Number Grids

Introduction

Nature is full of patterns, and many patterns can be found to have a corresponding mathematical pattern which describes them. Mathematical expressions can be found which describe the pattern, or mathematical rules or algorithms can be used to write down them pattern in the same way that nature has constructed them. A very famous example is in the pattern made by sunflower seeds, which is all bound up with the Fibonnacci sequence and the golden ratio. It seems that nature uses this sequence as a way of packing the seeds most closely.
In this piece of coursework we are going to investigate the pattern in number grids made by drawing squares in them, multiplying together numbers in opposite corners and then subtracting the answers. It may be that we can find an expression to explain our results and predict the answer for any size square drawn in any size grid.



































Investigation

In the 10 by 10 grid I draw squares 2 by 2 squares and calculate the differences between the products of the numbers in opposite corners.


1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 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 99 100



22x13-12x23=10
82x73-72x83=10
68x59-58x69=10

The difference is 10 each time. Can we prove it is always 10?
If we label the corners of a completely general square as shown,


N N+1
N+10 N+11

then the difference is (N+10)(N+1)-N(N+11)=10

In the 11 by 11 grid I draw squares 2 by 2 squares and calculate the differences between the products of the numbers in opposite corners


1 2 3 4 5 6 7 8 9 10 11
12 13 14 15 16 17 18 19 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
100 101 102 103 104 105 106 107 108 109 110
111 112 113 114 115 116 117 118 119 120 121
122 123 124 125 126 127 128 129 130 131 132
24x14-13x25=11
93x83-82x94=11
130x120-119x131=11

The difference is 11 each time. Can we prove it is always 11?
If we label the corners of a completely general square as shown,


N N+1
N+11 N+12

then the difference is (N+11)(N+1)-N(N+12)=11

In the 12 by 12 grid I draw squares 2 by 2 squares and calculate the differences between the products of the numbers in opposite corners


1 2 3 4 5 6 7 8 9 10 11 12
13 14 15 16 17 18 19 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 99 100 101 102 103 104 105 106 107 108
109 110 111 112 113 114 115 116 117 118 119 120
121 122 123 124 125 126 127 128 129 130 131 132
133 134 135 136 137 138 139 140 141 142 143 144
145 146 147 148 149 150 151 152 153 154 155 156

26x15-14x27=12
104x93-92x105=12
134x123-122x135=12.

Can we prove it is always 12?
If we label the corners of a completely general square as shown,


N N+1
N+12 N+13

then the difference is (N+12)(N+1)-N(N+13)=12




It seems that for a 2 by 2 square in a g by g grid, the difference is just g. What about larger size squares?


1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

In the number grid above, the highlighted square with corners 7, 8, 12,13 gives result = 12x8-7x13=5. The square with corners 19, 20, 24, 25 gives result = 24x20-19x25=5.

If we draw a general 2 by 2 square in a 5 by 5 grid then in general our square will look like this.


N N+1
N+5 N+6

If now we multiply corners together and subtract we obtain

(N+5)(N+1)-N(N+6)= 5

Hence for a 2 by 2 square in a 5 by 5 grid, we have proved the general case. What about a 3 by 3 square in a 5 by 5 grid?


1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

If we multiply the corners together now for top left grid and subtract the answer is = 11x3-1x13=20. For the grid at bottom right the answer = 23x15-13x25 =20

If we label the corners in the same way as before we obtain:


N N+1 N+2
N+5 N+6 N+7
N+10 N+11 N+12

Multiplying together the numbers now and subtracting give answer = (N+10)(N+2)-N(N+12)=20.

If we draw a 4 by 4 square in a 5 by 5 grid we obtain







1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

Multiplying together and subtracting the corners gives 16x4-1x19=45


N N+1 N+2 N+3
N+5 N+6 N+7 N+8
N+10 N+11 N+12 N+13
N+15 N+16 N+17 N+18

Ditto for the labelled square gives (N+15)(N+3)-N(N+18)=45


1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

Multiplying together and subtracting the corners for the 5 by 5 square gives 21x5-1x25=80.

