did earlier long hand, and therefore I conclude that my formula is an accurate way to calculate grids of this type.
Out curiosity I wonder whether my formula will work for other sized main grids.
So, my formula goes
(n-1)² x y
what if in this case y = 5, to work out a 3x3 grid from a 5x5 grid would equal:
(3-1)² x 5 = 20
the long hand version of a 5x5 grid:
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9
0
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9
20
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25
And the 3x3 grid
2
3
6
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8
1
2
3
3 x 11 = 33
x 13 = 13
33 - 13 = 20
So from this I also conclude that the formula (n-1)² x y is a universal formula for working out various grids from various larger grids. Note, it is not possible to work out the larger grid from a smaller grid.
Conclusion
My final conclusion, many of my methods proved accurate, the relationship with the 8 times table was very helpful in creating the formulas. This lead me onto a pattern which gave me all the results that I need to work on the formulas, and also shows the relations in a clearer view. The 2 times table also came into play, it was this table that helped me plot where the results were likely to be. The formula to work out the next grid (3x3 = 8 + 8 x 3) reference did work and was useful, though it required previous diagonal difference value from smaller grids and I couldn't manipulate it so that It incorporated a more universal use. Instead I came up with ((n-1)²xy) this formula did work, and It also works for other main grid sizes. Overall, my investigation was a success, and I did complete my Aim.
Rough Notes
These are my notes which include a graph; these notes were created based on my background Knowledge which helped me lead up to the final formula.
A Graph to Show the Diagonal Difference of nxn squares
My graph shows my results clearly, the x-axis (horizontal) shows the grid sizes from 2x2 - 8x8, the Y-axis (vertical) shows the diagonal difference from 0 up to 500, this also relates to the 8 times table (page 5). From graph you can see how the gap in-between the values increases as you move up to the bigger grids like 8x8.
Results Table
2x2 Grids
8
3x3 Grids
32
4x4 Grids
72
5x5 Grids
28
6x6 Grids
200
7x7 Grids
288
8x8 Grids
392
These are the same results gathered from all my methods, they all proved consistent.
Maths Course Work - 02/05/2007 - By Andrew Barnes.
