Number Grids Coursework
Introduction
In the following piece of coursework, I intend to investigate taking a square of numbers from a 10 x 10 grid, multiplying the opposite corners and then finding the difference between the two products.
I was first asked to take a 2 x 2 square from a 10 x 10 grid, multiply the opposite corners and then find the difference. This is the result I received;
2x2 squares
5
6
25
26
Square 1
5 x 26 = 390
6 x 25 = 400
Difference = 400 - 390 = 10
Here I found that the difference was 10, to find out if the difference was the same for every 2 x 2 square, I decided to test another 3 squares, and here are the results I received;
Square 2
8
9
28
29
8 x 29 = 522
9 x 28 = 532
Difference = 532 - 522 = 10
Square 3
32
33
42
43
32 x 43 = 1376
33 x 42 = 1386
Difference = 1386 - 1376 = 10
35
36
45
46
Square 4
35 x 46 = 1610
36 x 45 = 1620
Difference = 1620 - 1610 = 10
After calculating the differences between the products of four 2 x 2 square grids, I can conclude that the difference is always 10, and I predict that the difference will always remain 10 for 2 x 2 square grids. I then decided to look at a 3 x 3 square on a 10 x 10 grid, to see if the difference would be 10, or anything different, and here is the result;
32
33
34
42
43
44
52
53
54
3 x 3 squares
Square 1
32 x 54 = 1728
34 x 52 = 1768
Difference = 1768 - 1728 = 40
Here, however I found that the difference was 40 to find out if the difference was the same for every 3 x 3 square, I decided to test another 3 squares, and here are the results I received;
35
36
37
45
46
47
55
56
57
Square 2
35 x 57 = 1995
37 x 55 = 2035
Difference = 2035 - 1995 = 40
38
39
40
48
49
50
58
59
60
Square 3
38 x 60 = 2280
40 x 58 = 2320
Difference = 2320 - 2280 = 40
62
63
64
72
73
74
82
83
84
Square 4
62 x 84 = 5208
64 x 82 = 5248
Difference = 5248 - 5208 = 40
After calculating the differences between the products of four 3 x 3 square grids, I can conclude that the difference is always 40, and I predict that the difference will always remain 40 for all 3 x 3 square grids. I then decided to look at 4 x 4 squares on a 10 x 10 grid to find their differences; I also decided to look at 5 x 5 squares to see if there was any particular pattern, and here are the results
2
3
4
1
2
3
4
21
22
23
24
31
32
33
34
4 x 4 squares
Square 1
x 34 = 34
4 x 31 = 124
Difference = 124 - 34 = 90
5
6
7
8
5
6
7
8
25
26
27
28
35
36
37
38
Square 2
5 x 38 = 190
8 x 35 = 280
Difference = 280 - 190 = 90
5
6
7
8
5
6
7
8
25
26
27
28
35
36
...
This is a preview of the whole essay
4 x 31 = 124
Difference = 124 - 34 = 90
5
6
7
8
5
6
7
8
25
26
27
28
35
36
37
38
Square 2
5 x 38 = 190
8 x 35 = 280
Difference = 280 - 190 = 90
5
6
7
8
5
6
7
8
25
26
27
28
35
36
37
38
Square 3
41 x 74 = 3034
44 x 71 = 3124
Difference = 3124 - 3034 = 90
45
46
47
48
55
56
57
58
65
66
67
68
75
76
77
78
Square 4
45 x 78 = 3510
48 x 75 = 3600
Difference = 3600 - 3510 = 90
After calculating the differences between the products of four 4 x 4 square grids, I can conclude that the difference is always 90, and I predict that the difference will always remain 90 for all 4 x 4 square grids.
2
3
4
5
1
2
3
4
5
21
22
23
24
25
31
32
33
34
35
41
42
43
44
45
5 x 5 squares
Square 1
x 45 = 45
5 x 41 = 205
Difference = 205 - 45 = 160
6
7
8
9
0
6
7
8
9
20
26
27
28
29
30
36
37
38
39
40
46
47
48
49
50
Square 2
6 x 50 = 300
0 x 46 = 460
Difference = 460 - 300 = 160
51
52
53
54
55
61
62
63
64
65
71
72
73
74
75
81
82
83
84
85
91
92
93
94
95
Square 3
51 x 95 = 4845
55 x 91 = 5005
Difference = 5005 - 4845 = 160
Square 4
56
57
58
59
60
66
67
68
69
70
76
77
78
78
80
86
87
88
89
90
96
97
98
99
00
56 x 100 = 5600
60 x 96 = 5760
Difference = 5760 - 5600 = 160
After calculating the differences between the products of four 5 x 5 square grids, I can conclude that the difference is always 160, and I predict that the difference will always remain 160 for all 5 x 5 square grids. I then decided to put the results in a table to see if I could distinguish any particular pattern.
