If I multiply this out if I were doing what I was doing with the numbers in the grid I would,
(X+2)(X+20)…which is = X2+22X+40…and…
X(X+22)…which is = X2 +22X
[X2+22X+40]-[X2 +22X] = 40
X can be any number and will always equal 40!
4 x 4 square
If I multiply this out if I were doing what I was doing with the numbers in the grid I would,
(X+3)(X+30)…which is = X2+33X+90…and…
X(X+33)…which is = X2 +33X
[X2+33X+90]-[X2 +33X] = 90
X can be any number and will always equal 90!
5 x 5 square
If I multiply this out if I were doing what I was doing with the numbers in the grid I would,
(X+4)(X+40)…which is = X2+44X+160…and…
X(X+44)…which is = X2 +44X
[X2+44X+160]-[X2 +44X] = 160
X can be any number and will always equal 160!
Working out a formula for any square size on a 10 by 10 number grid
The sequence I get is 10, 40, 90 and then 160.
So I can do the use a square to find out a formula that works out N for any square size on a 10 by 10 number grid.
I do what I do with these figures if they were numbers:
1. [X+(S-1)][X+10(S-1)
I get X2+11X(S-1)+10(S-1)2
And…
2. X[X+11(S-1)]
I get X2+ 11X(S-1)
I then subtract 2 from 1…
[X2+11X(S-1)+10(S-1) 2] - [X2+ 11X(S-1)]
And whatever that is left over must be the formula, so therefore the formula to work out N for any square size in a 10 by 10 grid square is…
N= 10(S-1)2
Test
When the square size is 2…
Using the formula, (2-1) 2 = 1.. x 10 = 10
When the square size is 5…
Using the formula, (5-1) 2 = 16… x 10 = 160
The formula works!
What happens if I change the grid size?
I think that I have come to the stage where I can use algebra to work out what the number each time will be. So, I don’t have to show examples with numbers all the time.
A 6 by 6 grid square on a 2x2 square
If I multiply this out if I were doing what I was doing with the numbers in the grid I would,
(X+1)(X+6)…which is = X2+7X+6…and…
X(X+7)…which is = X2 +7X
[X2+7X+6]-[X2 +7X]= 6
X can be any number and will always equal 6 on a 2 by 2 square on a 6 by 6 grid.
An 8 by 8 grid square on a 2x2 square
If I multiply this out if I were doing what I was doing with the numbers in the grid I would,
(X+1)(X+8)…which is = X2+9X+8…and…
X(X+9)…which is = X2 +9X
[X2+9X+8]-[X2 +9X] = 8
X can be any number and will always equal 8 on a 2 by 2 square on a 8 by 8 grid.
A 9 by 9 grid square on a 2x2 square
If I multiply this out if I were doing what I was doing with the numbers in the grid I would,
(X+1)(X+9)…which is = X2+10X+9…and…
X(X+10)…which is = X2 +10X
[X2+10X+9]-[X2 +10X] = 9
X can be any number and will always equal 9 on a 2 by 2 square on a 9 by 9 grid.
Summary
For a 2x2 square on any grid size I get N equalling the grid size number!
Working out the formula for any square size in any grid size.
So I can do the use a square to find out a formula that works out N for any square size on a 10 by 10 number grid.
I do what I do with these figures if they were numbers:
1. [X+(S-1)][X+g(S-1)]
I get X2+gX(S-1)+X(S-1)+g(s-1)2
And…
2. X[X+X(g+1)(S-1)]
I get X2+ X(S-1) (g+1)
I then subtract 2 from 1…
[X2+gX(S-1)+X(S-1)+g(s-1)2] - [X2+ X(S-1) (g+1)]
Using common sense I know that g(s-1)2 should be left over so I need to prove that;
gX(S-1)+X(S-1) and X(S-1) (g+1) are the same.
If I rearrange these two so they look the same I will know that they cancel each other out.
X(S-1) (g+1) can rearrange to make (g+1)s(X-1)
gX(S-1)+X(S-1) can rearrange to make gs(X-1)+1S(X-1)
And, these are the same so they cancel out..
The only thing left over is g(s-1)2
And whatever that is left over must be the formula, so therefore the formula to work out N for any square size in a g by g grid square is…
N= g(S-1)2
Test
When the square size is 2 and the grid size is 10
Using the formula, (2-1) 2 = 1.. x 10 = 10
When the square size is 3 and the grid size is 6
Using the formula, (3-1) 2 = 4… x 6 = 24
The formula works!
Final formula: for any size square/rectangle in any grid square
Common sense tells me that, this is basically replacing the S by the length and width of any square/rectangle.
I do what I do with these figures if they were numbers:
1. [X+(w-1)][X+g(L-1)]
I get X2+gX(L-1)+X(w-1)+g(w-1)(L-1)
And…
2. X[X+g(L-1)+(w-1)]
I get X2+ gX(L-1)+X(w-1)
I then subtract 2 from 1…
[X2+gX(L-1)+X(w-1)+g(w-1)(L-1)] - [X2+ gX(L-1)+X(w-1)]
The only thing left over is g(w-1)(L-1)
And whatever that is left over must be the formula, so therefore the formula to work out N for any square/rectangle in a g by g grid square is…
N= g(w-1)(L-1)
Test
When the grid size is 10 and the length and width are 2
Using the formula, 10(2-1)(2-1) = 10x1x1= 10
When the grid size is 5, length is 2 and width is 4
Using the formula, 5(4-1)(2-1) = 5x3x1 = 15
The formula works!
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