• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16
17. 17
17
18. 18
18
19. 19
19
• Level: GCSE
• Subject: Maths
• Word count: 3601

# Maths Gridwork

Extracts from this document...

Introduction

Victor Truong 9Gr Maths Coursework Page  of

Introduction

In this investigation, I have been asked to investigate on a number grid that is 10 wide and 10 descending. We have been asked to test the equation (Top left x Bottom right) – (Top right x Bottom left) on grids varying in size, starting at 2x2, then on to 3x3 and so on. I will describe the constraints of the equation and explain the algebraic rule that determines the end outcome of the grid. I will then relate my new formula and describe how it can be related with rectangles. I will then find a formula that will suit a Master grid. A diagram of the number grid is shown below:

 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

2*2 squares

Equation: (TL*BR)-(TR*BL)

Example 1

 22 23 32 33

(22*33)- (23*32) = -10

Example 2

 37 38 47 48

(37*48)- (38*47) = -10

Example 3

 57 58 67 68

(57*68)- (58*67) = -10

Example 4

 1 2 12 11

(1*12)- (2*11) = -10

I predict that with all 2*2 grid squares the equation will always produce an answer of -10

Example5

 56 57 66 67

(56*67)- (57*66) = -10

I will now use algebra to prove my theory.

Algebraic equation:           (x)*(x+11) – (x+1)*(x+10)

X²+11x - x²+x+10x+10

+11x - x²+11x+10

λ    - Λ +10 = -10

Therefore my hypothesis was true.

3*3 squares

Equation: (TL*BR)-(TR*BL)

Example 1

 43 44 45 53 54 55 63 64 65

(43*65)- (45*63) = -40

Example 2

 56 57 58 66 67 68 76 77 78

(56*78)- (58*76) = -40

Example 3

 1 2 3 11 12 13 21 22 23

(1*23)- (3*21) = -40

Example 4

 8 9 10 18 19 20 28 29 30

(8*30)- (10*28) = -40

Middle

65

66

67

68

69

(25*69)- (29*65) = -160

I will now use algebra to try an prove my result

Algebraic equation

(x)*(x+44) – (x+4)*(x+40)

X² + 44x     – x² + 44x + 160

λ         –   (λ + 160) = -160

Predicting a 6*6 square

I can now see a pattern emerge from the results:

So I predict that a 6*6 square will give the result -250. I noticed that all the results were negative square numbers multiplied by ten. I made this prediction using the tabulate above using the previous results.

Considering the results are all consecutive square number it gave me reason to believe that the 6*6 square would give an answer on -250. I can also prove this:

The tabulate shows me that the results are all multiples of (-10) that are multiplied by consecutive sq numbers. So there is further evidence that the 6*6 square will give -250

Example 1

 2 3 4 5 6 7 12 13 14 15 16 17 22 23 24 25 26 27 32 33 34 35 36 37 42 43 44 45 46 47 52 53 54 55 56 57

(2*57)- (7*52) = -250

Example 2

 45 46 47 48 49 50 55 56 57 58 59 60 65 66 67 68 69 70 75 76 77 78 79 80 85 86 87 88 89 90 95 96 97 98 99 100

(45*100)- (50*95) = -250

Those examples proves me theory was correct. I will try to use algebra to prove me theory.

Algebraic equation

(x)*(x + 55) – (x + 5)*(x + 50)

X² + 55x      - x² + 55x +250

λ             – (λ + 250) = -250

My Prediction for an ‘N*N’ square:

My Prediction for an ‘N*N’ square: -10(n-1)². This equation can only suit the given rule if 2>n<10.

This equation is only used for predicting the result of the square.

Example 1

With a 7*7 grid I can predict using the N*N equation that the result will be -360.

