# Number Grid Aim: The aim of this investigation is to formulate an algebraic equation that works out the product of multiplying diagonally opposite corners of a particular shape and finding the difference between the results

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Introduction

Amanda Gaber

March 2006

Mathematics Coursework

Number Grid

Aim: The aim of this investigation is to formulate an algebraic equation that works out the product of multiplying diagonally opposite corners of a particular shape and finding the difference between the results. I will start off by working this out on a 2x2 square in a 10 x 10 grid and then will investigate varying the widths and lengths of squares and rectangles.

Method: In order to simplify the process, the investigation has been divided into sections according to the size of squares in the grids.

2 x 2 Squares

To begin with, I used a 10 x 10 grid, looking at 2 x 2 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 | 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 |

Then I took 2 x 2 squares from this grid and multiplied the opposing corners to calculate the difference between the two products.

1 | 2 |

11 | 12 |

1 x 12 = 12

2 x 11 = 22

∴ 22 – 12 = 10

So the difference between the answers is 10. I then took another 2 x 2 box from the above 10 x 10 grid:

47 | 48 |

57 | 58 |

47 x 58 = 2726

48 x 57 = 2736

∴ 2736- 2726 = 10

The difference is 10 again. Perhaps this means that because it is a 10 x 10 grid, that all the differences would be 10. I would still like to further investigate this theory.

The grid below is once again a 2 x 2 box derived from the original 10 x 10 grid.

89 | 90 |

99 | 100 |

89 x 100 = 8900

90 x 99 = 8910

∴ 8910 – 8900 = 10

This once again confirms what I stated; that the difference between the products of cross-multiplied boxes will always equal 10 in a 10 x 10 grid.

I would like to determine if this is definitely correct, so I am going to do it again twice.

27 | 28 |

37 | 38 |

27 x 38 = 1026

28 x 37 = 1036

∴ 1036 – 1026 = 10.

43 | 44 |

53 | 54 |

43 x 54 = 2322

44 x 53 = 2332

∴ 2332 – 2322 = 10.

Middle

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3620

As before, the highlighted boxes are the ones I am going to calculate, to see if the difference between the product of multiplying the opposite corners, equals the size of the grid. In this case, I am looking for that number to be 19.

- 1 x 21 = 21

2 x 20 = 40

∴ 40 – 21 = 19

- 32 x 52 = 1664

33 x 51 = 1683

∴1683 – 1664 = 19

- 177 x 197 = 34869

178 x 196 = 34888

∴ 34888 – 34869 = 19

- 302 x 322 = 97244

303 x 321 = 97263

∴ 97263 – 97244 = 19

- 305 x 325 = 99125

306x 324 = 99144

∴ 91444 – 99125 = 19

Once again, I can conclude that the difference between the cross multiplied products is the size of the grid; 19.

To confirm this, the number 19 has been inserted into the formula to prove that this is correct.

n | n + 1 |

n + g | n + g + 1 |

n(n + g + 1) = n2 + ng + n

n + 1(n + g) = n2 + ng + n + g

∴ [n2 + ng + n +g] – [n2 + ng + n ] = g

where g = 19;

n(n + 19 + 1) = n2 + 19n + n

n + 1(n + 19) = n2 + 19n + n + 19

∴ [n2 + 19n + n +19] – [n2 + 19n + n] = 19

My prediction of what the difference would be was correct. So for any 2x2 square taken from any size grid, the difference will be the number of the grid size.

I can now work out the difference of any 2x2 square as long as I know the grid size.

5 x 5 grid = Difference of 5

19 x 19 grid = Difference of 19

100 x 100 grid = Difference of 100

31 x 31 grid = Difference of 31

58 x 58 grid = Difference of 58

1000x1000 grid = Difference of 1000

3 X 3 Squares

Having studied 2 x 2 boxes within different sized grids, I would like to study 3 x3 boxes within grids of varying sizes to see if there is an emerging pattern.

I have decided to look at an 8 x 8 grid first to see if there is a pattern.

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 |

I am going to leave out all of the numbers in the middle of the 3 x 3 box and will concentrate on the four corners of this box.

Firstly, I need to work out the difference of the products for the highlighted 3 x 3 box.

