I have proved that the above results hold for any 3 by 2 rectangle.
This shows that for any 3 by 2 rectangle the difference will always be equal to 20.
I will now change the size of the rectangle to 4 by 2.
66 x 79 = 5214
69 x 76 = 5244
The difference between the above two products is equal to 30.
I am now going to try to prove that this holds for any 4 by 2 rectangle using algebra.
Let n be the top left number and n + 13 be the bottom right number.
n(n+13) = n²+13n
Let n + 3 be the top right number and n + 10 be the bottom left number.
(n+3)(n+10) = n²+3n+10n+30
= n²+13n+30
I will now subtract the two algebraic products to get an overall product.
n²+13n+30
n²+13n -
+30
This proves that the difference is always 30
I have proved that the above results hold for any 4 by 2 rectangle.
This shows that for any 4 by 2 rectangle the difference is always 30.
I will now change the size of the rectangle to 5 by 2.
76 x 90 = 6840
80 x 86 = 6880
The difference between the above two products is equal to 40.
I am now going to try to prove that this holds for any 5 by 2 rectangle using algebra.
Let n be the top left number and n + 14 be the bottom right number.
n(n+14) = n²+14n
Let n + 4 be the top right number and n + 10 be the bottom left number.
(n+4)(n+10) = n²+4n+10n+40
= n²+14n+40
I will now subtract the two algebraic products to get an overall product.
n²+14n+40
n²+14n -
+40
This proves that the difference is always 40
I have proved that the above results hold for any 5 by 2 rectangle.
This shows that for any 5 by 2 rectangle the difference is always equal to 40.
I will now put my results for the rectangles into a table.
P is equal to the length of rectangle.
I can tell from this table that the difference increases by 10, when the size of the rectangle increases by 1 along one side.
I will now try to find a general solution for 2 by p.
The difference for 2 by p can be shown using the following algebra.
(p-1) x 10
I will show that this works by substituting p with the number 7.
(7-1) x 10 = 60
This shows that the difference is 60, I will use a 2 by 7 rectangle to see if I am correct.
1 x 17 = 17
7 x 11 = 77
The difference between the above two products is equal to 60, I was correct.
I will now use algebra to try and prove that this holds for any 2 by p rectangle.
Let n be the top left number and n + p + 9 be the bottom right number.
n(n+p)-9 = n²+np+9n
Let n + p - 1 be the top right number and n + 10 be the bottom left number.
(n+10)(n+p-1) = n²+np-9n+10P-10
I will now subtract the two algebraic products to get an overall product.
n²+np-9n+10P-10
n²+np+9n -
10p-10
This proves that the difference is always 10P-10
The above algebra tells me that for any 2 by p square the difference is always 10P-10.
To extend the investigation further I am going to investigate the effect of changing the size of the square, I will begin with a 3 by 3 square.
35 x 57 = 1995
37 x 55 = 2035
The difference between the above two products is equal to 40.
I am now going to try to prove that this holds for any 3 by 3 square using algebra.
Let n be the top left number and n + 22 be the bottom right number.
n(n+22) = n²+22n
Let n + 2 be the top right number and n + 20 be the bottom left number.
(n+2)(n+20) = n²+2n+20n+40
= n²+22n+40
I will now subtract the two algebraic products to get an overall product.
n²+22n+40
n²+22n -
+40
This proves that the difference is always 40
I have proved that the above results hold for any 3 by 3 square.
I will now change the size of the square to 4 by 4.
41 x 74 = 3034
44 x 71 = 3124
The difference between the above two products is equal to 90.
I am now going to try to prove that this holds for any 4 by 4 square using algebra.
Let n be the top left number and n + 33 be the bottom right number.
n(n+33) = n²+33n
Let n + 3 be the top right number and n + 30 be the bottom left number.
(n+3)(n+30) = n²+3n+30n+90
= n²+33n+90
I will now subtract the two algebraic products to get an overall product.
n²+33n+90
n²+33n -
+90
This proves that the difference is always 90
I have proved that the above results hold for any 4 by 4 square
This shows that the difference for any 4 by 4 square will always be equal to 90.
I will now change the size of the square to 5 by 5.
6 x 50 = 300
10 x 46 = 460
The difference between the above two products is equal to 160.
I am now going to try to prove that this holds for any 5 by 5 square using algebra.
Let n be the top left number and n + 44 be the bottom right number.
n(n+44) = n²+44n
Let n + 4 be the top right number and n + 40 be the bottom left number.
(n+4)(n+40) = n²+4n+40n+160
= n²+44n+160
I will now subtract the two algebraic products to get an overall product.
n²+33n+160
n²+33n -
+160
This proves that the difference is always 160
I have proved that the above results hold for any 5 by 5 square
This shows that for any 5 by 5 square the difference will always be equal to 160.
I will now put my results for the squares into a table.
P is equal to the length of the square size.
10 (P-1)² holds for any m by m square.
The above table shows that the difference between the 2 products increases by 20 each time.
I can tell from this table that the difference between the differences increases by 20 each time.
I will now prove that this holds for any m by m using algebra.
Let n be the top left number and n + 11 (m-1) be the bottom right number.
n(n+11(m-1) = n²+11nm+11n
Let n + 4 be the top right number and n + 40 be the bottom left number.
(n+(m-1) (n+10(m-1) = n²+11nm+11n+10m²-20m+10
I will now subtract the two algebraic products to get an overall product.
n²+11nm+11n
n²+11nm+11n+10m²-20m+10 -
10m²-20m+10
I have proved that the above results hold for any M by M square
This shows that for any M by M square the difference will always be equal to 10m²-20m+10.
I will now substitute n and m with numbers to prove that this holds for any m by m square.
(20x2²-20x2+10) = 10, this is what I expected, as shown within my results within the above table, showing the results for a 2 by 2 square.
To extend the investigation further I am going to investigate the effect of changing the size of the overall number grid.
I will begin by looking at a 4 by 25 grid.
I will now look at a 5 by 20 grid.
I will now put my results into a table to look for a recurring pattern.
I have noticed that the difference between the 2 products for any 2 by 2 square is equal to the grid size.
I will now use Algebra to give a general answer, showing that this holds for any n x w grid.
Let n be the top left number and n + w + 1 be the bottom right number.
n(n+nw)+1 = n²+nw+1
Let n + 1 be the top right number and n + w be the bottom left number.
(n+1)(n+w) = n²+nw+w+1
I will now subtract the two algebraic products to get an overall product.
n²+nw+w+1
n²+nw+1 -
+w
This proves that the difference is always equal to w
I have proved that the above results hold for any n by w square. W is equal to the size of the overall number grid.
It shows that the rule applies to any combination of letters or numbers, within its limits, there are certain limitations which mean that the solution will not hold for every number, it will not hold for any grid which contains 100 or more numbers, (in other words greater than 10 by 10), this shows that although that the solution does hold for most possibilities certain limitations mean that it cannot fit for absolutely every possibility.
I will now co-ordinate all three of my variables to see if there is any link, when I changed the size of square I discovered that the difference of any m by m square was equal to 10 (p-1)², when I changed 10 (p-1), already I can spot that the only difference between these two differences is that the square difference is squared, and the rectangle difference isn’t.
In conclusion, I have learnt that the by simply exploring and investigating different variables you can discover that there are certain links and connected outcomes, from my investigations I have discovered that the differences that I have found, hold for that particular shape or grid size, I have proved this algebraically, providing the general solution which shows that it holds for any number or letter.