A 2 x 2 square would be 9(n-1)2 = 9(2-1)2 = 9 x 1 = 9.
A 3 x 3 square would be 9(n-1)2 = 9(3-1)2 = 9 x 4 = 36.
A 4 x 4 square would be 9(n-1)2 = 9(4-1)2 = 9 x 9 = 81.
A 5 x 5 square would be 9(n-1)2 = 9(5-1)2 = 9 x 16 = 144.
Now that the formula has been established, I am going to use it to predict the difference for a 7 x 7 square. 9(n-1)2 = 9(7-1)2 = 9 x 62 = 9 x 36 = 324.
My prediction was correct. To make sure the rule works all the time, I am going to prove it algebraically:
“x” is the number in the top-left corner. The number to the right of it is “x+ (n-1)” because it is “x” added to the length of the square (take away one). Because the square is on a 9 x 9 grid, the formula for the number underneath the square is “x” added to (9 multiplied by “the squares length”-1). The bottom-right number is “x+ 10(n-1)” because it is “x+ 9(n-1) added to another (n-1), so therefore this becomes “x+ 10(n-1).
[x + (n-1)][x + 9(n-1)]= x2 + 9x(n-1) + x(n-1) + 9(n-1)2.
x[x+10(n-1)] = x2 + 10x(n-1).
This becomes: x2 + 9x(n-1) + x(n-1) + 9(n-1)2 – x2 – 10x(n-1).
The x2 and –x2 cancel each other out, while the 9x(n-1) and x(n-1) add together to cancel out the -10x(n-1). This leaves 9(n-1)2, which is the overall formula for a 9 x 9 grid.
11 x 11 grids
2 x 2 squares:
40 x 50 = 2000
39 x 51 = 1989
2000-1989 = 11
3 x 13 = 39
2 x 14 = 28
39 – 28 = 11
3 x 3 squares:
92 x 112 = 10304
90 x 114 = 10260
10304 – 10260 = 44
11 x 31 = 343
9 x 33 = 299
343 – 299 = 44
4 x 4 squares:
86 x 116 = 9976
83 x 119 = 9877
9976 – 9877 = 99
40 x 70 = 2800
37 x 73 = 2701
2800 – 2701 = 99
5 x 5 squares:
8 x 48 = 384
4 x 52 = 208
384 – 208 = 176
77 x 117 = 9009
73 x 121 = 8833
9009 – 8833 = 176
After getting the results, I put them into a table:
A 10 x 10 grid’s formula is 10(n-1)2, and a 9 x 9 grid is 9(n-1)2. An 11 x 11 grid could be 11(n-1)2. All the differences in the table are multiples of 11. If the formula is 11(n-1)2, a 2 x 2 square will be will 11(2-1)2 = 11 X 12 = 11, which it is. This means a 3 x 3 square will be 11(3-1)2 = 11 X 22 = 11 X 4 = 44. These are both correct. 99 ÷ 11 = 9, which is the equivalent of 32. 176 ÷ 11 = 16, which is 42. This means that the formula must be correct. To make sure it is, I will use it to predict the difference for a 9 x 9 square on an 11 x 11 grid. The formula for a 9 x 9 square is 11(9-1)2 = 11 X 82 = 11 X 64 = 704. The grid I chose is as follows:
32 x 112 = 3584
24 x 120 = 2880
3584 – 2880 = 704. The formula was correct.
I am now going to prove the formula algebraically.
“x” is the number in the top-left corner. The number to the right of it is “x+ (n-1)” because it is “x” added to the length of the square (take away one). Because the square is on an 11 x 11 grid, the formula for the number underneath the square is “x” added to (11 multiplied by “the squares length”-1). The bottom-right number is “x+ 12(n-1)” because it is “x+ 11(n-1) added to another (n-1), so therefore this becomes “x+ 10(n-1).
[x + (n-1)][x + 11(n-1)]= x2 + 11x(n-1) + x(n-1) + 11(n-1)2.
x[x+12(n-1)] = x2 + 12x(n-1).
This becomes: x2 + 11x(n-1) + x(n-1) + 11(n-1)2 – x2 – 12x(n-1).
The x2 and –x2 cancel each other out, while the 11x(n-1) and x(n-1) add together to cancel out the -12x(n-1). This leaves 11(n-1)2, which is the overall formula for an 11 x 11 grid.
A 10 x 10 grid’s formula is 10(n-1)2, a 9 x 9 grid is 9(n-1)2 and an 11 x 11 grid is
11(n-1)2. The size of the grid corresponds directly to the first digit of the formula.
This means the formulae can be re-written as g(n-1)2 (this is the overall rule for any square on any grid).
I am now going to predict differences for squares on a 14 x 14 grid, using the formula g(n-1)2. First, I am going to predict a 12 x 12 square on a 14 x 14 grid. The formula, with the numbers in place, is 14(12-1)2 = 14 X 112 = 14 X 121 = 1694.
I chose the following square:
40 x 183 = 7320.
29 x 194 = 5626
7320 – 5626 = 1694.
