NUMBER GRID
Look at this number grid:
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
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
00
0 x 10 grid
* A box is drawn round four numbers.
* Find the product of the top left number and the bottom right number in this box.
* Do the same with the top right and bottom left numbers.
* Calculate the difference between these products.
Investigate further.
The purpose of this investigation is to prove or disprove that there is a correlation between the products of corner numbers of any size box within any size grid. I shall calculate the diagonal difference (d) for a box. There are two ways to calculate the difference between the products:
V
W
Y
Z
i. W x Y - V x Z
ii. V x Z - W x Y
I shall begin with the 10 x 10 grid, as shown above, and a 2 x 2 box.
2 x 2 box #1
I have substituted numbers for the letters in my two formulae.
2
1
2
i. d = 2 x 11 - 1 x 12 = 22 - 12 = 10
ii. d =1 x 12 - 2 x 11 = 12 - 22 = -10
d = +/-10
I will use formula i, because formula ii creates negative numbers, which could make my calculations more complex than necessary.
2 x 2 box #2
2
3
22
23
d = 13 x 22 - 12 x 23
d = 286 - 276
d = 10
I think that d = 10 for any 2 x 2 box in a 10 x 10 grid. I will test this theory again, using the highest numbers possible in the grid, to check that it is not a coincidence.
2 x 2 box #3
89
90
99
00
d = 90 x 99 - 89 x 100
d = 8910 - 8900
d = 10
Algebraic proof: formula #1:
+1
+10
a
a + 1
a + 10
a + 10 + 1
d = (a + 1) (a + 10) - a (a + 10 + 1)
= a2 + 10a + a + 10 - a2 - 10a - a
= 10
I have noticed that d = 10, and the grid is 10 columns (G) wide. To test if this is a coincidence, I have amended my formula.
Algebraic proof: formula #2:
+1
+G
a
a + 1
a + G
a + G + 1
d = (a + 1) (a + G) - a (a + G + 1)
= a2 + aG+ a + c - a2 - aG - a
= G, where G = 10, d = 10
I will test formula #2 using a 2 x 2 box on a different size grid. I will use a 5 x 5 grid.
2 x 2 box in a 5 x 5 grid
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
21
22
23
24
25
5 x 5 grid
G = 5
2 x 2 box #4
2
6
7
d = 2 x 6 - 1 x 7
d = 12 - 7
d = 5
2 x 2 box #5
7
8
2
3
d = 8 x 12 - 7 x 13
d = 96 - 91
d = 5
2 x 2 box #6
9
20
24
25
d = 20 x 24 - 19 x 25
d = 480 - 475
d = 5
The results show that d = G. I will use formula #2 from now on.
I shall go back to using a 10 x 10 grid. I shall test the formula using different sized boxes. I shall draw up a table to record my results. I shall compare the results to see if there is a pattern forming. This would prove correlation.
NB: To save time I shall only indicate the corner numbers in each box. The other numbers in the box are irrelevant for these calculations.
3 x 3 box #1
3
...
This is a preview of the whole essay
The results show that d = G. I will use formula #2 from now on.
I shall go back to using a 10 x 10 grid. I shall test the formula using different sized boxes. I shall draw up a table to record my results. I shall compare the results to see if there is a pattern forming. This would prove correlation.
NB: To save time I shall only indicate the corner numbers in each box. The other numbers in the box are irrelevant for these calculations.
3 x 3 box #1
3
21
23
d = 3 x 21 - 1 x 23
d = 63 - 23
d = 40
3 x 3 box #2
2
4
32
34
d = 14 x 32 - 12 x 34
d = 448 - 408
d = 40
3 x 3 box #3
78
80
98
00
d = 80 x 98 - 78 x 100
d = 7840 - 7800
d = 40
Algebraic proof:
+2
+2G
a
a + 2
a + 2G
a + 2G +2
d = (a + 2) (a + 2G) - a (a + 2G + 2)
= a2 + 2aG + 2a + 4G - a2 - 2aG - 2a
= 4G, where G = 10, d = 40
So far my theory is correct. I shall continue to test it on boxes of increasing in size. I feel it is only necessary to show one numeric and one algebraic calculation for these boxes.
