Number Grid Investigation

So in this case: (22x31) - (21x32) = 10

I would also like to investigate if the position in the 10 x 10 grid affects the resulting difference because if it does, then I would have to further advance my investigation to make sure I can truly see the different outcomes you can achieve.

2 x 2 Squares

3 x 3 Squares

4 x 4 Squares

5 x 5 Square

Results

Algebra

D = (B x C) – (A x D)

D =  x + (n-1)    X       x + 10(n-1)

   

     -  x                  X       x + 11(n-1)

        

D = (x + n +1) (x +10n -10)

    -  x (x +11n -11)

D = (x²+ 11nx - 11x + 10n² - 20n +10)

   

• (x² +11nx - 11x)

D = 10n² - 20n + 10

D = 10 (n² - 2n +1)

D = 10 (n – 1) ²

The difference is therefore 10 (n – 1)²

We cannot be 100% sure that this formula is correct without testing it. So, to test this formula, we have previously established that the difference in a 3 x 3 Square is 40 so we can therefore test this with our equation for squares.

So when n = 3

10 (n – 1)² = 40  

10 (3 - 1)²

= 10 x 2²

= 10 x 4

= 40

The formula worked.

Now we can test it again so that we are sure that it was not simply coincidence that we achieved the correct number:

So when n = 5

10 (n – 1)² = 160

10 (5 - 1)²

= 10 x 4²

= 10 x 16

= 160

The formula worked again. So I will therefore conclude that it is correct for this investigation and can be used as a valid result of my research.

Rectangles

The length and the Width of a rectangle are different so I cannot use the same formula for rectangles as squares. Instead I will have to alter the formula but use the same basic methods to devise a new one. Firstly, I investigated the correlation between width (n), height (m) and the difference (D) between the products of the top left number and the bottom right number of the square and the product of the bottom left and top right numbers.

2 x 3 Rectangle

2 x 4 Rectangle

2 x 5 Rectangle

3 x 4 Rectangle

3 x 5 Rectangle

4 x 5 Rectangle

5 x 6 Rectangle

Results

From these results, I can clearly see that there is a correlation between:

The Overall Difference we have been trying to calculate from the start of the experiment.

= D 

And the difference between the product of the length (m) and width (n)

and the sum of the length (m) and width (n)

= (n x m) – (n + m)

I can expand upon this to devise a formula that:

= 10 x [1 +(n x m) – (n + m)]

= 10(n x m)(n – 1)

D = 10(n x m)(n – 1)

This formula has only been devised by testing many different sized rectangles and can only be proved with the use of algebra.

Algebra

D = (B x C) – (A x D)

D =   x + (n-1)    X       x + 10(m-1)

   

     -        x           X       x + 10(m-1) + (n-1)

        

D = (x + n - 1) (x +10m -10)

    -  x (x +10m -10 + n -1)

D = (x² + 10mx - 10x + nx - 10mn – 10n - x -10m + 10)

   

- (x² + 10mx - 10x + nx - x)

D = 10mn - 10n – 10m +10

D = 10(mn – m – n + 1)

D = 10 (n – 1) (m - 1)

The difference is therefore 10 (n – 1) (m – 1)

We cannot be 100% sure that this formula is correct without testing it. So, to test this formula, we have previously established that the difference in a 5 x 4 Rectangle is 120 so we can therefore test this with our equation for squares.

So when:

n = 5

m = 4

10 (n – 1) (m – 1) = 120

10 (5 - 1)(4 – 1)

= 10(4 x 3)

= 10 x 12

= 120

The formula worked.

Grid Size

By changing the size of the overall grid used, the numbers should all be shifted in a different direction than before, depending on the size of the grid I use, previously I had used a 10 x 10 square.

In this case, the Grid size (G) is 9 as it is an alternative number to the previous recurring grid size used, 10, which was used in the previous equations to calculate the various factors of my investigation. Hopefully this change should have an effect on the difference produced in the squares I choose.