The formula worked. Just to make sure it works universally, I will choose another square, however this time it will be 7 x 7.
A 7 x 7 square on the grid will have the formula 14(7-1)2 = 14 X 62 = 14 X 36 = 504. I chose the following square:
70 X 148 = 10360
64 X 154 = 9856
10360 – 9856 = 504.
I have found that the formula g(n-1)2 works, and I will now prove it algebraically.
[x + (n-1)][x + g(n-1)] = x2 + xg(n-1) + x(n-1) + g(n-1)2
x[x + g(n-1) + (n-1)] = x2 + xg(n-1) + x(n-1).
This becomes x2 + xg(n-1) + x(n-1) + g(n-1)2 – x2 – xg(n-1) – x(n-1).
After cancelling out, only g(n-1)2 is left, which proves that it is the overall formula, not only for squares of different sizes, but also for different sized grids.
Now that I have found the rule for squares, I am going to try and find a rule for rectangles. To start off, I am going to use rectangles on a 10 x 10 gird.
I am going to use the smallest possible rectangles to start off with (2 x 3 rectangles):
13 x 21 = 273
11 x 23 = 253
273 – 253 = 20.
68 x 76 = 5168
66 x 78 = 5148
5168 – 5148 = 20.
34 x 53 = 1802
33 x 54 = 1782
1802 – 1782 = 20.
8 x 27 = 216
7 x 28 = 196
216 – 196 = 20.
All 2 x 3 rectangles have a difference of 20, regardless of whether they are bigger horizontally or vertically.
I will now get the differences for other sized rectangles:
3 x 4
25 x 42 = 1050
22 x 45 = 990
1050 – 990 = 60.
3 x 4
60 x 88 = 5280
58 x 90 = 5220
5280 – 5220 = 60.
4 x 5
5 x 42 = 210
2 x 45 = 90
210 – 90 = 120.
4 x 5
70 x 96 = 6720
66 x 100 = 6600
6720 – 6600 = 120.
5 x 6
56 x 91 = 5096
51 x 96 = 4896
5096 – 4896 = 200.
5 x 6
20x 66 = 1320
16 x 70 = 1120
1320 – 1120 = 200.
6 x 7
39 x 94 = 3666
34 x 99 = 3366
3666 – 3366 = 300.
After trying out rectangles of 2 x 3, 3 x 4, 4 x 5, 5 x 6 and 6 x 7 size, I assembled the differences and put it into a table:
The formula for squares was g(n-1)2 which is equal to g(n-1)(n-1). The g represented the grid size, and the n represented the width and height of the square. It could be re-written as g(h-1)(w-1), with h representing the height and w representing the width. This means the overall rule could be g(h-1)(w-1). I am going to test it in several ways. First, I am going to try it with rectangles I have already investigated:
For a 2 x 3 rectangle on a 10 x 10 grid, the rule would be 10(2-1)(3-1) which is 10 x 1 x 2 = 10 x 2 = 20. This is correct: I will now try the rule with a bigger rectangle.
A 5 x 6 rectangle is 10(5-1)(6-1), which become 10 x 4 x 5, which is 40 x 5 = 200. This is also correct. However, there is a possibility that this rule is not universal and that it only applies to rectangles with 1 square difference between the height and width.
I am now going to investigate the differences for other rectangles (i.e. rectangles with more than 1 square difference between the height and width (e.g. 2 x 4, 3 x 6 etc.)).
26 x 33 = 858
23 x 36 = 828
858 – 828 = 30.
2 x 31 = 62
1 x 32 = 32
62 – 32 = 30
40 x 56 = 2240
36 x 60 = 2160
2240 – 2160 = 80.
The formula for the 2 x 4 rectangle is 10(2-1)(4-1) = 10 x 1 x 3 = 30. This is correct.
The formula for the 3 x 5 rectangle is 10(3-1)(5-1) = 10 x 2 x 4 = 80. This is also correct. This means that the formula must work for all rectangles and squares. I am now going to predict the difference for a 2 x 5 rectangle:
The formula for a 2 x 5 rectangle is 10(2-1)(5-1) = 10 x 1 x 4 = 40.
34 x 73 = 2482
33 x 74 = 2442
2482 – 2442 = 40.
The formula is correct!
I will also predict it for a 4 x 9 rectangle:
The formula for a 4 x 9 rectangle is 10(4-1)(9-1) = 10 x 3 x 8 = 30 x 8 = 240.
59 x 81 = 4779
51 x 89 = 4539
4779 – 4539 = 240.
This means the formula 10(h-1)(w-1) is correct. I will now prove it algebraically:
[x + (w-1)][x + 10(h-1)] = x2 + 10x(h-1) + x(w-1) + 10(h-1)(w-1)
x[x + 10(h-1) + (w-1)] = x2 + 10x(h-1) + x(w-1)
This becomes: x2 + 10x(h-1) + x(w-1) + 10(h-1)(w-1) – x2 – 10x(h-1) – x(w-1).
The x2’s cancel out each other, along with the 10x(h-1) and the x(w-1).