Box size
Difference
2 x 2
0
3 x 3
40
4 x 4
90
5 x 5
60
From looking at the results I can tell that the differences are all square numbers which are multiplied by 10, for e.g. 1 is a square number and if I multiplied it by 10 I would get 10, similarly 4 is also a square number and multiplied by 10 I get 40. I also noticed that to get the squared numbers I could minus 1 from the box size, and then square it, I then decided to see if I could find a formula.
If I used b as the box size then my formula would be as follows;
(b-1)² x 10 as I noticed that if I multiplied the square number by 10 then I would get the differences. Now let's see if it works for a 6 x 6 square on a 10 x 10 grid.
2
3
4
5
6
1
2
3
4
5
6
21
22
23
24
25
26
31
32
33
34
35
36
41
42
43
44
45
46
51
52
53
54
55
56
6 - 1= 5
5² = 25
25 x 10 = 250
This is the result I predict using the formula, but let's see if it actually works.
6 x 6 square
x 56 = 56
6 x 51 = 306
Difference = 306 - 56 = 250
This proves that my formula was correct as I predicted (using the formula) that the difference would be 250, and I was correct. I have now proved how to work out the differences of any square in a 10 x 10 grid but I will now try to prove it using algebra.
2 x 2 squares
To find the end algebraic expression I am going to give each corner in the square an expression that would work for that corner in any 2 x 2 square.
I am going to call the first corner in the 2 x 2 square 's'.
s
This is the 2 x 2 square so far:
To get the top-right corner I am going to look at a 2 x 2 square taken from a grid:
5
6 is one more than 15
25
26
From this I have seen that the top right corner is one more than the top left corner. This is the same in any other 2 x 2 square. So using the top left corner 's' I can say that the top right corner is s + 1. I now have the two to expressions
s
S + 1
Again to find like the top right corner, I am going to have to look at some example of a 2 x 2 square to find the bottom two corners expressions.
5
6 is one more than 15
25 is ten more than 15
26 is eleven more than 15
From this I have seen that the bottom left corner is ten more than the top left corner, so I can say that the bottom left corner is s + 15, and from the bottom right corner I can see that it is eleven more than the top left corner, so I can say that the bottom right corner is s + 11, and now I have a four complete expressions for a 2 x 2 square, which now looks like this;
S
S + 1
S + 11
S + 11
Now I just have to multiply the corners, and find the differences like this;
s( s + 11) = s² + 11s
(s + 1)(s + 10) = s² + 10s + s + 10 = s² + 11s + 10
Difference = (s² + 11s + 10) - ( s² + 11s) = 10
From looking at this expression, I can tell I have proved (using algebra) that the difference will always be 10.
Now let's try to find an expression for a 3 x 3 square in a 10 x 10 grid and see if the difference is always 40 using algebra.
First I will again call my top left corner 's'. Then I will look at an example of a 3 x 3 square on a 10 x 10 grid.
