[x² + 11xn – 11x + 10(n-1)²] – [x² + 11nx - 11x] = 10(n-1)²
Therefore:
{[x+(n-1)]*[x+10(n-1)]} – {x*[x+10(n-1)+(n-1)]} = 10(n-1)²
The difference for any sized square on the number grid is 10(n-1)² when n is the length of the side of the square.
Rectangles
I am now going to investigate rectangles on the number grid
2x3 Rectangles
1*13 = 13
3*11 = 33
33-13 = 20
36*48 = 1728
38*46 = 1748
1748-1728 = 20
The difference is ALWAYS 20
Generalisation:
x*(x+12) = x²+12x
(x+2)*(x+10) = x²+10x+2x+20 = x²+12x+20
(x²+12x+20) – (x²+12x) = 20
Therefore:
[(x+2)*(x+10)] – [x*(x+12)] = 20
3x6 Rectangles
1*26 = 26
6*21 = 126
126-26 = 100
75*100 = 7500
80*95 = 7600
7600-7500 = 100
The difference is ALWAYS 100
Generalisation:
x*(x+25) = x²+25x
(x+5)*(x+20) = x²+20x+5x+100 = x²+25x+100
(x²+25x+100) – (x²+25x) = 100
Therefore:
[(x+5)*(x+20)] - [x*(x+25)] = 100
2x4 Rectangles
22*53=1166
23*52=1196
1196-1166=30
56*87=4872
57*86=4902
4902-4872=30
The difference is ALWAYS 30
Generalisation:
x*(x+31) = x²+31x
(x+1)*(x+30) = x²+31x+30
(x²+31x+30) – (x²+31x) = 30
Therefore:
[(x+1)*(x+30)]-[x+(x+31)] = 30
axb rectangles
x*[x+10(b+1)+(a-1)] = x*[x+10b-10+(a-1)] = x²+10bx+10x+ax-x = x²+10bx–11x+ax
[x+(a-1)]*[x+10(b-1)] = [x+(a-1)]*[x+10b-10] = x²+10bx-10x+ax-x+10ba-10b-10a+10
= x²+10bx-11x+ax+10ba-10b-10a+10
(x²+10bx-11x+ax+10ba-10b-10a+10) - (x²+10bx–11x+ax) = 10ba-10b-10a+10
Therefore:
{[x+(a-1)]*[x+10(b-1)]} – {x*[x+10(b+1)+(a-1)]} = 10ba-10b-10a+10
The difference for any rectangle is 10ba-10b-10a+10 when a is equal to the horizontal length and b is equal to the vertical length. This is also true when using squares. It is true because a square is just a rectangle with equal sides, and the formula is not affected if a and b are equal.
I am now going to continue the investigation with a 10x10 number grid numbered 1–200 with the numbers increasing by 2 instead of 1 each time.
3x2 rectangles
24*48=1152
28*44=1232
1232-1152 = 80
Difference = 80
2x6 rectangles
52*154=8008
54*152=8208
8208-8008 = 200
Difference = 200
4x3 rectangles
124*170 = 21080
130*164 = 21320
21320-21080 = 240
Difference = 240
axb rectangles
x*[x+2(a-1)+20(b-1)] = x*(x+2a-2+20b-20) = x*(x+2a+20b-22) = x²+2ax+20bx-22x
[x+2(a-1)]*[x+20(b-1)] = (x+2a-2)*(x+20b-20) = x²+20bx-20x+2ax+40ab-40a-2x-40b+40
= x²+20bx-22x+2ax+40ab-40a-40b+40
(x²+2ax+20bx-22x+40ab-40a-40b+40) – (x²+2ax+20bx-22x) = 40ab-40a-40b+40
Therefore:
{x*[x+2(a-1)+20(b-1)]} – {[x+2(a-1)]*[x+20(b-1)]} = 40ab-40a-40b+40
On any rectangle (or square) on this number grid, the difference is 40ab-40a-40b+40, where a is the horizontal length, and b is the vertical length. This is 2x the difference for the number grid where the numbers increase by 1 each time.
I will now attempt to find the difference for any sized rectangle on a 10x10 number grid with any increment between the numbers.
Let i equal the increment between the numbers on the number grid
When i = 1
Difference = 10ab-10b-10a+10
When i = 2
Difference = 40ab-40a-40b+40
Therefore
Difference = 10i²ab-10i²a-10i²b+10i²
x*[x+i(a-1)+10i(b-1)] = x*(x+ia-i+10ib-10i) = x²+iax-ix+10ibx-10ix
[x+i(a-1)]*[x+10i(b-1)] = (x+ia-i)*(x+10ib-10i) = x²+10ibx-10ix+iax+10i²ab-10i²a-ix-10i²b+10i²
(x²+10ibx-10ix+iax+10i²ab-10i²a-ix-10i²b+10i²) – (x²+iax-ix+10ibx-10ix) = 10i²ab-10i²a-10i²b+10i²
Therefore:
{[x+i(a-1)]*[x+10i(b-1)]} – {[x+i(a-1)]*[x+10i(b-1)]} = 10i²ab-10i²a-10i²b+10i²
I will now continue to the investigation with a number grid with 6 numbers per row instead of 10, an increment of 1, and numbers 1-36.