This leaves 10(h-1)(w-1), which is the universal rule for any square and rectangle on a 10 x 10 grid.
I know that this formula is correct: I have just found that the formula 10(h-1)(w-1) for 10 x 10 grids is correct: now I need to find the rule for all sized grids. I am going to use a 13 x 13 grid with which to confirm the rule.
13 x 13 Grids
I am first going to start with some 2 x 3 grids, then 3 x 4, and then 4 x 5.
2 x 3 grids
3 x 14 = 42
1 x 16 = 16
42 – 16 = 26.
32 x 57 = 1824
31 x 58 = 1798
1824 – 1798 = 26.
3 x 4 grids
8 x 31 = 248
5 x 34 = 170
248 – 170 = 78.
56 x 93 = 5208
54 x 95 = 5130
5208 – 5130 = 78.
4 x 5 grids
103 x 152 = 15656
100 x 155 = 15500
15656 – 15500 = 156.
5 x 6 grids
39 x 86 = 3354
34 x 91 = 3094
3354 – 3094 = 260.
6 x 7 grids
85 x 158 = 13430
80 x 163 = 13040
13430 – 13040 = 390.
I will now put the values into a table:
Seeing as the rule for the 10 x 10 grid was 10(h-1)(w-1), it is logical that the rule for the 13 x 13 grid could be 13(h-1)(w-1). To check, I am going to divide each of the “differences” by 13, to get the value of (h-1)(w-1):
Assuming the formula is correct, the differences (÷ 13) will be the multiple of the size-1 (i.e. if the size is 2 x 3, the difference (÷ 13) is (2-1) x (3-1) = 1 x 2 = 2. This is correct. For a size of 5 x 6, the difference (÷ 13) is (5-1) x (6-1), which is equal to 4 x 5 which is 20.
This means that the rule for a 13 x 13 grid is most probably 13(h-1)(w-1).
I am now going to prove It algebraically:
[x + (w-1)][x + 13(h-1)] = x2 + 13x(h-1) + x(w-1) + 13(h-1)(w-1)
x[x + 13(h-1) + (w-1)] = x2 + 13x(h-1) + x(w-1)
This becomes: x2 + 13x(h-1) + x(w-1) + 13(h-1)(w-1) – x2 – 13x(h-1) – x(w-1).
The x2’s cancel out each other, along with the 13x(h-1) and the x(w-1).
This leaves 13(h-1)(w-1), which is the universal rule for any square and rectangle on a 13 x 13 grid.
Now I have proven that on a 10 x 10 grid, the formula is 10(h-1)(w-1), and that the formula is 13(h-1)(w-1) for a 13 x 13 grid. This coincides with my prediction that the universal formula is g(h-1)(w-1).
Using the rule, I will predict the difference for a 7 x 12 rectangle on a 14 x 14 grid. The formula is 14(7-1)(12-1) = 14 x 6 x 11 = 154 x 6 = 924.
35 x 183 = 6405
29 x 189 = 5481
6405 – 5481 = 924.
The formula worked on a big square on a big grid. I will try the formula on a smaller square on a smaller grid: if the rule is correct then I know the formula works universally.
I will predict for a 2 x 6 rectangle on a 7 x 7 grid.
7(2-1)(6-1) = 7 x 1 x 5 = 35.
27 x 29 = 783
22 x 34 = 748
783 – 748 = 35.
The formula worked, meaning my prediction was correct. These results show the formula g(h-1)(w-1) is universal, and works algebraically as well.
I know that the formula works so far on certain sized grids: and now I will prove it algebraically:
[x+(w-1)][x+g(h-1)] = x2 + x(w-1) + xg(h-1) + g(h-1)(w-1).
x[x+g(h-1)+(w-1)] = x2 + xg(h-1) + x(w-1).
This becomes x2 + x(w-1) + xg(h-1) + g(h-1)(w-1) - x2 - xg(h-1) - x(w-1).
The x2’s cancel each other out, along with the xg(h-1)’s and the x(w-1)’s.
This leave g(h-1)(w-1), which is the universal rule for any square or rectangle on any size grid.
Matrices
A Matrix is a rectangular order of numbers (it can also be a square), i.e. in this case, it is the squares that are found on the grids (that are used to find the differences).
There are different things that can be done with matrices of different sizes. These include addition, subtraction, and multiplication.
Addition of a matrix is adding up the corresponding values of the matrices.
e.g. 2 matrixes on a 10 x 10 grid are:
and
The addition of these matrices would be:
…which in turn would become:
This is basic addition for matrices.
Matrix addition can be applied to any square (or rectangle for that matter) of any size, as I am going to show with a 5 x 8 rectangle on an 11 x 11 grid.
And
This becomes
I am going to show how algebraically, Matrix addition works:
Let a = top-left number of first matrix.
Written algebraically, the first matrix can be written as
There is a “g” in the cells: “g” stands for the size of the grid on which the squares are.
The second matrix would be exactly the same as this one, but with “b” replacing every “a”.