Number Grid Investigation.

   [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.