4 x 4 box #1
4
31
34
d = 4 x 31 - 1 x 34
d = 124 - 34
d = 90
Algebraic proof:
+3
+3G
a
a + 3
a + 3G
a + 3G +3
d = (a + 3) (a + 3G) - a (a + 3G + 3)
= a2 + 3aG + 3a + 9G - a2 - 3aG - 3a
= 9G, where G = 10, d = 90
5 x 5 box #1
5
41
45
d = 5 x 41 - 1 x 45
d = 205 - 45
d = 160
Algebraic proof:
+4
+4G
a
a + + 4
a + 4G
a + 4G +4
d = (a + 4) (a + 4G) - a (a + 4G + 4)
= a2 + 4aG + 4a + 16G - a2 - 4aG - 4a
= 16G, where G = 10, d = 160
The results so far are:
10, 40, 90, 160
There is a pattern, which is easier to recognise if written:
x x x
x 1
2 x 2
3 x 3
4 x 4
5 x 5
d
0
40
90
60
G
4G
9G
6G
2G
22G
32G
42G
The pattern shows consecutive square numbers. They are also the square of one less than the square size (x) that produced them.
So:
d = G (x - 1) 2
I have extended the results table and added my predicted results, in red, for the remaining box sizes.
x x x
x 1
2 x 2
3 x 3
4 x 4
5 x 5
6 x 6
7 x 7
8 x 8
9 x 9
0 x 10
d
0
40
90
60
250
360
490
640
810
G
4G
9G
6G
25G
36G
49G
64G
81G
2G
22G
32G
42G
52G
62G
72G
82G
92G
I shall now see if my predicted results are correct by testing three of these box sizes.
6 x 6 box #1
6
51
56
d = G (x - 1) 2
= 10 (6 - 1) 2
= 10 (5) 2
= 10 (25)
= 250
d = 6 x 51 - 1 x 56
d = 306 - 56
d = 250
8 x 8 box #1
8
71
78
d = G (x - 1) 2
= 10 (8 - 1) 2
= 10 (7) 2
= 10 (49)
= 490
d = 8 x 71 - 1 x 78
d = 568 - 78
d = 490
0 x 10 box #1
0
91
00
d = G (x - 1) 2
= 10 (10 - 1) 2
= 10 (9) 2
= 10 (81)
= 810
d = 10 x 91 - 1 x 100
d = 910 - 100
d = 810
The tests proved my theory correct. I conclude that this investigation has proved that there is correlation between square boxes on a grid. The master formula is:
d = G (x - 1) 2
To investigate further I need to change one of the variables.
* Grid size, from 10 x 10, to e.g. 5 x 5, 7 x 7, 9 x 9.
* Shape, from a square, to e.g. a rectangle, a cross, a T-shape.
I have chosen to investigate rectangle boxes on a 10 x 10 grid. I think that there will be a master formula for any size rectangle in any size grid. I think that it will be more complex because I will have to take into account the fact that a rectangle has sides of differing lengths, opposed to a square having sides all the same length.
Rectangles can be drawn in two ways:
Rectangle 2
Rectangle 1
I will call rectangle 1 a 4 x 2 ( 4 columns by 2 rows) and rectangle 2 a 2 x 4 ( 2 columns by 4 rows).
There are many different possible sizes of rectangle within a 10 x 10 grid:
2 x 10
2 x 9
2 x 8
2 x 7
2 x 6
2 x 5
2 x 4
2 x 3
3 x 10
3 x 9
3 x 8
3 x 7
3 x 6
3 x 5
3 x 4
4 x 10
4 x 9
4 x 8
4 x 7
4 x 6
4 x 5
5 x 10
5 x 9
5 x 8
5 x 7
5 x 6
6 x 10
6 x 9
6 x 8
6 x 7
7 x 10
7 x 9
7 x 8
8 x 10
8 x 9
9 x 10
This list shows 36 possible sizes. If the orientation of the rectangle produces a different value for d, then there will be 72 possible sizes.
My first task is to ascertain if the orientation of the rectangle affects the diagonal difference.