4x2 rectangles
14*23=322
17*20=340
437-340=18
27*36=972
30*36=990
990-972=18
Difference = 18
3x3 rectangles
6*16=96
4*18=72
96-72=24
Difference = 24
axb rectangles
x*[x+(a-1)+6(b-1)] = x*[x+(a-1)+6b-6] = x²+ax-x+6bx-6x = x²+ax+6bx-6x-x = x²+ax+6bx-7x
[x+(a-1)][x+6(b-1)] = [x+(a-1)][x+6b-6] = x²+ax+6bx-6x-x+6ab-6b-6a+6 = x²+ax+6bx-7x+6ab-6b-6a+6
(x²+ax+6bx-7x+6ab-6b-6a+6) - (x²+ax+6bx-7x) = 6ab-6a-6b+6
Therefore:
{[x+(a-1)][x+6(b-1)]} - {x*[x+(a-1)+6(b-1)]} = 6ab-6b-6a+6
On a number grid with rows 10 long, the rule for the difference in rectangles is 10ab-10a-10b+10
On a number grid with rows 6 long, the rule is 6ab-6a-6b+6
This shows that the co-efficient and the number added at the end of the formulae are equal to the length of the rows in the number grid.
Let l equal the length of the row of in the number grid.
Therefore, the difference of an axb rectangle when a is the horizontal length and b is the vertical length is abl-al-bl+l
x*[x+(a-1)+l(b-1)] = x*[x+(a-1)+lb-l] = x²+ax-x+blx-lx
[x+(a-1)]*[x+l(b-1)] = [x+(a-1)]*(x+lb-l) = x²+blx-lx+ax-x+abl-bl-al+l
(x²+blx-lx+ax-x+abl-bl-al+l) - (x²+ax-x+blx-lx) = abl-bl-al+l
Therefore:
{[x+(a-1)]*[x+l(b-1)]} - {x*[x+(a-1)+l(b-1)]} = abl-bl-al+l
I will now use a 6 row grid where i=2. I will find a formula for this, then attempt to find a formula for any grid with any increment.
3x2 rectangles
2*18 = 36
6*14 = 84
84-36 = 48
Difference = 48
4x3 rectangles
30*72 = 2160
36*66 = 2376
2376-2160 = 216
Difference = 216
axb rectangles
x*[x+2(a-1)+12(b-1)] = x[x+2a-2+12b-12] = x²+2ax-14x+12bx
[x+2(a-1)]*[x+12(b-1)] = (x+2a-2)*(x+12b-12) = x²+12bx-14x+2ax+24ab-24a-24b+24
(x²+12bx-14x+2ax+24ab-24a-24b+24) - (x²+2ax-14x+12bx) = 24ab-24a-24b+24
Therefore:
{x*[x+2(a-1)+12(b-1)]} – {[x+2(a-1)]*[x+12(b-1)]} = 24ab-24a-24b+24
On a 6 row grid where i=1, the rule is {[x+(a-1)][x+6(b-1)]} - {x*[x+(a-1)+6(b-1)]}
= 6ab-6b-6a+6
On a 6 row grid where i=2, the rule is {x*[x+2(a-1)+12(b-1)]} – {[x+2(a-1)]*[x+12(b-1)]} = 24ab-24a-24b+24
Therefore, the general rule for a grid where l=6 for any value of i is 6i²ab-6²a-6i²b+6i²
Generalisation:
The co-efficient of each part of the rule is the same as the value of l. This means that the general rule will be li²ab-l²a-li²b+li²
x*[x+i(a-1)+il(b-1)] = x*(x+ia-i+ilb-il) = x²+iax-ix+ilbx-ilx
[x+i(a-1)]*[x+il(b-1)] = (x+ia-i)*(x+ilb-il) = x²+ilbx-ilx+iax+li²ab-li²a-ix-li²b+li²
(x²+ilbx-ilx+iax+li²ab-li²a-ix-li²b+li²) – (x²+iax-ix+ilbx-ilx) = li²ab-li²a-li²b+li²
Therefore:
{[x+i(a-1)]*[x+il(b-1)]} – {x*[x+i(a-1)+il(b-1)]} - li²ab-li²a-li²b+li²
On any number grid with any length of sides and any increment, the difference between the products of the numbers in opposite corners in any rectangle is li²ab-li²a-li²b+li² when a=the horizontal length of the rectangle, b=the vertical length of the rectangle, i=the increment between numbers on the number grid, and l=the length of the rows on the grid.