(n + 7) x (n + 1) – n x (n + 8) = d
or
n2 + n + 7 x n + 7 – (n2 + 7 x n + n) = d
d = difference
eg:
4 x 12 = 48
5 x 11 = 55
55 – 48 = 7
(4 + 7) x (4 + 1) – 4 x (4 + 8) = 7
42 + 4 + 7 x 4 + 7 – (42 + 7 x 4 + 4) = 7
or
20 x 28 = 560
21 x 27 = 567
567 – 560 = 7
(20 + 7) x (20 + 1) – 20 x (20 + 8) = 7
202 + 20 + 7 x 20 + 7 – (202 + 7 x 20 + 20) = 7
I had discovered the algebraic representation that enabled me to determine the patterns result with just the root number. This formula made me realise that the slight modification of it would allow me to predict the result of any pattern using the root number and the number of columns in its grid. This led me to my first variable; changing the number of columns.
I discarded the calender format and formulated two different grids: one with eight columns and one with six. If my earlier prediction stating that the number of columns was equal to the difference presented by the pattern then should not my equation work if I merely exchanged the seven in my equation with the column length of the grid I was using?
I have established that on a grid with seven columns:
d = (n + 7) x (n + 1) – n x (n + 8)
or
n2 + n + 7 x n + 7 – (n2 + 7 x n + n)
I constructed an algebraic representation of a box in a grid of eight columns first to form a foundation for the modification of my equation:
So, according to my ‘algebra box’, on a grid with eight columns the equation should be:
d = (n + 8) x (n + 1) – n x (n + 9)
or
n2 + n + 8 x n + 8 – (n2 + 8 x n + n)
I verified this with two examples from my eight columned grid:
4 x 13 = 52
5 x 12 = 60
60 – 52 = 8
(4 + 8) x (4 + 1) – 4 x (4 + 9) = 8
42 + 4 + 8 x 4 + 8 – (42 + 8 x 4 + 4) = 8
9 x 18 = 162
10 x 17 = 170
170 – 162 = 8
(9 + 8) x (9 + 1) – 9 x (9 + 9) = 8
92 + 9 + 8 x 9 + 8 – (92 + 8 x 9 + 9) = 8
This had worked but in order to firmly establish my theory I would now have to try a grid length smaller than seven. Initially I contrived another ‘algebra box’ for a grid with six columns:
My algebraic formula would now be:
d = (n + 6) x (n + 1) – n x (n + 7)
or
n2 + n + 7 x n + 7 – (n2 + 7 x n + n)
I repeated the process and examined two examples from a grid with six columns:
4 x 11 = 44
5 x 10 = 50
50 – 44 = 6
(4 + 6) x (4 + 1) – 4 x (4 + 7) = 6
42 + 4 + 7 x 4 + 7 – (42 + 7 x 4 + 4) = 6
13 x 20 = 260
14 x 19 = 266
266 – 260 = 6
(13 + 6) x (13 + 1) – 13 x (13 + 7) = 6
132 + 13 + 7 x 13 + 7 – (132 + 7 x 13 + 13) = 6
The confirmation of the efficiency of my equation in determining any patterns result with the root number and the number of columns was easy enough to convert into a penultimate formula that would work out any difference with those two sets of data:
n2 + 7n + c – (n2 + 7n) = d
c = number of columns
eg:
4 x 17 = 68
5 x 16 = 80
80 – 68 = 12
42 + 7 x 4 + 12 – (42 + 7 x 4) = 12
13 x 26 = 338
14 x 25 = 350
350 – 338 = 12
132 + 7 x 13 + 12 – (132 + 7 x 13) = 12
The final equation that I cogitated, which expressed this pattern far more simply, was:
d = 1c
or
difference always = number of columns
I decided now to modify the actual box that determined the result of the pattern, I altered the size of the box from two by two into two (horizontal) by three (vertical). The same rule will still apply; the opposite corners will be multiplied together and the smaller number subtracted from the larger. I hope that this will produce a pattern, although I have no supposition as to what that pattern could be. My first task will be to investigate this new method and attempt to identify a pattern within it:
August
4 x 19 = 76
5 x 18 = 90
90 – 76 = 14
8 x 23 = 184
9 x 22 = 198
198 – 184 = 14
Again I am only in the preliminary stages of the investigation yet I already recognise a potential link. The number appears to have doubled from seven to fourteen; could it be that changing the box size to two by three makes the difference become double the column length? If so does a further increase in the height of the box administer a supplementary doubling? I have decided to change the length of the grid before investigating an algebraic equation for the pattern:
4 x 23 = 92
5 x 22 = 110
110 – 92 = 18
16 x 35 = 560
17 x 34 = 578
578 – 560 = 18
It does appear that the column length has doubled, so I have already determined a pattern and link between the previous investigation and this one. I will continue to elaborate upon this occurrence after I have determined algebraic representations of this pattern. To be gin with I will construct an ‘algebra box’ for the seven columned grid:
Judging by this box the equation for the root times its diagonal would be:
n x (n + 15)
4 x 19 = 76
4 x (4 + 15) = 76
Similarly, the equation for the root plus one times its diagonal would be:
(n + 14) x (n + 1)
5 x 18 = 90
(4 + 14) x (4 + 1) = 90
As before I can now combine these two formulas and assemble an equation for ascertaining the difference with just the root number and the grid length:
(n + 2c) x (n + 1) – (n x (n + 2c + 1)) = d
eg:
10 x 27 = 270
11 x 26 = 286
286 – 270 = 16
(10 + 2 x 8) x (10 + 1) – (10 x (10 + 2 x 8 + 1)) = 16
6 x 15 = 90
7 x 14 = 98
98 – 90 = 8
(6 + 2 x 4) x (6 + 1) – (6 x (6 + 2 x 4 + 1)) = 8
My prediction, it seemed, proved to be correct. I have now established that when using a box with the height of three the difference of the opposing sums becomes double the column length or difference of a box with a height of two. I can express this as:
d = 2c
or
difference always = double the number of columns
Before continuing onto changing my final variable I have decided to investigate an additional question which I asked myself earlier; does a further increase in the height of the box administer a supplementary doubling? I inserted another calender grid and tested this theory on it:
August
4 x 26 = 104
5 x 25 = 125
125 – 104 = 21
8 x 30 = 240
9 x 29 = 261
261-240 = 21
21 / 7 = 3
Unfortunately the figure did not double as expected, but it did increase again by seven. It now seemed that as the height of the box increased by one the difference increased by the length of the grid. I produced a universal algebraic formula for this four by two box:
(n + 3c) x (n + 1) – (n x (n + 3c + 1)) = d
4 x 29 = 116
5 x 28 = 140
140 – 116 = 24
(4 + 3 x 8) x (4 + 1) – (4 x (4 + 3 x 8 + 1)) = 24
10 x 23 = 230
11 x 22 = 242
242 – 230 = 12
(10 + 3 x 4) x (10 + 1) – (10 x (10 + 3 x 4 + 1)) = 12
I am certain now that as the height of the box increases the difference does as well. I can express this universally as:
d = ch
or
difference always = number of columns times height of box
h = height of box
I now intend to investigate how the width of the box can affect the result of the pattern. To do this I am going to keep my earlier modification of the height of the box and increase the width by one as well. The box now measures three by three. Once again my first task will be to locate and initial pattern within the sum:
August
4 x 20 = 80
6 x 18 = 108
108 – 80 = 28
11 x 31 = 341
13 x 29 = 377
377 – 341 = 36
It appears that this box produces a difference that is four times the number of columns. To prove this I made an algebraic representation for this process:
(n + 4c) x (n + 1) – (n x (n + 4c + 1)) = d
eg:
7 x 19 = 133
9 x 17 = 153
153 – 133 = 20
(7 + 4 x 5) x (7 + 1) – (7 x (7 + 4 x 5 + 1)) = 20
15 x 41 = 615
17 x 39 = 663
663 – 615 = 48
(15 + 4 x 12) x (15 + 1) – (15 x (15 + 4 x 12 + 1)) = 48
The inclusion and functionality of ‘4c’ within the equation proves that my earlier thought; as with the height of the box the difference increases by the number of columns when the width increases by one. I can equate this to:
d = cw
or
difference always = number of columns times width of box
w = width of box
In the course of this investigation I have discovered three main things:
∙ When the width and the height of the box increase by one the difference is squared.
∙ When the width or height of the box is individually increased by one the number of columns is added to the difference.
∙ The number of columns is directly proportional to the difference presented by the pattern.
A question that perplexed me throughout the investigation was; why did the number of columns on the grid have a connection with and influence the difference whereas the number of rows was irrelevant? I discovered eventually that this is, obviously, because the length of the grid changes the numeric structure within the box; the number of rows has no effect since it is the height of the box that has a relation to the rows and changes the equation within it.
The investigation I feel was a success, I have recognised and explored relationships and factors that effect the result of the pattern and have proven nearly all my predictions to be correct. I would have liked to have simplified and shortened some of the algebraic equations as they are far too long, however I found that this took too much time and tended to confuse me.
