# An investigation into Number Grids.

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Introduction

## An investigation into Number Grids

We are investigating the effect of number grids and to try and explain the pattern between a 2x2 selection grid on a 10x10 grid. We found that if you multiply the bottom left number and the top right number and the bottom right number and top left number, then subtract one from the other you will find that the difference is 10. As is shown in the below table and number 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 | 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 |

If we multiply 3x 14 we get 42 and if we multiply the sum of 4x13 we get 52. 42 subtracted from 52 equals 10. This occurs for any selection square sized grids ie 3x3, 4x4 etc in a 10x10 grid as is shown in the table below.(Blue minus Red)

Due to the fact that the difference is always 10 we have decided to show this in a table algebraically. If we multiply the bottom left digit by the top right digit and the top left digit by the bottom right digit we can see that the N’s disappear to leave just 10 which is the difference. As is shown below.

N(n+1)=n2+1 =see below

N(n+21)=n2+21 =N2+21-N2+1 =10

For a two by two grid we can see that there is an algebraic table This is shown below also:

N | N+1 |

N+10 | N+11 |

N2+1

N2+21

When subtracted the n cancels out leaving the number 10 which is actually the sized number grid I am working with.

First number multiplication (blue) | Second number multiplication (red) | Difference |

13x4 | 3x14 | 10 |

98x89 | 88x99 | 10 |

54x45 | 44x55 | 10 |

69x60 | 69x80 | 10 |

17x8 | 7x18 | 10 |

47x38 | 37x48 | 10 |

We can see that for every multiplication we carry out the difference will always be 10. We shall carry this on and see if this works for different sized selection grids i.e. a 3x3, 4x4 etc.

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 |

Note: when one number is both blue + red it is being used in separate number grid selections.

Below is the table for a three by three grid and the algebraic formula:

N | N+2 |

N+20 | N+22 |

N2+2

N2+42

When subtracted I get 40 which is my difference.

First number multiplication(blue) | Second number multiplication | Difference |

23x5 | 3x25 | 40 |

94x76 | 74x96 | 40 |

61x43 | 41x63 | 40 |

98x80 | 78x100 | 40 |

28x10 | 8x30 | 40 |

Middle

250

90

7x7

360

110

NxN

N+p.d+20(previous Differnce)

N+20

* If I wanted to find out what the next difference would be I would add this to the difference previously. For instance take 40 to be the difference of a 3x3-selection grid. I want to find out what a 4x4 grids difference will be, I shall add 40 plus the p.d which happens to be 30 (difference between 2x2 and 3x3) and then add 20 to the result(because this is the p.d difference which increases by 20 each time) which gives me 90.

The difference between each distance increases by 20 on the previous one each time. Therefore this can be shown as n+20 where n is the difference previously for example If I wanted to find the difference for a 3x3 selection grid and a 4x4 selection grid I could find the difference between the 2x2 grid and 3x3 grid and call this n. The difference between the 3x3 and 4x4 would be n+20. The difference of the difference is basically the difference of this selection grid difference and the previous one.

I shall now see if this works on a difference-sized grid i.e. an 8x8 and 9x9.

I shall do e

Exactly the same as the previous grid-10x10 but on this 8x8. Also I shall do the equations in algebraic form.

N | N+1 |

N+8 | N+9 |

N2+1

N2+17 This equals 16 when subtracted. When divided by 2 we get 8 which is the size of the grid. Below is the table full of our results for the 8x8 grids and different sized selection grids. The above seems only to work when the selection grid size is a 2x2.(this to find the sized grid we are working from)

Top row | Bottom row | Difference |

N2+1 | N2+17 | 16 |

N2+99 | N2+115 | 16 |

N2+79 | N2+95 | 16 |

The difference is always 16. If I divide 16 by 2 I get 8 which is the number sized grid we are working with. As with my previous calculations I multiplied diagonally. I chose not to do this as it was harder when algebraic numbers appeared, however if I had done that this time I would have found the difference to be 8 initially. As is shown in the singular example below.

56 are my numbers I shall work with.

1314

13x6=78

5x14=70

78-70=8.

We tried the 8x8 grid with various different selection grid sizes and below are my results.

Grid Selection sizes | Difference | Difference of the difference each time |

2x2 | 16 | |

3x3 | 32 | 16 |

4x4 | 48 | 16 |

5x5 | 64 | 16 |

NxN | N+p.d(previous difference) | N+16 |

N being the difference of the previous result for instance if I wanted to find the difference for a 5x5 grid then I would add 16 to 48.

I shall now produce a table for the 9x9 grid to show our findings from this.