3 x 2 rectangle #1
2
3
1
2
3
d = 3 x 11 - 1 x 13
d = 33 - 13
d = 20
3 x 2 rectangle #2
2
3
4
22
23
24
d = 14 x 22 - 12 x 24
d = 308 - 288
d = 20
3 x 2 rectangle #3
88
89
90
98
99
00
d = 90 x 98 - 88 x 100
d = 8820 - 8800
d = 20
2 x 3 rectangle #1
2
1
2
21
22
d = 2 x 21 - 1 x 22
d = 42 - 22
d = 20
2 x 3 rectangle #2
2
3
22
23
32
33
d = 2 13 x 32 - 12 x 33
d = 416 - 396
d = 20
2 x 3 rectangle #3
79
80
89
90
99
00
d = 80 x 99 - 79 x 100
d = 7920 - 7900
d = 20
The results indicate that the diagonal difference is not dependant on the orientation of the rectangle. I shall test this theory on a set of larger rectangles.
NB As the rectangles are getting larger, I shall depict them as a smaller standard rectangle, showing just the corner numbers.
4 x 3 rectangle #1
4
21
24
d = 4 x 21 - 1 x 24
d = 84 - 24
d = 60
4 x 3 rectangle #2
2
5
32
35
d = 15 x 32 - 12 x 35
d = 480 - 420
d = 60
4 x 3 rectangle #3
77
80
97
00
d = 80 x 97 - 77 x 100
d = 7760 - 7700
d = 60
3 x 4 rectangle #1
3
31
33
d = 3 x 31 - 1 x 33
d = 93 x 33
d = 60
3 x 4 rectangle #2
2
4
42
44
d = 14 x 42 - 12 x 44
d = 588 - 528
d = 60
3 x 4 rectangle 3
68
70
98
00
d = 70 x 98 - 68 x 100
d = 6860 - 6800
d = 60
The results still show that the diagonal difference is not dependent on orientation. This means I have a choice of 36 different sized rectangles to use in my investigation. I shall use the rectangles highlighted on the list on page 6 as it gives an adequate variety of sizes.
So far I have found that d = 20 on a 3 x 3 rectangle and d = 60 on a 4 x 3 rectangle.
Algebraic proof:
3 x 2 rectangle
+2
+G
a
a + 2
a + G
a + G + 2
d = (a + 2) (a + G) - a (a + G + 2)
= a2 + aG + 2a + 2G - a2 - aG - 2a
= 2G, where G = 10, d = 20
4 x 3 rectangle
+3
+2G
a
a + 3
a + 2G
a + 2G + 3
d = (a + 3) (a + 2G) - a (a + 2G + 3)
= a2 + 2aG + 3a + 6G - a2 - 2aG - 3a
= 6G, where G = 10, d = 60
5 x 4 rectangle
5
31
35
d = 5 x 31 - 1 x 35
d = 155 x 35
d = 120
+4
+3G
a
a + 4
a + 3G
a + 3G + 4
d = (a + 4) (a + 3G) - a (a + 3G + 4)
= a2 + 3aG + 4a + 12G - a2 - 3aG - 4a
= 12G, where G = 10, d = 120
6 x 5 rectangle
6
41
46
d = 6 x 41 - 1 x 45
d = 246 - 46
d = 200
+5
+4G
a
a + 5
a + 4G
a + 4G + 5
d = (a + 5) (a + 4G) - a (a + 4G + 5)
= a2 + 4aG + 5a + 20G - a2 - 4aG - 5a
= 20G, where G = 10, d = 200
7 x 6 rectangle
7
51
57
d = 7 x 51 - 1 x 57
d = 357 - 57
d = 300
+6
+5G
a
a + 6
a + 5G
a + 5G + 6
d = (a + 6) (a + 5G) - a (a + 5G + 6)
= a2 + 5aG + 6a + 30G - a2 - 5aG - 6a
= 30G, where G = 10, d = 300
8 x 7 rectangle
8
61
68
d = 8 x 61 - 1 x 68
d = 488 - 68
d = 420
+7
+6G
a
a + 7
a + 6G
a + 6G + 7
d = (a + 7) (a + 6G) - a (a + 6G + 7)
= a2 + 6aG + 7a + 42G - a2 - 6aG - 7a
= 42G, where G = 10, d = 420
9 x 8 rectangle
9
71
79
d = 9 x 71 - 1 x 79
d = 639 - 79
d = 560
+8
+7G
a
a + 8
a + 7G
a + 7G + 8
d = (a + 8) (a + 7G) - a (a + 7G + 8)
= a2 + 7aG + 8a + 56G - a2 - 7aG - 8a
= 56G, where G = 10, d = 560
0 x 9 rectangle
0
81
90
d = 10 x 81 - 1 x 90
d = 810 - 90
d = 720
+9
+8G
a
a + 9
a + 8G
a + 8G + 9
d = (a + 9) (a + 8G) - a (a + 8G + 9)
= a2 + 8aG + 9a + 72G - a2 - 8aG - 9a
= 72G, where G = 10, d = 720
Two-way table showing the differences of
all possible box sizes on o 10 x 10 grid.
