Number Grid Investigation.
Number Grid Investigation
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Task.
My task is to use the box above to do the following things:
* Draw a box around 4 numbers.
* Find the product of the top left number and the right number within this box.
* Do the same with the top right and the bottom left numbers.
* Calculate the difference between these products.
* INVESTIGATE FURTHER...
This assignment is concerned with product differences in different size matrices for different sized grids.
Above is a 10 X 10 grid. We are trying to investigate product differences within this 10 X 10 grid. As you can see I have drawn a 2 X 2 grid inside the 10 X 10 matrix. To find the product difference I do:
Top left number X bottom right - top right X bottom left.
TL X BR - TR X BL
TL = Top left
BR = Bottom right
TR = Top right
BL = Bottom left.
The product difference I found was 10. This was anywhere on the grid where a 2 X 2 grid could be drawn.
To prove this I will take a 2 X 2 grid and use the above formula to work out the product difference. Below is the result of me picking a 2 X 2 grid from this 10 X 10 grid and calculating the product difference.
67
68
77
78
I did (67 X 78) - (68 X 77) = 10
This shows that the product difference is 10.
Below are two more examples to prove it will work anywhere on the grid.
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43
52
53
I did (42 X 53) - (43 X 52) = 10
TL X BR - TR X BL
2
3
22
23
I did (12 X 23) - (13 X 22) = 10
TL X BR - TR X BL
I am now going to prove algebraically that in a 10 X 10 grid, with a 2 X 2 square the difference will always be 10.
X
X+1
X+10
X+11
( (x + 1) (x + 10) ) - x ( x + 11)
(x² + x + 10x + 10) - (x² + 11x)
(x² + 11x + 10) - (x² + 11x)
x² + 11x + 10 - x² - 11x
Difference = 10
This proves algebraically that the difference in a 2 X 2 square is 10.
What can be done now?
Brainstorm
* Vary the shape of box (square, rectangle)
* Vary size of grid (width, shape)
* Try different calculations.
* Use a different pattern of numbers.
Plan.
The above brainstorm explains the different approaches that can be taken in this investigation. Firstly I will vary the size of the box inside the 10 X 10 grid. I will then move my way through the different approaches constantly comparing and making notes on my findings.
Vary the shape of box
We have already worked out the Product difference of a 2 X 2 square within a 10 wide grid. This was always 10, I have used algebra to prove the 'always'.
I am now going to see what happens when I change the box size to a 3 X 3 square.
Here is a 3 X 3 square taken from a 10 wide grid:
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57
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59
So, the calculation that I did was: (TL X BR) - (TR X BL)
(37 X 59) - (39 X 57) = 40.
With the above calculation I discovered that for a 3 X 3 square within a 10 wide grid the product difference is 40.
I will now do another two examples to see if it is the same anywhere in the grid as the 2 X 2 was.
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38
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40
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(28 X 50) - (30 X 48) = 40. This shows again that the product difference is 40.
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(16 X 38) - (18 X 36) = 40.
The above examples show that the product difference of a 3 X 3 grid will always be 40 no matter where it is on the grid.
To prove this above statement I am now going to prove algebraically that the product difference will always be 40.
x
x +
x + 20
x + 22
((x + 2)(x + 20)) - x(x + 22)
(x² + 2x + 20x + 40) - (x² + 22x)
(x² + 22x + 40) - (x² + 22x)
Difference = 40.
Mini Conclusion
So far I have discovered that in a 2 X 2 square the product difference is always 10 and in a 3 X 3 square the product difference is always 40.
I will now try a 4 X 4 square within a 10 wide grid.
0 wide grids with a 4 X 4 square.
Again, I am going to use the formula: (TL X BR) - (TR X BL)
This is an example of a 4 X 4 box taken from a 10 wide grid:
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67
I then did:
(TL X BR) - (TR X BL)
(34 X 67) - (37 X 64) = 90.
90 is the product difference in a 4 X 4 square.
Mini prediction
I am presuming from my above predictions that any 4 X 4 square taken from this 10 wide grid will also show a product difference of 90.
Let's see...
I will now do another two examples to see if my prediction is correct.
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(24 X 57) - (27 X 54) = 90.
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(66 X 99) - (69 X 96) = 90.
Both calculations show a product difference of 90.
I will now prove the above findings algebraically.
x
x + 3
x + 30
x + 33
((x + 3)(x + 30)) - x(x + 33)
(x² + 3x + 30x + 90) - (x² + 33x)
(x² + 33x + 90) - (x² + 33x)
Difference = 90.