 34 35 36 37 38 39 40 44 45 46 47 48 49 50 54 55 56 57 58 59 60 64 65 66 67 68 69 70 74 75 76 77 78 79 80 84 85 86 87 88 89 90 94 95 96 97 98 99 100

(34*100)- (40*94) = -360

I worked the formula out by placing the results of the other grid into a diagram

With my previous outcomes I can see that the equation should produce a multiple of -10. The previous products all are square number. This gives reason to say that it is (n-1²).

Algebraic working:

‘X’ being the given number and ‘n’ being the square of the length of one of the squares.

Example1

 x x + (n+1) x + 10(n-1) x + 11(n-1)

Using the given rule, this gives us

X[X + 11(n-1)] - [(X + n-1) (X+10(n-1)]

Taking the out ‘[ ]’s leave…

X(X+11N-11) – (X+N-1) (X+10N-10)

Equals…

X²+11NX-10x – (X²+11NX-11X- +10N²-10N-10N+10)

Simplifies to…

-10n²+20n-10

And again simplifies to…

-10(n²+2n-1)

This finishes as…

-10(n-1)²

This proves my hypothesis.

This works because the product of top right multiplied by the bottom left will always have a common difference to the top left multiplied by the bottom right in squares. This means that you can calculate a ‘n’ equation by applying the above grid. As we can simplify the equation this tells that with all 2x2 – 10x10 on a 10*10 Master grid, the equation to apply is -10(n-1)² when we it is (Top left x Bottom right) – (Top right x Bottom left)

Rectangles

I am about to relate my formula used with squares to fit with a rectangle. I will only increase a variable slowly so that I can read the difference easily and state a hypothesis. I will then prove the hypothesis using algebra and the use the same method for the other variable. The first variable will be the width of the rectangle, which I will increase by 1 for each example until I can see a common difference.

2 down*3 wide grids

Example 1

 1 2 3 11 12 13

(1*13)- (3*11) = -20

Example 2

 24 25 26 34 35 36

(24*36)- (26*34) = -20

2 down*4 wide grids

Example 1

 1 2 3 4 11 12 13 14

(1*14)- (4*11) = -30

Example 2

 55 56 57 58 65 66 67 68

(55*68)- (58*65) = -30

2 down*5 wide grids

Example 1

 1 2 3 4 5 11 12 13 14 15

(1*15)- (5*11) = -40

Example 2

 66 67 68 69 70 76 77 78 79 80

(66*80)- (70*76) = -40

2 down* 6 wide grids

Example1

 1 2 3 4 5 6 11 12 13 14 15 16

(1*16)- (6*11) = -50

Example 2

 73 74 75 76 77 78 83 84 85 86 87 88

(73*88)- (78*83) = -50

Observing these worked examples, the variable- width has a common difference and can be seen in the diagram

As the width increases by 1, the difference goes up by -10

So I predict that in the next 2x7 grid the difference will go up by -10

2 down * 7 wide

 4 5 6 7 8 9 10 14 15 16 17 18 19 20

(4*20)- (10*14) = -60

Example 2

 33 34 35 36 37 38 39 43 44 45 46 47 48 49

(33*49)- (39*43) = -60

So my prediction was right, I will now try and prove my theory using algebra.

Algebraic equation

‘X’ here represents a number starting the rectangular grid.

 x x+1 x+2 x+3 x+4 x+5 x+6 x+10 x+11 x+12 x+13 x+14 x+15 x+16

(X * [x+16])- ([x+6] * [x+10])

(X²+16x)- (x²+16x+60)

X²+16x- x²+16x+60

16x – 16x+60

=-60

As we have now proved it with an algebraic equation for the width we can now move on and work out the formula for the length down.