x values
2
3
4
5
6
7
8
9
0
2
0
20
30
40
50
60
70
80
90
3
20
40
60
80
00
20
40
60
80
4
30
60
90
20
50
80
210
240
270
5
40
80
20
60
200
240
280
320
360
6
50
00
50
200
250
300
350
400
450
7
60
20
80
240
300
360
420
480
540
8
70
40
210
280
350
420
490
560
630
9
80
60
240
320
400
480
560
640
720
0
90
80
270
360
450
540
630
720
810
y values
I have drawn up a two-way table to record my results. I have written the rectangle differences in black. I notice that they have left a gap where the square box differences would go. As I have already investigated square boxes and have the results, I have recorded them in the table in red. I can see patterns in the results so far, and have predicted the differences for the rest of the rectangles. I have recorded these in the table in green. I will test six predicted differences (as highlighted) to confirm my predictions.
9 x 3 rectangle
9
21
29
d = 9 x 21 - 1 x 29
d = 189 - 29
d = 160
7 x 4 rectangle
7
31
37
d = 7 x 31 - 1 x 37
d = 217 - 37
d = 180
5 x 6 rectangle
5
51
55
d = 5 x 51 - 1 x 55
d = 255 - 55
d = 200
9 x 7 rectangle
9
61
69
d = 9 x 61 - 1 x 69
d = 549 - 69
d = 480
2 x 8 rectangle
2
71
72
d = 2 x 71 - 1 x 72
d = 142 - 72
d = 70
7 x 10 rectangle
7
91
97
d = 7 x 91 - 1 x 97
d = 637 - 97
d = 540
This proves that my predictions were correct.
Generating a formula
x
x
y
x
I have labelled the sides of a rectangle and square as shown. This relates to the labels on the two-way table.
A rectangle is defined thus:
x x y , where x = y
A square is defined thus:
x2
The master formula for any square on any grid is:
d = G (x - 1) 2
or
d = G(x - 1) (x - 1)
As there are two different values for the sides of a rectangle, I think that the formula would be:
d = G(x - 1) (y - 1)
To test this formula, I have calculated the difference for three different sized boxes on three different sized grids.
3 x 8 rectangle on a 9 x 9 grid
3
64
66
d = 3 x 64 - 1 x 66
d = 192 - 66
d = 126
+2
+7G
a
a + 2
a + 7G
a +7G + 2
d = (a + 2) (a + 7G) - a (a + 7G + 2)
= a2 + 7aG + 2a + 14G - a2 - 7aG - 2a
= 14G, where G = 9, d =126
4 x 6 rectangle on a 7 x 7 grid
4
36
39
d = 4 x 36 - 1 x 39
d = 144 - 39
d = 105
+3
+5G
a
a + 3
a + 3G
a + 5G + 3
d = (a + 3) (a + 5G) - a (a + 5G + 3)
= a2 + 5aG + 3a + 15G - a2 - 5aG - 3a
= 15G, where G = 7, d = 105
2 x 4 rectangle 0n a 5 x 5 grid
2
6
7
d = 2 x 16 - 1 x 17
d = 32 - 17
d = 15
+1
+3G
a
a + 1
a + 3G
a + 3G + 1
d = (a + 1) (a + 3G) - a (a + 3G + 1)
= a2 + 3aG + a + 3G - a2 - 3aG -a
= 3G, where G = 5, d = 15
To conclude, I have found a master formula to find the difference of any sized box which is defined thus:
x x y , where x can be equal to y
on grid with any number of columns (G).
d = G(x - 1) (y - 1)
- 1 -