What have we noticed so far?
Within a 10 wide grid:
2 X 2 squares product difference is 10.
3 X 3 squares product difference is 40.
4 X 4 squares product difference is 90.
Size of Square
2 X 2
3 X 3
4 X 4
Product Difference
0
40
90
I now need to attempt finding a formula connecting the size of the square inside the 10 wide grid.
Finding a formula for size of square inside 10 wide grid
Firstly I noticed that if you take the '0' of the product difference you are left with square numbers.
, 4, 9 ...
So, I tried to find a way you could get the square numbers.
(2 - 1 = 1) X 10 = 10
The formula would be 10(width of square - 1)
I then tried, using this formula, in a 3 X 3 square.
10(3 - 1) = 20.
This was wrong as my intended answer was 40, not 20. I need to multiply the 10 by 4 somehow.
So, I squared the (3 - 1) to get 4.
10 (3 - 1)²
This worked. The ...
This is a preview of the whole essay
, 4, 9 ...
So, I tried to find a way you could get the square numbers.
(2 - 1 = 1) X 10 = 10
The formula would be 10(width of square - 1)
I then tried, using this formula, in a 3 X 3 square.
10(3 - 1) = 20.
This was wrong as my intended answer was 40, not 20. I need to multiply the 10 by 4 somehow.
So, I squared the (3 - 1) to get 4.
10 (3 - 1)²
This worked. The formula for a 10 wide grid is:
10(width of square - 1)²
I will now try this formula in a 5 X 5 grid. Something I have not yet investigated.
0(5 - 1)² = 160.
Let's see if this is correct.
Here is a 5 X 5 square taken from a 10 wide grid:
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(TL X BR) - (TR X BL)
(22 X 66) - (26 X 62) = 160.
The formula has proven to be correct. I will now check it in another 5 X 5 square just to be sure.
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(54 X 98) - (58 X 94) = 160.
Again the formula is correct.
We can now predict using the formula:
10(width of square - 1)²
the product difference of any size square capable of fitting inside a 10 wide grid.
Size of Square
2 X 2
3 X 3
4 X 4
5 X 5
6 X 6
7 X 7
8 X 8
Product Difference
0
40
90
60
250
360
490
I then noticed that there is a number pattern within the results.
10 40 90 160 250
\ / \ / \ / \ / \
30 50 70 90
\ / \ / \ / \
20 20 20 and so on...
We could use this technique to calculate answers as well as the formula
What next?
Sticking with a 10 wide grid, I am now going to change the shape of the box within it. At the minute I have been using squares, 2 X 2, 3 X 3, 4 X 4 etc. I am now going to change this to say 3 X 2, 4 X 2, 5 X 2 etc to see what effect this has on my formula:
10(width of box - 1)²
As mentioned in my plan, I am working my way down the different approaches and making notes on my findings.
I am now going to use the abbreviation 'n' for 'width of box' in my formula.
10(width of box - 1)² = 10(n-1)²
3 X 2 square inside 10 wide grid.
Below are a few examples of 3 X 2 squares taken from a 10 wide grid.
I am still using the formula:
(TL X BR) - (TR X BL)
Example 1.
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(36 X 48) - (38 X 46) = 20.
Product difference = 20.
My results in a 3 X 3 box showed a product difference of 40.
In a 3 X 2 it is 20.
I will now try this another two times to see if it is the same on any square taken from the grid.
Example 2.
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(22 X 34) - (24 X 32) = 20.
Product difference = 20.
Example 3.
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(87 X 99) - (89 X 97) = 20.
Product difference = 20.
Again, both times the Product difference of a 3 X 2 square is 20.
I will now prove algebraically that the product difference will always be 20 with a 3 X 2 square.
x
x + 2
x 10
x + 12
X(x + 12) - (x + 2)(x + 10) =
(X² + 12x) - (x² + 12x + 20) = 20.
I will now see what the product difference is in a 4 X 2, 5 X 2 etc to see if I can find a general formula and pattern from my results gathered.
4 X 2 squares inside a 10 wide grid
(TL X BR) - (TR X BL)
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(16 X 29) - (19 X 26) = 30.
Product difference = 30.
I will now see if I gain the same product difference from other 4 X 2 squares.
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(60 X 73) - (63 X 70) = 30.
Product difference = 30.
