# Number Grid Coursework

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Introduction

Maths Investigation 1

Higher Tier Task - Number Grid

## Section 1: 2x2 Box on Width 10 Grid

1) Introduction

I was given a number grid, like the one in Fig 1.1. The task was to, in the 2x2 box, find the product of top-left (TL) and bottom-right (BR) numbers, and the product of the top-right (TR) and bottom-left (BL) numbers and then to calculate the difference of these two products.

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 |

Fig 1.1

This calculation rule was to be followed throughout the investigation. Having done this, I found the difference of the two products to be 10 and I wondered what would happen if the box was placed in other locations on the grid.

2) Method

To discover this, I will calculate the difference of the two products for 5 different random locations of the box within the grid. As it would be impractical (and impossible if the grid extended to more than 10 rows) to do all possible calculations, 5 should be enough to display any patterns that may lie therein.

3) Data Collection

Here are the results of the 5 calculations for 2x2 Box on Width 10 Grid:

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

15 | 390 | 400 | 10 |

21 | 672 | 682 | 10 |

44 | 2420 | 2430 | 10 |

57 | 3876 | 3886 | 10 |

83 | 7802 | 7812 | 10 |

4) Data Analysis

From the table, it is very easy to see that on all tested locations of the box, the difference of the two products was 10.

5) Generalisation

Using this apparently constant number, it can be assumed that for all possible locations of the 2x2 box on the width 10 grid, that the difference is always 10. Therefore, the following equation should be satisfied with any real value of a, where:

a is the top-left number in the box;

(a + 1) is the top-right number in the box, because it is always “1 more” than a;

(a + 10) is the bottom-left number in the box, because it is always “10 more” than a;

(a + 11)

Middle

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Fig 2.5

Fig 2.1 to Fig 2.5 are the grids used for the varying values of z. An example of the 2x2 box has been highlighted on each one.

(a) Here are the results of the 5 calculations for 2x2 Box on Width 11 Grid (Fig 2.1):

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

12 | 288 | 299 | 11 |

25 | 925 | 936 | 11 |

42 | 2268 | 2279 | 11 |

63 | 4725 | 4735 | 11 |

65 | 5005 | 5016 | 11 |

(b) Here are the results of the 5 calculations for 2x2 Box on Width 12 Grid (Fig 2.2):

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

8 | 168 | 180 | 12 |

26 | 1014 | 1026 | 12 |

43 | 2408 | 2420 | 12 |

51 | 3264 | 3276 | 12 |

67 | 5360 | 5372 | 12 |

(c) Here are the results of the 5 calculations for 2x2 Box on Width 13 Grid (Fig 2.3):

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

2 | 32 | 45 | 13 |

28 | 1176 | 1189 | 13 |

36 | 1800 | 1813 | 13 |

48 | 2976 | 2989 | 13 |

69 | 5727 | 5740 | 13 |

(d) Here are the results of the 5 calculations for 2x2 Box on Width 14 Grid (Fig 2.4):

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

15 | 465 | 479 | 14 |

32 | 1504 | 1518 | 14 |

54 | 3726 | 3740 | 14 |

66 | 5346 | 5360 | 14 |

82 | 7954 | 7968 | 14 |

(e) Here are the results of the 5 calculations for 2x2 Box on Width 15 Grid (Fig 2.5):

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

5 | 105 | 120 | 15 |

19 | 665 | 680 | 15 |

39 | 2145 | 2160 | 15 |

48 | 3072 | 3087 | 15 |

70 | 6020 | 6035 | 15 |

From these results, it is possible to take the calculated difference of the two products, and plot this against the width of the grid (z):

Fig 2.6

4) Data Analysis

From the tables (a)-(e), it is possible to see that with a 2x2 box, the difference of the two products always equals z: the width of the grid. When plotted on a graph (Fig 2.6), the relationship is clearly visible as a perfect positive correlation. A graph is a good choice to show this relationship however, a straight line drawn between these would not be correct because with this particular problem, the values to be inputted into equations must be natural numbers i.e. Integers > 0.

5) Generalisation

Using this apparent relationship, it can be assumed that, when a 2x2 box is placed anywhere on the grid, the difference of the two products will be z for all possible widths. Therefore, the following equation should be satisfied with any real value of a and any real value of z where:

a is the top-left number in the box;

(a + 1) is the top-right number in the box, because it is always “1 more” than a;

(a + z) is the bottom-left number in the box, because it is always “the grid width (z)” more than a;

(a + z + 1) is the bottom-right number in the box, because it is always “the grid width (z) plus 1” more than a.

(a + 1)(a + z) - a(a + z + 1) = z

a | a+1 |

a+z | a+z+1 |

This means that I predict that with a 2x2 box on a width z grid, the difference of the two products will always be z, the width of the grid.

6) Testing

My formula works as shown with the following, previously unused values:

1) Where a = 9, and z = 16

Difference = | (a + 1)(a + z) - a(a + z + 1) (9 + 1)(9 + 16) - 9(9 + 16 + 1) 10 x 25 – 9 x 26 250 – 234 16 (N.B. also = z) |

2) Where a = 72 and z = 17

Difference = | (a + 1)(a + z) - a(a + z + 1) (72 + 1)(72 + 17) - 72(72 + 17 + 1) 73 x 89 – 72 x 90 6497 - 6480 17 (N.B. also = z) |

7) Justification

The formula can be proven to work with the following algebra:

Difference = | (a + 1)(a + z) - a(a + z + 1) a2 + az + a + z - {a2 + az + a} a2 + az + a + z - a2 – az – a a2 - a2 + az – az + a – a + z z |

The formula works because the “z” term is only produced in the expansion of the two brackets on the left of the minus sign, and not from the more simply factorised “a(a + z + 1)” term. The a2 term is present on both sides of the minus sign, as are the az and a terms. Therefore, they cancel each other out to leave z.

