I will use the formula first
to make sure it works.
According to the formula +40 is the correct answer.
(n+2)(n+20)=n?+20n+40
So to make sure that 40 is the correct answer I will check using the standard method.
31x53=1643
33x51=1683
1683-1643= 40
It is confirmed and the formula does work.
I am now going to try a 4x4 grid using the same principles as the other two grids I have tried.
The answer for the grid using the basic method is-
7x40=280
10x37=370
370-280=90. so 90 is the difference for the 4x4 grid but does the formula agree?
(n+3)(n+30)=n?+30n+3n+90. So the answer for any 4x4 grid the difference will be 90.
As the formula has proven it has worked four times I will skip 5x5 and go on to 6x6 sized grids.
45x100=4500
50x95=4750
4750-4500=250
Now I’m going to see if the formula agrees with the standard method.
(n+5)(n+50)=n?+50n+5n+250 again the formula confirms that the answer is 250.
Now that I have several differences let me put it into a table to see if I am able to find a pattern.
On the left, I have the differences I have gained so far and below that I have divided the numbers by 10 to remove the 0. I have noticed that the numbers with the 0 removed they are all square numbers.
Size difference (/10)
On the table Above the first column represents the grid size, the second is the length of sizes multiplied together, and the third shows the difference of the grid divided by 10,(unlined figures are predicted). There is an interesting link between the second and third columns, as the result of the numbers multiplied column equals the differences.
As I have not done the 5x5 grid, the table predicts that the difference will be 16 by multiplying this number by 10 I got 160.
Using the formula I can easily find out the difference-
(n+4)(n+40)=n?+40n+4n+160
However there is an easier way to gather information, by using a different formula that eliminates the making of an algebra grid. The formula is:
10(n-1)?. (n refers to the length of one of the grid size)
Let’s test the formula on a 7x7 grid.
10(7-1)?
=10(6)?
=360
It claims that 360 is the correct answer this is also correct by the table I have done earlier but to make sure I am going to test it using the algebra equation.
(n+6)(n+60)=n?+60n+6n+360
So the formula proves that the smaller formula works and it shows the answer is 360. I now want to see if the formulas are able to work with a different shape like a rectangle.
These are both rectangles but I am going to
start by looking at regular pattern of rectangle size.
2x3, 3x4, 4x5 and so on.
I am going to start with a 2x3 grid. I’m going to use the standard method first to find the difference.
1x13=13
3x11=11
33-11-22 this says the difference is 22 but will the formula work with it? The smaller 10(n-1) will not work as it does not take into account the two different lengths of the sides. However the larger formula will work.
(n+2)(n+10)=n?10n+2n+22
So I am able to use this method to find out the new formula.
I am now going to try my formula on a 3x4 grid. This is to make sure that the formula still works on a larger scaled grid.
1x24=24
4x21=84
84-24=60
Now I’m going to use the formula again:
(n+3)(n+20)=n?20n+3n+60
4x5 grids
(n+4)(N+30)=n?30n+4n+120 so 120 is the difference for all 4x5 grids.
5x6 grids
(n+5)(N+40)n?40n+5+200 so 200 is the difference for all 5x6 sized grids. One more example should be enough to find out a simpler formula.
6x7 grids
(n+6)(n+50)=n?+50n+6n+300 so the formula did not work as it did not take into account two different sides of lengths.
We need replace the (n-1)? With a new piece. W for width and L for length.
This will make: (w-1)(l-1) 10
So let’s test the formula.
Above we have an 8x7 grid,
(8-1)(7-1)10
=7x6x10
=420
Now I’ve got an answer from my new Formula I will check it with the more conventional method.
(n+7)(n+6)=n?+60n+7n+420
the formula works, but would the formula work for a rectangle which is not in my pattern?
Below is a 3x7 grid, I am going to see if the formula is able to find a difference.
(3-1)(7-1)10
=2x6x10
=120
The formula claims that 120 is the difference but would the conventional methods agree?
(n+29n+60)=n?+60n+2n+120
It is confirmed now that we have the formula for any shaped rectangle in a 10x10 grid. But so far all of the formulas have been for 10x10 grids. This is mad apparent by the number 10 recurring in them.
10(n-1)?, and (w-1)(l-1)10. The number 10 in these formulas refers to the grid size, so if I was to change the grid size they would not work.
Conclusion
Through out my work I have found three main formulas.
For squares a 10x10 grid the formula is 10(n-1), can be used
For rectangles (also works with squares) in a 10x10 grid the formula (w-1)(l-1) 10 can be used.
With the third of these three formulas it is possible to work out the difference on any sized square or rectangle in any sized grid. In a way, it is the concluding formula. With out moving into different or three dimensional shapes.
(n+1)(n+12)=n?+12n+1n+10
