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Introduction

Mark Johnson 10MR

Number Grid Investigation

In this investigation, I am using a 10x10 number grid, with numbers 1 to 100.  I am going to take 2x2 sections of this grid, and calculate the differences between the products of the top-left and bottom-right numbers, and the bottom-right and top-left numbers.  Once I have found a pattern, I will attempt to write a generalisation using algebra.  After I have done this, I investigate larger sections taken from the grid, 3x3, 4x4 and so on.  I will then try to find a general rule for the numbers in any size grid using algebra.

2x2 Sections

12*23=276

13*22=286

286-276=10

65*76=4940

66*75=4950

4950-4940=10

29*40=1160

30*39=1170

1170-1160=10

The difference is ALWAYS 10

Generalisation:

x

x+1

## x+10

x+11

x*(x+11) = x²+11x

(x+1)(x+10) = x²+11x+10

(x²+11x+10) – (x²+11x) = 10

Therefore:

(x+1)(x+10) – x(x+11) = 10

3x3 Sections

13*35=455

15*33=495

495-455=40

27*49=1323

29*47=1363

1363-1323=40

65*87=5655

67*85=5695

5695-5655=40

The difference is ALWAYS 40

Generalisation:

## x

x+1

x+2

x+10

x+11

x+12

x+20

x+21

x+22

x(x+22) = x²+22x

(x+2)(x+20) = x²+22x+40

(x²+22x+40) – (x²+22x) = 40

Therefore:

(x+2)(x+20) - x(x+22) = 40

4x4 Sections

11*44=484

14*41=574

574-484=90

26*59=1534

29*56=1624

1624-1534=90

The difference is ALWAYS 90

Generalisation:

X

x+1

## x+2

x+3

x+10

x+11

x+12

x+13

x+20

x+21

x+22

x+23

x+30

x+31

x+32

x+33

x*(x+33) = x²+33x

(x+3)(x+30) = x²+33x+90

(x²+33x+90)

Middle

x+31

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+(a-1) x+10(b-1) x+10(b-1)+(a-1) ## Difference

3x2

20

6x3

100

2x4

30

a x b

10ba-10b-10a+10

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

Conclusion

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

## Difference

10

10i²ab-10i²a-10i²b+10i²

6

6i²ab-6²a-6i²b+6i²

l

li²ab-li²a-li²b+li² z z+i (z+i)+i x x+i(a-1) b x+il(b-1) x+i(a-1)+il(b-1) 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.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

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