8) Conclusion

After this justification, it can now be said that for every 2x2 box on a Width z Grid, the difference of the two products will always be z.

9) Extension

Having done this, I saw that my formula would only work for 2x2 boxes on a Width z grid. To improve the usefulness of my formula, I wondered what would happen to the difference of the two products if I varied the length of the box i.e. made it a 3x3 or 4x4.

## Section 3: “p x p” Box on Width 10 Grid

1) Introduction

Throughout this section, the variable p will be used to represent the length of the square box. The variable a will continue to be used for the top-left number in the box i.e. the location of the box upon the grid.

2) Method

Varying values of p will be tested to give different lengths of sides for the boxes. The lengths will range from 3 to 7. With these boxes, in 5 different random locations on the width 10 grid, the differences of the two products will be calculated. Again, I believe 5 calculations are enough to display any patterns.

3) Data Collection

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 |

Fig 3.1

Fig 3.1 is the width 10 grid, with 3x3 to 7x7 example boxes.

a) Here are the results of the 5 calculations for 3x3 Box on Width 10 Grid:

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

5 | 135 | 175 | 40 |

23 | 1035 | 1075 | 40 |

37 | 2183 | 2223 | 40 |

52 | 3848 | 3888 | 40 |

77 | 7623 | 7663 | 40 |

b) Here are the results of the 5 calculations for 4x4 Box on Width 10 Grid:

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

6 | 234 | 324 | 90 |

35 | 2380 | 2470 | 90 |

55 | 4840 | 4930 | 90 |

57 | 5130 | 5220 | 90 |

62 | 5890 | 5980 | 90 |

c) Here are the results of the 5 calculations for 5x5 Box on Width 10 Grid:

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

2 | 92 | 252 | 160 |

34 | 2652 | 2812 | 160 |

41 | 3485 | 3645 | 160 |

52 | 4992 | 5152 | 160 |

56 | 5600 | 5760 | 160 |

d) Here are the results of the 5 calculations for 6x6 Box on Width 10 Grid:

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

1 | 56 | 306 | 250 |

13 | 884 | 1134 | 250 |

21 | 1596 | 1846 | 250 |

32 | 2784 | 3034 | 250 |

43 | 4214 | 4464 | 250 |

e) Here are the results of the 5 calculations for 7x7 Box on Width 10 Grid:

Top-Left Number | Product 1 (TL x BR) | Product 2 (TR x BL) | Difference (P’duct 2 - P’duct 1) |

3 | 207 | 567 | 360 |

11 | 847 | 1207 | 360 |

23 | 2047 | 2407 | 360 |

31 | 3007 | 3367 | 360 |

34 | 3400 | 3760 | 360 |

4) Data Analysis

From tables (a)-(e), it is possible to see that all the differences of the products tested are multiples of 10. It also seems that there is a pattern, where the difference equals “the length of the box minus 1” squared multiplied by 10.

5) Generalisation

Using this apparent rule, it can be assumed that for all possible locations of a box of side p on a width 10 grid, the difference of the two products is always “the length of the box minus 1” squared, multiplied by 10. Therefore the following equation should be satisfied with any real value of a, and any real value of p where:

a is the top-left number in the box;

(a + [p – 1]) is the top-right number in the box, because it is always “[p – 1] more” than a;

(a + 10[p – 1]) is the bottom-left number in the box, because it is always “10 multiplied by [p – 1]” more than a;

(a + 10[p – 1] + [p – 1]) is the bottom-right number in the box, because it is always “[p – 1] plus 10 multiplied by [p – 1]” more than a.

For simplicity, d = [p – 1]:

(a + d)(a + 10d) - a(a + 10d + d) = 10d2

a | a+[p-1] | |

Conclusion

8) Conclusion

After this justification, it can now be said that for every possible p x q box on a Width z Grid, the difference of the two products will always be z(p – 1)(q – 1).

#### Investigation Conclusion and Evaluation

From the simple study of a 2x2 square box on a width 10 grid, I have been able to progress all the way to the formula to find the difference of the two products on any rectangular box on any width grid. The 2x2 box given in the project brief was useful because, now, I can see that it gives the number “1” in (p – 1)(q – 1). This is, therefore, very useful for spotting patterns i.e. the differences with different grid widths. However, when advancing into more complex areas of the project, like the last section where three different variables were altered, spotting patterns became more difficult and knowledge from the prior sections was required to find the formula.

Overall, the best way of presenting the results was in tables because there was an obvious pattern of values in the “Difference” column, because they were all in one line. A graph or diagram would not have been as suitable because the patterns would not have been as apparent. However, a graph was useful in proving that the grid width has a bearing on the difference with a 2x2 box. It showed a perfect positive correlation which meant that the formula was visibly true for all real grid width values. (N.B. “Real”, in this investigation, meant a number that could be used within the problem itself – as it happens, these were natural numbers.)

As a result of this investigation, I can now present a formula which has been proven to work for all real values inputted:

z(p – 1)(q – 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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