D = 40 D = 40
n(n+22) = n2 + 22n
(n+2)(n+20) = n2 + 22n + 40
Squares of equal sizes always have the same difference.
E.g. 2 x 2, D=10, 3 x 3, D=40.
From now on I will only use n grids in my diagrams as they show the numbers that produce the difference.
4 x 4
Length of side = X
n(n+33) = n2 + 33n
(n+3)(n+30) = n2 + 33n +90
For 4x4, D = 90
From this table I can see that to find the difference in a square the formula is 10(X-1) 2 OR 10X2 –20X +10
This works for all 2 x 2 squares on 10 x 10 grids.
Now I will look at rectangles.
X+1
n(n+12) = n2 + 12n
X (n+2)(n+10) = n2 + 12n +20
D = 20
n(n+23) = n2 + 23n
(n+3)(n+20)= n2 + 23n + 60
D = 60
n(n+34)= n2 + 34n
(n+4)(n+30)=n2+34n+120
D = 120
This is a table of my results.
2 x 10 = 20 From this I can tell that the formula for
3 x 20 = 60 finding X(X+1) rectangles is
4 x 30 = 120 10X(X-1) OR 10X2 –10X
This number is always 10(X-1)
Now I will look at rectangles where one pair of sides are two units longer than the other pair of sides.
X(X+2)
n(n+13)= n2 + 13n (n+3)(n+10)= n2 +13n+30
D = 30
n(n+24)=n2 +24n (n+4)(n+20)=n2 +24n+80
D = 80
n(n+35)=n2 +35n
(n+5)(n+30)=n2 +35n+150
D = 150
From this I can tell that the formula for finding X(X+2) rectangles is 10(X2-1) OR 10X2 –10
Now I will look at rectangles where one pair of sides are three units longer than the other pair of sides.
X(X+3)
n(n+14)=n2 +14n
(n+4)(n+10)=n2 +14n+40
D = 40
n(n+25)=n2 +14n
(n+5)(n+20)=n2 +25n+100
D = 100
n(n+36)=n2 +36n
(n+6)(n+30)=n2 +36n+180
D = 180
From this I can tell that the formula for finding X(X+3) rectangles is 10(X2 +X -2) OR 10X2 +10X –20
From my formulae form X(X+1), X(X+2) and X(X+3), I can now work out the formula for all rectangles.
The formulae for all rectangles are quadratic equations. In
X(X+1) and X(X+2) one of the terms has 0 as its multiplier.
I will call this number Y.
X(X+1) X(X+2) X(X+3)
The table shows:
-
That 10X2 is constant, it is even present in the formula for squares. Because all the rectangles and squares were on a 10 x 10 grid, the formula always contained a 10X2 term. I predict that on a 9 x 9 grid the formula will contain a 9X2 term.
- All the X terms of the formulae have a multiplier. This multiplier is going up by 10 for each addition to Y, e.g. –10, 0, 10 etc.
- The formulae also have a number term that decreases by 10 as Y increases, e.g. 0, -10, -20.
From these terms I can predict that the formula to find the difference for X(X+4).
The formula will start with 10X2, the next number in the X term series will be +20X and the next number in the number term series will be –30. So the formula will be 10X2 + 20X –30
To work out the X term the formula is 10X(Y-2) OR 10XY –20X
To work out the Number term the formula is
–10(Y-1) OR –10Y +10
Putting all these formulae together (+10x2) will give me the formula for all rectangles on a 10 x 10 grid.
10x2 + 10XY –20X –10Y +10
This can be simplified to
10(X2 +XY –Y –2X +1)
Now I will try squares on a 9 x 9 grid.
n(n+10) = n2 + 10n
(n+1)(n+9) = n2 + 10n + 9
D = 9
n(n+20) = n2 + 20n
(n+2)(n+18) = n2 + 20n + 36
D = 36
n(n+30) = n2 + 30n
(n+3)(n+27) = n2 + 30n + 81
D = 81
I will call the number of columns in a grid Z.
From this I can see that the formula is 9(X-1)2 OR 9X2 –18X +9
The formula for a square on a 10x10 grid is 10X2 –20X +10
By comparing these two formulae I can tell that the formula for any square on any grid size is ZX2 –2ZX +Z
I predict that for the formula for rectangles I will have to substitute Z in to some of the values.
10(X2 +XY –Y –2X +1) Z(X2 +XY –Y –2X +1)
For example if Z=9, X=4 and Y=3 the difference will be 162.
20 x 53 = 1060
26 x 47 = 1222
D = 162
Now I have the formulae for any square or any rectangle on any size of grid. Given more time I would have found formulae for more irregular shapes like a T or a Pyramid.
