# number grid

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Introduction

Khadiya Ahmed

GCSE math coursework

Tutor: Ronnie Fraser

Aim: investigate the product difference of different size boxes and produce an algebraic formula.

First I will draw a box around four numbers and find the product of the numbers cross opposite side of the corners of the square.

The product is found by multiplying the top left number with the bottom right number in the box. Now do the same thing with the top right and the bottom left as shown in the diagram.

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 | 98 | 100 |

1 2 3

11 12 13

21 22 23

Product 1: 12 x 23 = 276

Product 2: 13 x 22 = 286

Now subtract the two products, to find the difference

286 – 276 = 10

I will use this method to calculate the difference of more 2 by 2 boxes, chosen randomly in the big 10 by 10 grid.

2 by 2 Boxes | Subtracting products | Difference |

1st | 2832 – 2842 | 10 |

2nd | 2530 – 2520 | 10 |

3rd | 70 – 60 | 10 |

4th | 682 – 672 | 10 |

5th | 5986 – 5976 | 10 |

Now I will prove that the difference is 10 algebraically. Select any 2 by 2 box in the 10 by 10 grid. Swap one of the numbers in the 2by2 box with N

12 | 13 |

22 | 23 |

N | N + 1 |

N + 10 | N + 11 |

→

Now multiply the top left box with the bottom right one e.g. N x (N + 11). Then multiply top right box with bottom left box e.g. (N +1) (N + 10). Do the same as with the arithmetic product when told to find the difference, by subtracting the N x (N + 11) from the (N +1) (N + 10) using the FOIL method.

N (N + 11) – (N + 10) (N + 1)

N2 +11N – N2 + 1N +10N + 10

N2 +11N – N2 + 11N + 10

Difference = 10

Cconclusion: the difference for a 2by2 squared box is always 10

Now I will investigate further, by finding the difference of the following squares sizes 3 by3, 4 by 4, 5 by 5, 6 by 6, 7 by 7, 8 by 8, 9 by 9 and 10 by 10.

Middle

12

13

14

15

21

22

23

24

25

31

32

33

34

35

N +40

42

43

44

N +44

1 | 2 | 3 | 4 | 5 |

11 | 12 | 13 | 14 | 15 |

21 | 22 | 23 | 24 | 25 |

31 | 32 | 33 | 34 | 35 |

41 | 42 | 43 | 44 | 45 |

Difference is found by subtracting the two products (bold numbers)

N (N + 44) – (N + 4 (N + 40)

N2 +44N – N2 + 4N +40N + 40

N2 +44N – N2 + 44N + 40

Difference = 160

Conclusion: the difference for a 5x5 squared box is always 160

The product difference for the different size square show pattern in their sequence

2x2 3x3 4x4 5x5 6x6

10 40 90 160 250 1st difference

30 50 70 90 2nd difference

20 20 20 3rd difference

The 3rd number difference is constant. Consequently now I can predict the next term or in this case the box number difference using the sequences

10 40 90 160 250

30 50 70 90

20 20 20

The next term can be predicted by adding 20 to the last term of the second sequence, which 90. Then add 90 to the last term in the first sequence, in this case it’s 160. Therefore 90 + 160 = 250.

This method can be used to predict the difference instead of the arithmetic way. I used this method to find the difference for the following by 7, 8 by 8, 9 by 9 and 10 by 10.

2x2 3x3 4x4 5x5 6x6 7x7 8x8 9x9 10x10

10 40 90 160 250 360 490 640 810 1st difference

30 50 70 90 110 130 150 170 2nd difference

20 20 20 20 20 20 20 3rd difference

I was provided with a formula to find the Nth an2 + bn + c. and all that was lift for me to do was to use this formula to find the Nth of my sequences.

c = 0 10 40 90 160 250 360 490 640 810 1st difference

a + b = 10 30 50 70 90 110 130 150 170 2nd difference

2 a = 20 20 20 20 20 20 20 20 3rd difference

First the value of a is found

2a = 20

a = 20 ÷ 2

a = 10

Then the value of b is found

a + b = 10

10 + b = 10

b = 10-10

b = 0

From the sequence I can already concluded that the value of c = 0. Now I will put the values in the formula of finding the nth term. The formula to find the Nth of a quadratic sequence is an2 + bn + c

a = 10

b = 0

c = 0

10n2 + 0n + 0 Formula cancels down to 10n2

I will check if my formula works

10 x 22 = 40 wrong answer

Since my formula didn’t work. When I substituted 2 in n’s place. I decided to do experimentations with my sequence and formula. I found out a way to get the answer but it involved the use of invented numbers, which didn’t include the sequence.