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(12 X 25) - (15 X 22) = 30.
Product difference = 30.
All three attempts show a product difference of 30 in a 4 X 2 square.
So, in a 3 X 2 square we have a product difference of 20. In a 4 X 2 square we have a product difference of 30.
I've noticed an increase of 10.
Mini prediction
I predict in a 5 X 2 square the product difference will be 40.
I will now try my prediction to see if it's correct.
5 X 2 square taken from 10 wide grid
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(43 X 57) - (47 X 53) = 40.
Product difference = 40.
My prediction is correct.
I will now try it again just to be sure.
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(83 X 97) - (87 X 93) = 40.
Again, the product difference is 40.
5 X 2 proven algebraically:
x
x + 4
x + 10
x + 14
X(x + 14) - (x + 4)(x + 10) =
(x² + 14x) - (x² +14x + 40) = 40.
So, information I have gathered:
Size of Square
3 X 2
4 X 2
5 X 2
...
6 X 2
7 X 2
...
Product Difference
20
30
40
...
50
60
...
There is an obvious difference of 10 as the sizes increase.
I now need to find a formula.
Let's try...
(n - 1) X 10
'n' being the width of the box.
3 X 2 =
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87
3 - 1 = 2
2 X 10 = 20
Mini Prediction
I predict using my formula that the Product difference of this 3 X 2 square will be 20.
Let's see...
(75 X 87) - (77 X 85) = 20.
My formula is correct.
I could use this formula to work out any size of square such as:
18 X 2 = 170
97 X 2 = 970 and so on...
Question ??
Will this still be the same if I changed the depth? Maybe if I did say, 3 X 4 or 3 X 6 for example.
Let's see...
3 X 4 square inside 10 wide grid.
Here is a 3 X 4 square taken from a 10 wide grid.
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(TL X BR) - (TR X BL)
(16 X 48) - (18 X 46) = 60
Is this Product difference the same for any 3 X 4 square?
Below is another example.
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40
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60
(TL X BR) - (TR X BL)
(28 X 60) - (30 X 58) = 60.
Yes it is.
What effect would it have on my results if I changed the square size to say 3 X 5?
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(57 X 99) - (59 X 97) = 80.
What about 3 X 6?
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(22 X 74) - (24 X 72) = 100.
The product difference increases by 20 as the depth of the square increases by 1.
Size of square
3 X 4
3 X 5
3 X 6
...
3 X 7
3 X 8
Product difference
60
80
00
...
20
40
I now need to find a new formula connecting the width of the grid, the width of the square and also the depth of the square.
Based on my original formulas:
0 (n - 1)² and 10 (n - 1)
I have now discovered another:
0 (N - 1) (d - 1)
d = depth of square.
N = width of square.
Let's try this formula in a calculation made earlier.
A 3 X 6 grid.
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0 (N - 1) (d - 1)
(N = 3) ... (D = 6)
N - 1 = 2
D - 1 = 5
5 X 2 = 10
0 X 10 = 100.
This formula is correct.
I will now try it again but in the 3 X 4 calculation.
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60
0 ( n - 1 ) ( d - 1 )
(N = 3)
(D = 4)
n - 1 = 2
d - 1 = 3
3 X 2 = 6
6 X 10 = 60.
This is again correct.
Formulas I have discovered so far
Based in a 10 wide grid...
. To calculate the product difference of a 2 X 2, 3 X 3, 4 X 4 etc. The following formula can be used...
10 ( n - 1 )² n = width of square.
2. To calculate 3 X 2, 4 X 2, 5 X 2 etc. The following formula can be used...
10 ( n - 1 )
3. To calculate any sized square inside a 10 wide grid. The following formula can be used...
10 ( n - 1 ) ( d - 1 )
d = depth of square.
To check the third formula I am going to take a random square size to check one more time.
5 X 7
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0 ( n - 1 ) ( d - 1 )
n = 5
d = 7
(n - 1 = 4)
(d - 1 = 6)
4 X 6 = 24
24 X 10 = 240.
Using the above formula, the product difference is correct.
What next?
So far I have changed:
* The size of the box: 2 X 2, 3 X 3, 4 X 4 etc and found a formula.
* The width of the box: 3 X 2, 4 X 2, 5 X 2 etc and found a formula.
* A combination of the two above and found a formula.
Will my formulas work in different sized grids?