Squares

2 x 2 Square

 

 

        

3 x 3 Square

4 x 4 Square

Algebra

D = (B x C) – (A x D)

D =  x + (n-1)     X       x + G(n-1)

   

     -  x                X       x + G(n-1) + (n-1)

D = (x² + xG(n - 1) - x(n - 1) + G(n - 1)² )

   

- (x² + xG(n - 1) - x(n - 1) )

D = G (n - 1)²

The difference is therefore G (n – 1)²

We know that the difference, when altering the Grid Size to 9, in a square that has a length and width of 3, is 36. So we can use this result that we have previously gathered from experimenting to test if this formula is correct.

If n = 3

  G = 9

Then G(n – 1) ² Should equal 36

D =    9(3 - 1) ²

   =     9 x (2)²

   =     9 x 4

   =     36

Rectangles

2 x 3 Rectangle

 

 

        

3 x 4 Rectangle

4 x 5 Rectangle

3 x 5 Rectangle

Algebra

D = (B x C) – (A x D)

D =   x + (n-1)     X      x + G(m-1)

   

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

D = (x² + xG(m - 1) + x(n – 1) + G(m - 1)(n – 1) )

   

    - (x² + xG(m – 1) + x(n - 1) )

D = G(m – 1)(n – 1)

The difference is therefore G(m – 1)(n – 1)

Using an example that G is 9, m = 3, n = 2,

We know that the difference of G(m – 1)(n – 1) Should equal 18

D = G(m - 1)(n - 1)

   =   9(3 - 1)(2 - 1)

   =     9 x (2 x 1)

   =     9 x 2

   =     18

Therefore, my formula is correct.

Number Gap

To further my investigation, I am going to bridge off and investigate if changing the space between the numbers in the overall gird will have any affect on the overall difference that I am trying to calculate.

In this case, the Number Gap (L) is 2 as it is an alternative number to the previous number gap that I have used in all of my experiments, 1, to calculate the various factors that I had been trying to find. I would like to hope that this change will have a big effect on the result I get ad that I can hopefully offer an explanation for this, using algebra and exploring the changes in depth.

Squares

2 x 2 Square

3 x 3 Square

4 x 4 Square

Algebra

D = (B x C) – (A x D)

D =   x + L(n - 1)    X      x + 10L(n  - 1)

   

     -        x              X       x + 11L(n - 1)        

D = (x² + x10L(n - 1) + xL(n – 1) + 10L²(n -1)² )

   

    - (x² +  x11L(n - 1) )

D = (x² + x11L(n - 1) + 10L²(n -1)² )

   

    - (x² +  x11L(n - 1) )

D = 10L²(n -1)²

The difference is therefore 10L²(n -1)²

Now we can test the formula that I appear to find to see if it is, in fact, correct.

 

So when:

L = 2

n = 4

10L²(n -1)² = 360

D  = 10 x 2² (4 - 1)²

     = 10 x 4  x 3²

     = 40 x 9

     = 360

 

The formula worked again. So I will therefore conclude that it is correct for this investigation and can be used as a valid result of my research.

Rectangles

2 x 3 Rectangle

 

 

        

3 x 4 Rectangle

4 x 5 Rectangle

3 x 5 Rectangle

Algebra

D = (B x C) – (A x D)

D =   x + L(n-1)     X      x + 10L(m - 1)

   

     -        x             X       x + 10L(m - 1) + L(n – 1)        

D = (x² + x10L(m - 1) + xL(n – 1) + 10L²(m -1)(n  - 1) )

   

    - (x² +  x10L(m - 1) + xL(n - 1) )

D = 10L²(m -1)(n - 1)

The difference is therefore 10L²(m -1)(n - 1)

I can test this now, using the results that I have previously gathered when investigating the affect of changing the Number Gap (L). I know that in a 3 x 5 Rectangle, the difference is 320.

So, in this case:

D = 10L²(m -1)(n - 1)

   =  10 x 2² x (5 - 1)(3 - 1)

   =     10 x 4 x (4 x 2)

   =     40 x 8

   =     320