32
33
34 is 2 more than 32
42
43
44
52 is 20 more than 32
53
54 is 22 more than 32
From looking at this 3 x 3 square I can tell that the top right corner is 2 more than 32, which means that a 4 x 4 square's top right corner will be 3 more than the top left corner, as it is going up by 1 each time, a 5 x 5 square will be 4 more than the top left corner and so on. I can also tell that the bottom left corner is 20 more than 32, which means a 4 x 4 square's bottom left corner will be 30 more than the top left as it is going up in 10's each time, and so on, and the bottom right corner I can tell is 22 more than 32, which means a 4 x 4 squares bottom right corner will be 33 as it is going up by 10 each time in each column, and so on. From all this I can come up with algebraic expressions for any square box in a
0 x 10 grid, like so;
3 x 3 squares
S
S + 2
S + 20
S + 22
s(s + 22) = s² + 22s
(s + 2)(s + 20) = s² + 20s + 2s + 40 = s² + 22s + 40
Difference = (s² + 22s + 40) - (s² + 22s) = 40
4 x 4 squares
S
S + 3
S + 30
S + 33
s(s + 33) = s² + 22s
(s + 3)(s + 30) = s² + 30s + 3s + 90 = s² + 33s + 90
Difference = (s² + 33s + 90) - (s² + 33s) = 90
5 x 5 squares
S
S + 4
S + 40
S + 44
s(s + 44) = s² + 44s
(s + 4)(s + 40) = s² + 40s + 4s + 160= s² + 44s + 160
Difference = (s² +44s) - ( s² + 44s + 160) = 160
Now that I am able to come up with algebraic expressions for any square box in a 10 x 10 grid, I shall try to come up with an algebraic expression that can be used on any sized square in any sized grid.
First I am going to look at a 2 x 2 square box on a 9 x 9 grid, so like I have done above I am going to give expressions for the corners of the 2 x 2 square. I will start by calling the top left corner 's'
s
To find the top left expression I am going to look back at what it was when working out the expressions for the 2 x 2, 3 x 3, 4 x 4, and 5 x 5
2 x 2 top right corner = s + 1
3 x 3 top right corner = s + 2
4 x 4 top right corner = s + 3
5 x 5 top right corner = s + 4
Looking at these I can see that the number added onto 's' is the square size minus 1. I f I call the square size 'J', then I can use it in my expression, for e.g. in a 6 x 6 square 'J' would be 6. I know that the number added onto 's' is the square size minus 1, so the expression will be,
s + (J - 1), like so;
s
S + (J - 1)
Now I need to look at the expressions in the bottom left corners of the squares. I found that the bottom left corner was always the grid size being added onto the top left corner for e.g. in a 10 x 10 grid the bottom left corners were always going up in 10's as that was the grid size like so;
2
3
1
2
3
21
22
23
As you can see as the grid size is 10 x 10 then the bottom left corners are always going up in 10's. Therefore if it was a 9 x 9 grid then the bottom left corners will always go up in 9's etc.
When I found out the top right corner expression I saw that the numbers added onto 's' were (J - 1). These numbers again are being used in these bottom left corners. In a 2 x 2 square
(J - 1) is 1, and 1 x g (Grid size) is g, which is being added onto 's'. In a 3 x 3 square (J - 1) is 2, and 2 x g is 2g, which is being added onto 's', and so on.
So the expression I can use in the bottom left corner is s + g( J - 1) like so;
s
S + (J - 1)
S + g(J - 1)
Then like I found expressions for the squares like 2 x 2 and 3 x 3, to get to the bottom right corner I will just add the same increase to the bottom left as I did to the top left to get the top right, so I added (J - 1), therefore my expression will be s + g(J - 1) + (J - 1) like so;
s
S + (J - 1)
S + g(J - 1)
S + g(J - 1) + (J - 1)
Now I will just multiply the corners, and then find the expression for the difference like so;
S (s + g(J - 1) + (J - 1) = s² + gs (J - 1) + s (J - 1)
(S + (J - 1) ) (S + g(J - 1) = s² + gs( J - 1) + s( J - 1) + g( J - 1) ²
Difference = (s² + gs (J - 1) + s (J - 1) ) - (s² + gs( J - 1) + s( J - 1) + g( J - 1) ²) = g(J - 1) ²
Now I have an algebraic expression that can be used on any sized square in any sized grid, to work out the difference of the answers, when the corners were multiplied. To prove that this equation is correct I am going to take a 3 x 3 square from an 8 x 8 grid, using the equation first.
2
3
9
0
1
7
8
9
Difference = 8( 3 - 1) ² = 8 x 2² = 8 x 4 = 32
So if my equation is correct then the difference will always we 32 of a 3 x 3 square taken from an 8 x 8 grid.
2
3
9
0
1
7
8
9
x 19 = 19
3 x 17 = 51
Difference = 51 - 19 = 32
So the difference is 32 of a 3 x 3 square taken from an 8 x 8 grid, which proves that my equation was correct, and it can be used on any sized square in any sized grid, to work out the difference of the answers, when the two opposite corners are multiplied.
Number Grids Coursework
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