So far, all my formulas have been gained from calculations within a 10 wide grid. I am now going to take a different approach in this investigation. I am going to investigate the effect on the formula when I change the grid size to say 12 X 12 or 8 X 8. I am going to take one number higher than ten and one lower to see what happens.
Starting with:
8 X 8 grid.
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First I will see what the Product difference of a 3 X 3 square is.
Here are 3 examples:
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3
9
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1
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9
(TL X BR) - (TR X BL)
(1 X 19) - (3 X 17) = 32.
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(6 X 24) - (8 X 22) = 32
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(41 X 59) - (43 X 57) = 32
Firstly we notice that any 3 X 3 square inside a 8 wide grid is 32.
I will now see what effect this will have if I change the square size to a 4 X 4 grid.
3 examples:
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3
4
9
0
1
2
7
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9
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(1 X 28) - (4 X 25) = 72.
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(5 X 32) - (8 X 29) = 72.
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(33 X 60) - (36 X 57) = 72.
The product difference in a 4 X 4 square is clearly 72. All three examples show this.
5 X 5 squares.
2 examples:
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(1 X 37) - (5 X 33) = 128.
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(28 X 64) - (32 X 60) = 128.
Product difference in a 5 X 5 square is 128.
I will now try and prove a few of the examples above for a 8 wide grid algebraically.
Algebraic proof
This shows why 32 is obtained for the diagonal product difference of a 3 X 3 grid.
x
x + 2
x + 16
x + 18
(TL X BR) - (TR X BL)
x (x + 18) = x² + 18x = 18x²
(x + 2) (x + 16) = x² + 16x + 2x + 32 = 18x² + 32
8 x² - 18x² + 32 = 32
Product Difference = 32.
This shows why 72 is obtained for the diagonal product difference of a 4 X 4 grid.
x
x + 3
x + 24
x + 27
X(x + 27) = x² + 27x = 27x²
(x + 3) (x + 24) = x² + 24x + 3x + 72
= 27x² + 72.
27x² - 27x² + 72 = 72.
Product difference = 72.
This shows why 128 is obtained as the diagonal product difference of a 5 X 5 grid.
x
x + 4
x + 32
x + 36
(TL X BR) - (TR X BL)
x (x + 36) = x² + 36x = 36x²
(x + 4) (x + 32) = x² + 32x + 4x + 128 =
36x² + 128
36x² - 36x² = 128 = 128.
Product difference = 128.
Justifying my results.
A pattern has emerged from my calculations.
Large Grid
Square size
Product difference
8 X 8
2 X 2
8
8 X 8
3 X 3
32
8 X 8
4 X 4
72
8 X 8
5 X 5
28
8 32 72 128 ?
\ / \ / \ / \ /
24 40 56 ?
\ / \ / \ /
16 16 ?
We see that 16 is added onto the 1st difference everytime.
I can now calculate:
56 + 16 = 72
28 + 72 = 200
Mini prediction
I predict that in a 6 X 6 square, the product difference will be 200.
Let's try...
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9
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(1 X 46) - (6 X 41) = 200
Product difference = 200.
My prediction is correct and my number pattern was successful.
Finding a formula for 8 wide grid.
Based on my formula from a 10 wide grid. I predict that the formula for an 8 wide grid will be:
8 (n-1)²
I will now test the above formula in a 7 X 7 square taken from the 8 wide grid.
I will firstly do the calculations:
7 - 1 = 6
n = 6
8 (6)² = 288.
I think the product difference should be 288 for a 7 X 7 grid.
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(1 X 55) - (7 X 49) = 288.
This calculation proves that my formula is correct.
Formula for 10 wide grid => 10(n-1)²
Formula for 8 wide grid => 8 (n-1)²
We see that (n-1)² has to be multiplied by the width of the overall grid.
Mini prediction.
I now predict that in a 12 wide grid the formula will be:
2(n-1)² to work out a 2 X 2, 3 X 3, 4 X 4 etc...
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Take a 4 X 4 square from the grid above.
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(20 X 59) - (23 X 56) = 108.
I predict to work this out using a formula it would be:
2 (n-1)²
2 (4-1)² = 108
My formula is correct.
Now I have established a formula that works for all sized grids as long as you change the first number accordingly.
Width of grid (width of box - 1)²
Only if squares are 2 X 2, 3 X 3, 4 X 4 etc
Say the grid was 108 wide, the formula would be
08 (n-1)²
What next?
I am now going to see if my 2nd and 3rd formulas work in the same way as mentioned before and above.