26 x 44 = 1144

28 x 42 =1176

∴ 1176 – 1144 =32

From this, I can see that the difference is not the size of the grid (g). However, I will formulate an expression to predict the size of the difference for any 3 x 3 box within an 8 x8 grid,

The first top left hand box is shown as ‘n’ and the size of the grid is still ‘g’. The top right hand corner of the box is 2 more than the top left hand corner, so this is n+ 2. The bottom left hand corner of the box is two rows down exactly, so this n + 2g and finally, the bottom right hand corner of the box is n + 2g + 2.

Therefore;

n | n + 2 |

n + 2g | n + 2g+ 2 |

After multiplying the diagonal corners of the box, the algebraic formula is as follows:

(n )( n + 2g + 2) = n2 + 2ng + 2n

(n + 2)(n + 2g) = n2 + 2ng +2n + 4g

Then subtract either sides of the equation from each other to confirm what the difference is.

(n2 + 2ng +2n + 4g) – (n2 + 2ng + 2n) = 4g

This is accurate because the 8 x8 grid had a difference between the products of 32, which is equal to 4 x 8. Therefore, I can confirm that, according to my calculations, the difference between the products of the diagonal corners is 4g.

I will now use a different 3 x 3 box within the same 8 x8 grid to see if this rule remains the same;

6 | 7 | 8 |

14 | 15 | 16 |

22 | 23 | 24 |

Here, I will also multiply the corners only so:

6 x 24 = 144

8 x 22 = 176

∴ 176 – 144 = 32.

This proves that my prediction is correct.

I will now investigate if this formula for 3 x 3 boxes works in a different sized 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 |

The box I have highlighted will be used to work out the product of the corners and the difference between them.

1 x 17 = 17

3 x 15 = 45

∴45 – 17 = 28.

This fits the pattern, as 4 x 7 is equal to 28, where 7 is the grid size.

4 X 4 Square

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 |

- 4 x 31 = 124

1 x 34 = 34

∴124 – 34 = 90

- 24 x 57 = 1368

27 x 54 = 1458

∴ 1458 – 1368 = 90

- 61 x 94 = 5734

64 x 91 = 5824

∴ 5824 – 5734 = 90

- 67 x 100 = 6700

70 x 97 = 6790

∴ 6790 – 6700 = 90

In a 10 x 10 grid, the difference between the diagonal corners of a 4 x 4 square is always 90.

2 x 3 Rectangle

As yet, I am unable to determine a general formula to predict the difference between multiplying the diagonals of a square or a rectangle in any sized grid. I will therefore investigate the difference between the products of the diagonal corners of a rectangle.

I am going to begin with a 2 x 3 rectangle in a 10 x 10 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 |

Conclusion

If the number in the top left hand corner is n then the following squares will be…

For a 2 x 3 rectangle when the first square is n the difference is:

n | n+2 |

n+G | n+2+G |

(n+2) (n+G) = n2 + nG + 2n + 2G

n (n+G+2) = n2 + nG + 2n

∴ (n+2) (n+G) - n (n+G+2) = 2G

The algebraic expression for the difference of a 2 x 3 rectangle is 2G, the width of the grid multiplied by two.

For a 2x4 rectangle when the first square is n the difference is:

n | n+3 |

n+G | n+3+G |

(n+3) (n+G) = n2 + nG + 3n + 3G

n (n+G+3) = n2 + nG + 3n

∴ (n+3) (n+G) - n (n+G+3) = 3G

The algebraic expression for the difference for this size of rectangle is 3G, the width of the grid multiplied by three.

From this information, I have been able to tabulate my results:

L | H | nG |

2 | 3 | 2 |

2 | 4 | 3 |

3 | 5 | 6 |

From the table above, it is apparent that between 2 x 3 and 2 x 4 rectangles, nincreases by 1. This could be due to the height increasing by 1. I think that the height is probably related in some way to the difference value.

Evaluation:

I have formulated the algebraic formula that can predict the difference for any sized rectangle or square for any sized grid. Having investigated this observation, I can only ascertain that this is correct and that;

L – 1 x G (H -1)

Perhaps further investigations on different grid sizes would be suitable. Also, given more time, I would have liked to look at different shapes in different sized grids and to determine a general formula for them. For example, I would like to know if the same formula fits a rhombus on a 10 x 10 grid.

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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## Here's what a teacher thought of this essay

This is a well written piece of work with only a couple of minor errors. This piece of work shows a good application of some algebraic techniques. There are specific strengths and improvements suggested throughout.

Marked by teacher Cornelia Bruce 18/04/2013