3 down *2 wide

 1 2 11 12 21 22

(1*22)- (2*21) = -20

Example 2

 26 27 36 37 46 47

(26*47)- (27*46) = -20

4 down *2 wide

 5 6 15 16 25 26 35 36

(5*36)- (6*35) = -30

Example 2

 8 9 18 19 28 29 38 39

(8*39)- (9*38) = -30

5 down * 2 wide

 24 25 34 35 44 45 54 55 64 65

(24*65)- (25*64) = -40

Example 2

 8 9 18 19 28 29 38 39 48 49

(8*49)- (9*48) = -40

6 down * 2 wide

 49 50 59 60 69 70 79 80 89 90 99 100

(49*100)- (50*99) = -50

Example 2

 5 6 15 16 25 26 35 36 45 46 55 56

Conclusion

As explained in the end of the rectangles section we can include an ‘m’ by removing an ‘n’. In this formula ‘m’ equals width, ’n’ equals the amount of rows down, ‘a’ equals the size of the Master grid and x equals the starting value on the grid.

 x X+(m-1) X+a(n-1) X+a(n-1)+(m-1)

= {x[x+ (m-1) +a (n-1)]} – {[x+ (m-1)] [x + a (n-1)]}

= [x(x+m-1+an-a)] – [(x+m-1) (x + an-a)]

= (x²-x+mx+anx-ax) – (x²+anx-ax+mx+anm-am-x-an+a)

= - (anm-am-an+a)

= - anm+am+an-a

= -a (nm-m-n+1)

= -a (n-1) (m-1)

This was achieve by replacing the 10, which corresponds to the size of the Master grid, with an ‘a’ using the working form rectangle. If we replace the 10 with the ‘a’s it doesn’t affect the formula it only means that everywhere a 10 appears we see an ‘a’ instead.

This formula is identical to the rectangular one on a 10*10 Master grid but with an ‘a’ instead of an -10. This is because it is the same but we have an ‘a’ representing the size of the master grid which in the 10*10 master grid was 10. But because on squares they have an equal length/width we don’t need an ‘m’ to represent the width. Leaving the formula as –a (n-1²) which means – (size of the Master grid) * (length-1²) which is similar to -10(n-1²).

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Number Stairs, Grids and Sequences essays

1. ## GCSE Maths Sequences Coursework

I have noticed that each shape is made up off sections, these sections are similar to the shapes that I got when I was investigating squares at the very start of the growing shapes investigation. For example stage 1 in the cube sequence is made up off two "start shapes" and stage 1 in the square sequence.

2. ## Number Grid Coursework

- a(a + z[q - 1] + [p - 1]) (12 + 3)(12 + 68) - 12(12 + 68 + 3) 15 x 80 - 12 x 83 1200 - 996 204 (N.B. also = 17 x 3 x 4 = z[p - 1][q - 1])

1. ## Number Grids Investigation Coursework

Differences between the Numbers in the Grid The next part of my investigation is to investigate the difference between the numbers in the grid; for example, in all the grids I have done so far the difference between the numbers in the grid has been 1 (it has gone up

2. ## Number Grid Investigation.

+ (TR X BL) Firstly in a 2 X 2... 31 32 41 42 (TL X BR) + (TR X BL) = 2614 70 71 80 81 Product difference = 11350. The product difference is not the same. I do not see that changing the calculations will have any effect on my formulas.

1. ## Investigation of diagonal difference.

investigation by only calculating the diagonal difference of my initial example for each size cutout i.e. the horizontally aligned cutout 2 x 5 cutouts 1 2 3 4 5 11 12 13 14 15 n n + 4 n + G n + G + 4 1 2 11 12

2. ## Maths-Number Grid

will continue this investigation by extending this coursework and examining the product difference between different sized rectangle grids. I will begin this investigation by starting off with a 2 � 4 rectangle grid. Method:- 1. Firstly, I will find a rectangle grid anywhere on the 10 � 10 square grid.

1. ## Number Grid Maths Coursework.

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

2. ## For other 3-step stairs, investigate the relationship between the stair total and the position ...

For example for a 10x10 numbered grid using a 6-step stair the formula is 21x-315 then we increase the grid size by 1, 11x11 and using the same 6-stepped stair approach the formula is 21x-350, etc. We can clearly see the constant number [35] is consistent every time the grid size increases by [1].

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to