Formula 2.
10 (n-1) for squares such as 3 X 2, 4 X 2 etc...
Will it be 7 (n-1) for a 3 X 2 square in a 7 wide grid?
Formula 3.
10 (n-1)(d-1)
Will this be 7 (n-1)(d-1) for random boxes in a 7 wide grid?
Let's see...
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38
39
40
41
42
43
44
45
46
47
48
49
Formula 2. 3 X 2 square taken from above.
6
7
8
23
24
25
7(n-1)
n=3 => (3-1) = 2
7 X 2 = 14. The product difference should be 14.
This is correct...
(16 X 25) - (18 X 23) = 14.
Formula 3. 3 X 5 box taken from above.
2
3
4
9
20
21
26
27
28
33
34
35
40
41
42
7 (n-1)(d-1)
2 X 4 = 8
7 X 8 = 56
Product difference should be 56.
This is correct.
(12 X 42) - (14 X 40) = 56.
Both formulas are correct.
You just have to change the first number accordingly to match the width of the grid.
I can now abbreviate my formulas.
Abbreviated formulas
'z' => width of grid.
'n' => width of square within grid.
'd' => depth of square.
For 2 X 2, 3 X 3, 4 X 4 etc...
. z ( n - 1 )²
For 3 X 2, 4 X 2, 5 X 2 etc...
2. z ( n - 1 )
For any random square inside any random width grid.
3. z ( n - 1 )( d - 1 )
I again need to take a different approach in this investigation.
What next?
I have so far varied size of square inside grid and changed the overall size of the grid.
I am now going to use the above formulas in varied grids but using different calculations.
Maybe: (TL X BR) + (TR X BL)
Or (TL X BR) / (TR X BL)
I will firstly try division.
0 wide grid
2 X 2 square.
8
9
28
29
(TL X BR) / (TR X BL)
(18 X 29) / (19 X 28) = 0.98
Let's try it in another box...
72
73
82
83
(72 X 83) / (73 X 82) = 0.1
Although the results are VERY similar, a formula cannot be formed.
I'm guessing that the product difference will not be the same with a 3 X 3 grid.
36
37
38
46
47
48
56
56
58
(TL X BR) / (TR X BL)
(36 X 58) / (38 X 56) = 0.98
56
57
58
66
67
68
76
77
78
(56 X 78) / (58 X 76) = 0.1
Again, although the results are VERY similar, a formula cannot be formed.
I will now try:
(TL X BR) + (TR X BL)
Firstly in a 2 X 2...
31
32
41
42
(TL X BR) + (TR X BL) = 2614
70
71
80
81
Product difference = 11350.
The product difference is not the same.
I do not see that changing the calculations will have any effect on my formulas.
I cannot gain any more information from this approach in the investigation.
Pattern of numbers
I will now try changing the pattern of numbers within the grid. Using formula 3. to see if it effects it in any way.
Formula 3 => Z (n - 1) (d - 1)
But still using the formula:
(TL X BR) - (TR X BL).
Multiples of 2
2
4
6
8
0
2
4
6
8
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
52
54
56
58
60
62
64
66
68
70
72
74
76
78
80
82
84
86
88
90
92
94
96
98
00
02
04
06
08
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
Random square size of 5 X 3.
46
48
50
52
54
66
68
70
72
74
86
88
90
92
94
Z (n-1)(d-1)
0 (5-1)(3-1) = 80
Using the above calculation and formula the Product difference should be 80.
Let's see if it is...
PD = (46 X 94) - (54 X 86) = 320.
The formula is not correct for this approach.
Let's try again:
Random 4 X 3 square.
88
90
92
94
08
10
12
14
28
30
32
34
Z (n-1)(d-1)
0 (4-1)(3-1) = 60.
Is this correct?? Let's see...
(88 X 134) - (94 X 128) = 240.
No.
Both times I've noticed that the product difference is 4 times the amount of the result when taken from formula 3.
Z(n-1)(d-1)
Therefore, the formula =>
4Z (n-1)(d-1) should work.
I will now try this formula in another square.
5 X 4.
4
6
8
0
2
24
26
28
30
32
44
46
48
50
52
64
66
68
70
72
4Z (n-1)(d-1)
40 (5-1)(4-1) = 480. Hopefully, the product difference should be 480.
(4 X 72) - (12 X 64) = 480. This is correct.
What next?
