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  • Level: GCSE
  • Subject: Maths
  • Word count: 3313

Number grid

Extracts from this document...

Introduction

Uzair

Diagonal differences

Introduction - In this casework I am going to examine the difference between the products of opposite corners. I am firstly going to start with a 2x2 box then go further into my investigation by working out formulas and using different box sizes and grids.

Part 1 (Square boxes -10 column grid)

I am going to find the diagonal difference of any 2x2 box in a 10 column grid. From this I will try to find out if there are any correlations and to find differences for other

2 x 2 Box-Numeric’simage00.png

1image12.pngimage01.png

image22.png

2

11

12

image33.png

65

66

75

76

image41.png

By looking at the 2 x 2 boxes in a 10 column grid, I have found out that if you place a   2 x 2 box, any where in the grid, the difference will always be 10.

Predict

Therefore I predict that, if I place any 2 x 2 box in a 10 column grid the difference will be 10.

27

28

37

38

image42.png

This estimate proves my aim, that if you place any 2 x2 box in a 10 column grid, the difference will always be 10

Difference in Algebra

Now I am going to show what the difference will be for any 2 x 2 box in a 10 column grid algebraically.image43.png

n

n+1

n+10

n+11

Using algebra shows that my prediction was correct that, for any 2x2 box in a 10 column grid the difference will always be 10.

...read more.

Middle

42=16

Therefore I can predict that the difference for a 5 x 5 grid will be 80. Now I am going to prove this prediction algebraically.

5 x 5 box - Proving algebraically

n

n+1

n+2

n+3

n+4

n+5

n+6

n+7

n+8

n+9

n+10

n+11

n+12

n+13

n+14

n+15

n+16

n+17

n+18

n+19

n+20

n+21

n+22

n+23

n+24

image11.png

This shows that I have proved the difference for a 5 x 5 box in a 5 column grid correctly.

Now to find out a general formula for the difference for any square box in a 5 column grid will be.

Box size ( b )

Formula

Difference

3 x 3

5 (3-1)2 =

20

5 x 5

5 (5-1)2 =

80

         b x b

5(b–1)2

Therefore, the overall formula for any square box in a 5 column grid is 5(n–1)2

I am now going to test the general formula by doing the differences for all square boxes in a 5 x 5 column grid.

2 x 2 = 5(2–1)2=5

3 x 3 = 5(3–1)2=20

4 x 4 = 5(4–1)2=45

5 x 5 = 5(5–1)2=80

7 x 7 Column Grid

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

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27

28

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30

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32

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42

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49

n

n+1

n+2

n+3

n+4

n+5

n+ 6

n+7

n+8

n+9

n+10

n+11

n+12

n+13

n+14

n+15

n+16

n+17

n+18

n+19

n+20

n+21

n+22

n+23

n+24

n+25

n+26

n+27

n+28

n+29

n+30

n+31

n+32

n+33

n+34

n+35

n+36

n+37

n+38

n+39

n+40

n+41

n+42

n+43

n+44

n+45

n+46

n+47

n+48

I am going to find the diagonal difference of any 2 x2 box in a 7 column grid. From this I will try to find out if there are any correlations.

2 x 2 Box-Numeric’s

1

2

8

9

image13.png

34

35

41

42

image14.png

By looking at the 2 x 2 boxes in a 7 column grid, I have found out that if you place a   2 x 2 box, any where in the grid, the difference is 7.

Predict

Therefore I predict that, if I place any 2 x 2 box in a 7 column grid, the difference will always be 7.

image15.png

29

30

36

37

This estimate proves my aim, that if you place any 2 x2 box in a 7 column grid, the difference will always be 7

Difference in Algebra

Now I am going to show what the difference will be for any 2 x 2 box in a 7 column grid algebraically

2 x 2 Boximage16.png

n

n+1

n+7

n+8

I am now going to increase the size of my boxes to 3 x 3, 4 x 4 etc and going to work out the difference in algebraic forms.

3 x 3 Box

n

n+1

n+2

n+7

n+8

n+9

n+14

n+15

n+16

image17.png

 4 x 4 Box

n

n+1

n+2

n+3

n+7

n+8

n+9

n+10

n+14

n+15

n+16

n+17

n+21

n+22

n+23

n+24

image18.png

I am going to display my results in a table, where I would try finding any patterns within the results I have.

7 x7 Gridimage10.png

Square Box sizes

Difference

Pattern

2 x 2

7 = 1x 7

12=1

3 x 3

28 = 4 x 7

22=4

4 x 4

63 = 9 x 7

32=9

Predict

7 x 7

252= 36 x 7

62=36

Therefore I can predict that the difference for a 7 x 7 grid will be 252. Now I am going to prove this prediction algebraically.

7 x 7 box - Proving algebraically

n

n+1

n+2

n+3

n+4

n+5

n+ 6

n+7

n+8

n+9

n+10

n+11

n+12

n+13

n+14

n+15

n+16

n+17

n+18

n+19

n+20

n+21

n+22

n+23

n+24

n+25

n+26

n+27

n+28

n+29

n+30

n+31

n+32

n+33

n+34

n+35

n+36

n+37

n+38

n+39

n+40

n+41

n+42

n+43

n+44

n+45

n+46

n+47

n+48

image19.png

This shows that I have proved the difference for a 7 x 7 box in a 7 column grid correctly.

Now to find out a general formula for the difference for any square box in a 7 column grid will be.

Box size ( b )

Formula

Difference

3 x 3

7(3-1)2 =

20

7 x 7

7(7-1)2 =

252

          b x b

7 (b–1)2

...read more.

Conclusion

Therefore, by using this method, I will use the formula 10(c-1)(r-1) which I gained from part 3, however instead of the grid size 10, I will use (g) to represent my random grid size.

I will now represent this in a table.

image40.png

Grid size ( g )

Formula

10 x 10

10(c-1)(r-1)

7 x 7

7(c-1)(r-1)

6 x 6

6 (c-1)(r-1)

g x g

g (c-1)(r-1)

Proving overall formula algebraically

I am now going to explain how I got this formula algebraically.

nimage39.pngimage38.png

n+(c-1)

n+g(r-1)

n+g(r-1)+(c-1)

1) n[n+g(r-1)+(c-1)]

=n2+gn(r-1)+n(c-1)

[n+(c-1)][n+g(r-1)]

=n2+gn(r-1)+n(c-1)+g(c-1)(r-1)

2) n2+gn(r-1)+n(c-1)+g(c-1)(r-1) - n2+gn(r-1)+n(c-1)

=g(c-1)(r-1)

Therefore this overall equation “g(c-1)(r-1)”explains the difference, for any size rectangular box in any size grid.

Conclusion - In this task I have successfully found out general formula for any square box in any size grid which was to be ‘g(b-1)2and the general formula any rectangle box in any size grid which was to be ‘g(c-1)(r-1)’. These formulas were proven algebraically above.

If I were to extend this task further, I would change the grid numbers i.e. rather than it going up in 1s, it would go in 2s or 3s.

...read more.

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