Theory:

1 2 3 4 5 6 7 8 9

2x2 3x3 4x4 5x5 6x6 7x7 8x8 9x9 10x10

10 40 90 160 250 360 490 640 810 1st difference

30 50 70 90 110 130 150 170 2nd difference

20 20 20 20 20 20 20 3rd difference

I decided instead of putting 2 in n’s place to find the difference for 2 by 2 boxes. I will put my first number of the bold number sequence. Thus to find the nth term, for 3 by 3 boxes you do 10 x 12.

Box size | Formula | Difference |

2x2 | 10 x 12 | 10 |

3x3 | 10 x 22 | 40 |

4x4 | 10 x 32 | 90 |

5x5 | 10 x 42 | 160 |

6x6 | 10 x 52 | 250 |

7x7 | 10 x 62 | 360 |

8x8 | 10 x 72 | 490 |

9x9 | 10 x 82 | 640 |

10x10 | 10 x 92 | 810 |

My formula did work as it showed the right answers. However I believed it wasn’t consistent enough. So I went back to my old theory about the 10n2 would equal the difference of the box, where n represent the square size, even though it did not work straight away. I believed that I should continue to use the formula just to see if it shows a pattern of any kind

Box size | Formula | Difference |

2x2 | 10 x 22 | 40 |

3x3 | 10 x 32 | 90 |

4x4 | 10 x 42 | 160 |

5x5 | 10 x 52 | 250 |

6x6 | 10 x 62 | 360 |

7x7 | 10 x 72 | 490 |

8x8 | 10 x 82 | 640 |

9x9 | 10 x 92 | 810 |

10x10 | 10 x 102 | 1000 |

Conclusion

I used this principle of adding the constant difference to the previous term in the sequence to find the next term in the sequence. I decided that tabulation was the best way presents my data.

Box size | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

2 | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |

3 | 20 | 40 | 60 | 80 | 100 | 120 | 140 | 160 | 180 |

4 | 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 |

5 | 40 | 80 | 120 | 160 | 200 | 240 | 280 | 320 | 360 |

6 | 50 | 100 | 150 | 200 | 250 | 300 | 350 | 400 | 450 |

7 | 60 | 120 | 180 | 240 | 300 | 360 | 420 | 480 | 540 |

8 | 70 | 140 | 210 | 280 | 350 | 420 | 490 | 560 | 630 |

9 | 80 | 160 | 240 | 320 | 400 | 480 | 560 | 640 | 720 |

10 | 90 | 180 | 270 | 360 | 450 | 540 | 630 | 720 | 810 |

Now I will find a constant formula for the rectangles boxes as I did for the squared boxes. Formula to find the nth for a quadratic sequence is an2 + bn + c. I decided to use my formula 10 (n-1)2

Even though, my sequence wasn’t a quadratic sequence. Looking carefully at the formula I noticed that it was squared. I didn’t need to square my numbers because they were not the same numbers in the columns and rows. Therefore by rewriting the formula I eliminated the square.

10 (n-1)2 = 10 (n-1) (n-1)

I tested if my new theory worked, by substituting the value of constant width of 2 (row) and a length of 3 (column) into my formula. 10 (2-1) (3-1) = 20. This the correct answer, because when I did the arithmetic work for 2x3 boxes I also got 20 as the product difference. The formula however needs to be modified, since n has two values, one constant number, and a changing number.

The formula for rectangle boxes is 10 (m-1) (n-1), where m represent the number of columns and n represent the number of rows. E.g. 2x3, the 2 would be the m and the 3 would be the n in the formula. To find the product difference for any size box in any size grid. You use this formula g (m-1) (n-1), where g represents the grid size and m represents the number of columns and n represent the number of rows.

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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