I am now going to find a general formula for a grid with multiples.
In the one above, multiples of 2, the formula was...
4Z(n-1)(d-1)
The multiple number, 2, is squared to give 4 (as in 4Z)
Mini prediction.
I predict that in a grid with multiples of 3, the formula will be
9Z (n-1)(d-1)
Let's try...
3
6
9
2
5
8
21
24
27
30
33
36
39
42
45
48
51
54
57
60
63
66
69
72
75
78
81
84
87
90
93
96
99
02
05
08
11
14
17
20
5 X 4 square taken from above.
2
5
8
21
24
42
45
48
51
54
72
75
78
81
84
02
05
08
11
14
(12 X 114) - (24 X 102) = 1080.
9Z (n-1)(d-1)
90 (5-1)(4-1) = 1080.
The formula is correct.
I can now work out the product differences of grids of all multiples with any sized square within them.
Multiples of:
4
5
6
7
8
5 X 4 square PD's:
920
3000
4320
5880
7680
I will now try some of the above just to double check.
5 X 4 taken from multiples of 5 grid:
20
25
30
35
40
70
75
80
85
90
20
25
30
35
40
70
75
80
85
90
5² = 25.
25Z(n-1)(d-1)
250(5-1)(4-1) = 3000. This is correct.
5 X 4 taken from multiples of 7 grid:
28
35
42
49
56
98
05
12
19
26
68
75
82
89
96
238
245
252
259
266
7² = 49
49Z(n-1)(d-1)
490(5-1)(4-1) = 5880. This is also correct.
I have now completed all the different approaches in this investigation.
My Final Tested Results.
All the formulas now stated below have been tested using algebra and calculation somewhere in this investigation.
'z' => width of grid.
'n' => width of square within grid.
'd' => depth of square.
For 2 X 2, 3 X 3, 4 X 4 etc...
z ( n - 1 )²
So, if for example we were given a 3 X 3 square taken from a 9 wide grid the calculation would be: 9(3-1)².
For 3 X 2, 4 X 2, 5 X 2 etc...
z ( n - 1 )
So, if for example we were given a 7 X 2 square taken from a 8 wide grid the calculation would be: 8(7-1)
For any random square inside any random width grid.
z ( n - 1 )( d - 1 )
So, if for example we were given a 9 X 6 square from a 14 wide grid the calculation would be: 14(9-1)(6-1)
Changing the pattern of numbers within the grid:
I have only tried this approach in a grid with a width of 10.
(Z X (multiple number²)) X (n-1)(d-1)
So, if for example we were given a 6 X 4 square taken from a grid with multiples of 4 in it the calculation would be: (10 X 4²) X (6-1)(4-1).
Based on my original brainstorm, I have carried out all the different approaches outlined. My initial plan simply said that I would take each approach and investigate it fully. I have done this. I therefore think that my investigation has come to an end, as there is nothing else I can test or investigate.
Now completed, final conclusion.
Limitations
I feel that this investigation is perhaps limited in a way. Thinking further into this investigation I've realised that I could have gone into greater depth in my different approaches. For example my final section, changing the numbers in the grid to multiples of a number. I only tried this approach in 10 wide grids. I may have found something different if I'd have tried it in say 8 wide grids or 14 wide grids. I may also have varied the shape of the grid. I could have had an 'L' shaped one like the one below:
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
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
This approach especially could have gone into greater depth. I could have tried cross shape, triangles etc...
I could have made my numbers run vertically rather than horizontal and used this approach in grid with multiples, grids of different shapes, grids of different sizes. The approaches are un-countable.
I also think it would be hard for me to present all my findings and investigations in an interesting way without losing the interest of the reader.
Final conclusion.
In conclusion to the project as a whole I think that I handled the given task very well. I have followed my plan and brainstorm accurately gathering different calculations and formulas as I worked my way down.
I have presented my results and formulas in an appropriate way so as to make sense to the reader. I have challenged my own thoughts by using mini predictions throughout the experiment and tested them using the phrase 'let's see...' I have used algebra where necessary to prove why the formula or calculation I have gathered has worked in the way that it has.
Unfortunately I have not been able to show my results in any form chart but I have used number patterns to make my results more easily understandable.
To gather a larger set of formulas and results and to improve this investigation I could have ventured into some of the approaches written above or gone into greater depth with my approaches carried out. This could prove difficult as presenting it in an interesting way would cause great problems.
Stuart Small.
1 - O D.
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