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  • Level: GCSE
  • Subject: Maths
  • Word count: 2293

number grid

Extracts from this document...

Introduction

12800/w2                                                                                                   Joanne Barton

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Number grid

Task

      My coursework task is to look at the number grid

  • A box is drawn round four numbers.
  • Find the product of the top left number and the bottom right number in this box.
  • Do the same with the top right and bottom left numbers.
  • Calculate the difference between these products.

Investigate further.

In this investigation I will look at the different number grids and different sized rectangles within these grids, I will explain the patterns and give algebraic equations for the results found during this investigation. I will also try to find a formula and prove my findings.

To find out the product difference I need to:-

Multiply the top left number with the bottom right and multiply the top right with the bottom left and then - the products from each other.  

I will first try with a 2 x 2 square.

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22

      23

a)

12 x 23 = 276   The difference of the two products are 10.

13 x 22 = 286   I worked the difference out by doing, 286-276 = 10

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48

49

b)      

38 x 49 = 1862

39 x 48 = 1872, I worked the difference out by 1872 - 1862 = 10

The product is again 10.

My theory is that all 2 x 2 squares will have the product of 10, I will show one more 2 x 2 square to prove the theory.

86

87

96

97

c)      

86 x 97 = 8342

87 x 96 = 8352, I worked the difference out by 8352 – 8342 = 10

The product is 10 also, so the theory works all 2 x2 squares = D 10

D = Difference.

I have drawn a table to show this clearer.

First sum answer

Second sum answer

difference

a)

8352

8342

10

b)

1862

1872

10

c)

276

286

10

...read more.

Middle

7104

7144

40

c)

1560

1600

40

I will again use algebra to prove this.

For a 3x3 square:

n = any number

  n

n +2

  n + 20

n +22

(n + 20) (n + 2) – n(n + 22)

                       n² +2n+20n+40-n² -22n

n²+22n+40 - n²- 22n = 40

From this algebraic calculation it shows that the answer is 40, which is the difference of the 3x3 square.

This formula could be used for any 3x3 square in a 10x10 grid.

I will now try with a 4x4 square in a 10 x 10 grid and my prediction is the same as above, the difference will be the same for all 3 4x4 squares.

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a)    

I will multiply the top right number and the bottom left and the top left and bottom right.

1x34 = 34

4 x31 =124

To find the difference I will subtract the numbers.

124 – 34 = 90

The difference is 90.

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b)      

42 x 75 = 3150

45 x 72 = 3240

3240 – 3150 = 90

The difference is 90.

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c)      

24 x 57 = 1368

27 x 54 = 1458

1458 – 1368 = 90

The difference is again 90.

The table below shows this.

1st sum answer

2nd sum answer

difference

a)

34

124

90

b)

3240

3150

90

c)

1368

1458

90

I will again prove this with algebraically.

n

n + 3

n + 30

n + 33

 (n + 3)(n + 30) – n(n+33)

  n² +30n+3n+90-n² -33n

  n²+33n+90 - n²- 33n = 90

My prediction was right any number in a 4x4 square within a 10x10 grid has a difference of 90.

I drew a table to show my results so far.

Square size

Difference

2x2

10

3x3

40

4x4

90

From the product differences I noticed a pattern emerging so I took the noughts away from the numbers, I would be left with square numbers i.e. 1, 4, and 9.

The next square number would be 16 so if I added the naught I would get 160, so I predict that a 5x5 square will have the difference of 160.

I will now test my prediction.

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a)

 .

1 x 45 = 45

5 x 41 = 205

205 – 45 = 160

The difference is 160 as predicted.

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b)

50 x 94 = 4700

54 x 90 = 4860

4860 – 4700 = 160

My prediction was right I will now prove this will algebra.

n

n + 4

n + 40

n + 44

               (n + 4)(n + 40) – n(n + 44)

                n² +40n+4n+160-n² -44n

                n² + 44n+160 - n² - 44n = 160


With this you can see the difference for any number in a 5x5 square on a 10x10 grid is 160, which proves my prediction.

Using the same method I predict that the 6x6 square on a 10x10 grid difference will be 250, because if I took the nought away it would be the next square number in the sequence, 25.

1, 4, 9, 16, and 25

I will test my theory using the same method I have used throughout.

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...read more.

Conclusion

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a)      

1 x 15 = 15

5 x 11 = 55

55 – 15 = 40

The difference is 40.

b)      

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65 x 79 = 5135

69 x 75 = 5175

5175 – 5135 = 40

My prediction was correct the difference was 40, I will prove this algebraically.

n

n + 4

  n + 10

n + 14

n(n + 14) – (n + 4)(n + 10) =

(n² + 14n) – (n² +14n + 40) = 40.

This proves the difference is 40 and the number has gone up by 10 again so you can then predict 2x6 difference of 60 and 2x7 difference of 70.

I am now going to find a formula.

I tried

(n – 1) X 10

n =the width of the box.

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7 x 2

                                      7 – 1 = 6

                                   6 x 10 = 60

I predict using my formula that the difference of this 7 x 2 square will be 60.

22 x 38 = 836

28 x 32 = 896

896 – 836 = 60

My prediction was correct.

I could use this formula to work out any size of rectangle.

        I.e. 18 X 2 = 170

Conclusion

I found 3 things out about rectangles in a 10x10 grid.

  • The rectangles I used differences increased by 10 every time.
  • The 2nd difference is always 10.
  • The formula for any rectangle is (n – 1) X 10

Evaluation

By doing lots of experiments and recording my results I found different patterns in a square and a rectangle by finding the difference of the corner numbers, I also used algebra to prove the differences. I found 2 formulas which can be used in any size square and any size rectangle in a 10x10 grid.

I could have expanded my coursework if I had more time and looked at different grid sizes and maybe different shapes like triangles or t shape patterns.

This coursework has helped me to understand number patterns and has developed my ability to introduce algebra when trying to spot patterns and I am happy with my results.

...read more.

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