We have a sequence: 5, 20, 45, 80. We can write them as
5x12,5x22,5x32,5x42. Hence for an k by k square in a 5 by 5 grid, the answer is
5(k-1)2. Can we do more? What about 6 by 6 grids or bigger grids than this?


1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36

7x2-1x8=6
11x6-5x12=6
21x16x15x22=6
31x26-25x32=6
35x30-29x36=6


N N+1
N+6 N+7

(N+6)(N+1)-N(N+7)=6








1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36

13x3-1x15=24
27x17-15x29=24


N N+1 N+2
N+6 N+7 N+8
N+12 N+13 N+14

(N+12)(N+2)-N(N+14)=24


1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36

19x4-1x22=54
33x18-15x36=54


N N+1 N+2 N+3
N+6 N+7 N+8 N+9
N+12 N+13 N+14 N+15
N+18 N+19 N+20 N+21

(N+18)(N+3)-N(N+21)=54










1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36

25x5-1x29=96
32x12-8x36=96


1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36

Multiplying opposite corners together and subtracting gives
31x6-36x1=150

We have another sequence: 6, 24,54,96,150.
We can write this as 6x12,6x22,6x32, 6x42,6x52


If we compare this with our previous result, the only difference is that 6 has been replaced by 5. The rule is therefore 6(k-1)2 .Is this coincidence?

Suppose we write down the rule, for a k by k square in a g by g grid, the answer is g(k-1)2 . We will test this for a 10 by 10 grid.

Our results should go
10(2-1)2 =10
10(3-1)2 =40
10(4-1)2 =90
10(5-1)2 =160















1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 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 99 100

2 by 2 11x2-12x1=10
3 by 3 77x59-57x79=40
4 by 4 35x8-5x38=90
5 by 5 81x45-41x85=160

As predicted.
So far we have only drawn squares but what happens for rectangles? If we draw a j by k rectangle what will the difference be then?


1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

For the 2 by 3 rectangle shown the difference is 7x4-2x9=10
For the 2 by for rectangle the difference is 22x20-17x25=15
Remember that for the two by two rectangle in a the difference was 5. It seems that for a 2 by k rectangle in a 5 by 5 grid the difference is 5 (k-1).


1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

For the 3 by 2 rectangle the difference is 11x2-1x12=10
For the 4 by 2 rectangle the difference is19x5-4x20= 15
Remember that for the 2 by two rectangle the difference was 5. It seems that in a j by 2 rectangle in a 5 by 5 grid the difference is 5(k-1).


1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

For the 3 by 4 table the difference is 11x4-1x14=30


1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25

For the 4 by 5 table the difference is 16x5-1x20=60
If we take out the factor of 5 because of the size of the table the last two results become 6 and 12 respectively.

6=(3-1)(4-1) and 12=(4-1)(5-1)

It seems that two find the difference for a j by k table in a 5 by 5 grid we only have to find 5(j-1)(k-1). To find the difference for a j by k table in a g by g grid would we just find g(j-1)(k-1)? Let’s try for a 9 by 9 grid.
I predict for the rectangle drawn below 9(2-1)(7-1)=108


1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
19 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

39x27-21x45=108. Our formula has worked.




We can explain our rule algebraically. I have draw the corners of a general j by k rectangle and ignored the interior. Each square to the right increases the entry by 1, so moving k-1 squares increases the total to n+k-1. One square moved down increases the total by g, so moving j-1 squares increases the total by (j-1)g.


n n+k-1
n+(j-1)g n+(j-1)g+k-1

The difference is (n+(j-1)g)(n+k-1)-n(n+(j-1)g+k-1)=g(j-1)(k-1). Our hypothesis is proved.


















Conclusion

We have found a rule explaining our results. For a k by k square in a g by g grid , the answer is g(k-1)2 . For a j by k rectangle in a g by g grid the difference is g(j-1)(k-1).
For the patterns we have found we have deduced a formula. This formula is completely general and applies to any formula in any size grid. This is typical of the relationship between maths and the world around us, in which things move according to the laws of physics, described in mathematical terms. Nature is full of patterns and the beauty of maths lies in compacting a whole phenomena into an equation as short as the one we have found.