N | N+1 |

N+9 | N+10 |

Below is the table for the differences of a basic 2x2 grid within this 9x9 grid. I could predict that the difference would be 9 on the diagonal multiplications and 18 on the top and bottom separate multiplications. Lets take a look to see if my results are correct in the table below. I shall give one example of the top and bottom multiplications and one of the diagonal ones.

10x2=20

1x11=11

20-11 = 9(as I predicted)

N + N+1= N2+1

N+9 + N+10=N2 +19

19-1=18 (as I predicted)

One doubles the other because one set is multiplied whilst the other is added.

Top row | Bottom row | Difference |

N+11 | N + 30 | 18 |

N+101 | N+119 | 18 |

N+51 | N+69 | 18 |

N+143 | N+161 | 18 |

N+17 | N+35 | 18 |

NxN | NxN | 18 |

I also changed the grid selection sizes for this grid below is the table for these.

Selection size | Difference | Differences of the differences |

2x2 | 18 | 18 |

3x3 | 36 | 18 |

4x4 | 54 | 18 |

5x5 | 72 | 18 |

6x6 | 90 | 18 |

NxN | N+p.d(previous differnce) | N+18 |

Looking at these results at a whole I can see a pattern and to bring this all together I shall develop a formulae.

Firstly, I noticed that if you took the zeros off the difference, then you get square numbers for 1, 2, 3 and 4. So I tried to find a way where you get square numbers only:

2(0)-1(0)=1 x10= 10

The formula would be 10(s-1)

So I tried using this formula for a 3 by 3 square:

10(3-1) = 20

This was wrong as I had to get 40 not 20, so I would have to multiply the 10 by 4 somehow. So I squared the (3-1) to get 4.

10(3-1)²

This worked and so I substituted the number of rows and columns for letters and so the formula became:

10(r-1)² where s is the selection sized grid

Lets see if this works on a 10 x 10 number grid and a 5x5 selection grid.

10 (which is the size of the number grid) (5( which is the size of the selection grid)-1)2 So it works out at 10(5-1)2=160 which is actually the correct result.

I was right about the formulas and so an overall formula for any sized square would be: G(r-1)² where G is the grid size and r is the number of rows.

## Furthering my Investigation

I have decided to further my investigation and to look into rectangles. I shall change both the grid size and the rectangle size to try and construct a formulae linking length, size and width.

I shall start by using a 2x3 rectangle within a 10x10 grid. Below are examples and results in a table.

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 |

N | N+2 |

N+10 | N+12 |

N + N+2=N2+2

N+10 + N+12=N2+22

N2+22 – N2+2= 20

Top row | Bottom row | difference |

N2+2 | N2+22 | 20 |

N2+42 | N2+62 | 20 |

N2+112 | N2+132 | 20 |

Conclusion

I decided to alter the equation by including some of the first part of the first equation by not actually including the multiplication or squaring of the numbers but to multiply one by the other. Also I need to include the size of the actual grid that I was to use.

My new formulae I was t try was:

G(r-1)x(c-1)

To see if it worked I tried it out on my example of a 4x6 rectangle placed within an 8x8 grid.

8(4-1)x(6-1)=120

This could actually be the difference we were hoping for. Below is the calculations for the difference before and after.

Grid-size | No. of rows | No. of Columns | Difference | Difference in-between |

8x8 | 4 | 5 | 96 | 24 |

8x8 | 4 | 6 | 120 | 24 |

8x8 | 4 | 7 | 144 | 24 |

This shows that this equation does work with the result being constant and the difference being a multiple of 8 which is the grid size.

This is a table to show the differences in rectangles for a 10 by 10 grid.

Row | Column | Difference | Formula |

2 | 3 | 20 | (10(r-1))x(c-1) |

2 | 4 | 30 | (10(r-1))x(c-1) |

2 | 5 | 40 | (10(r-1))x(c-1) |

3 | 3 | 40 | (10(r-1))x(c-1) |

3 | 4 | 60 | (10(r-1))x(c-1) |

3 | 5 | 80 | (10(r-1))x(c-1) |

4 | 3 | 60 | (10(r-1))x(c-1) |

4 | 4 | 90 | (10(r-1))x(c-1) |

4 | 5 | 120 | (10(r-1))x(c-1) |

5 | 3 | 80 | (10(r-1))x(c-1) |

5 | 4 | 120 | (10(r-1))x(c-1) |

5 | 5 | 160 | (10(r-1))x(c-1) |

From this I can produce my final equation that will find the difference in any rectangle to be:

G(r-1)x(c-1)

I have taken into account the length, width and grid size to produce an equation that gives precise differences at all times.

In this formula, r is the No. of Rows and c is the No. of columns.

Basically, you multiply (r-1) by 10 and then times your answer by (c-1).

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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