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  • 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:

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2*2 squares

Equation: (TL*BR)-(TR*BL)

Example 1

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

Example 2

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

Example 3

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

Example 4

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

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

Example5

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(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

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

Example 2

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

Example 3

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

Example 4

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

...read more.

Middle

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(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:

image00.png

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:

image01.png

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  

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

Example 2

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(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.

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

I worked the formula out by placing the results of the other grid into a diagramimage00.png

image01.png

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

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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

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

Example 2

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

2 down*4 wide grids

Example 1

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

Example 2

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

2 down*5 wide grids

Example 1

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

Example 2

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

2 down* 6 wide grids

Example1

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

Example 2

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

Observing these worked examples, the variable- width has a common difference and can be seen in the diagram image04.png

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

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

Example 2

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(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.

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(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

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

Example 2

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

4 down *2 wide

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

Example 2

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

5 down * 2 wide

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 (24*65)- (25*64) = -40

Example 2

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

6 down * 2 wide

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

Example 2

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...read more.

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).

...read more.

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