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Introduction

Algebraic Investigation 1: Square Boxes on a 10x10 Grid

In this first investigation, the difference in products of the alternate corners of a square, equal-sided box on a 10x10 gridsquare will be investigated. It is believed that the products and their differences should demonstrate a constant pattern no matter what dimensions are used; as long as they remain equal. In order to prove this, both a numeric and algebraic method will be used in order to calculate this difference. The numeric method will help establish a baseline set of numbers for testing, and to help in the establishment of a set of algebraic formulae for use on an n x n gridsquare.

In the example gridsquare below, the following method is used in order to calculate the difference between the products of opposite corners.

 (a) (b) (c) (d)

Stage A:        Top left number x Bottom right number =        (a) multiplied by (d)

Stage B:        Bottom left number x Top right number =         (c) multiplied by (b)

Stage B – Stage A:        (c)(b) - (a)(d)          =  The difference

The overall, 10 x 10 grid that is used for the first investigation will be a standard, cardinal gridsquare, which progresses in increments of 1. The formulae calculated will mainly be applicable to this grid, as other formats of gridsquares will require others formulae to provide valid results.

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

This first investigation will focus only on gridsquares with equal widths and heights, which will at this stage be represented by the universal, constant term ‘w’.

The top left number in the grid (letter (a) in the above example) will be represented by the term ‘n’, which will be referred to in this manner in all proceeding investigations also.

This is only the first section of the investigation.

Middle

75

The above example has further proved that the formula is correct and can be used to predict the difference values of larger, square boxes in a 10x10 grid.

Algebraic Investigation 2: Rectangular Boxes on a 10x10 Grid

The second investigation comprises of two main parts

Part A:                Changing the Width, ‘w’

In order to investigate the formulae and trends that occur when different dimension grids are used, it is necessary to examine various different widths of rectangular box to find suitable algebraic expressions for 2 x w, 3 x w and 4 x w boxes and the differences between the products of the applicable alternate corners. In order to do this, a similar method as before will be undertaking, using both an algebraic and numeric method of checking and validating formulae and results.

In this section of the investigation, additional variables (constant terms) will be used due to the change in the number of factors involved.

‘n’ will continue to represent the top left number in the grid. All other formulae in the grids will refer to n.

Due to the height and width of the boxes no longer being equal values, ‘w’ will represent the width of the number box, whereas ‘h’ will refer to the height,

Part B:                Changing the Height, ‘h’

To provide a comparison for finding the height formula, summary boxes will be included, showing the algebraic results of the last section.

As in this part of the investigation, additional factors are concerned; the overall formula will contain numerous terms. The rules of brackets and of ‘bidmas’ will always apply to the formulae, helping to create a workable equation.

It should be possible to create an algebra grid detailing the numbers to be expected on any height x width grid on a 10 x 10 gridsquare. After multiplying the alternate corners and subtracting, an overall formula should be gained that can be used to calculate the difference in any h x w box; always on a 10 x 10 grid.

Part A:2 x Width Rectangles

2x2 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 15 16 25 26
 n n+1 n+10 n+11

Stage A:        Top left number x Bottom right number =        n(n+11) =        n2+11n

Stage B:        Bottom left number x Top right number =         (n+10)(n+1)=        n2+1n+10n+10

=        n2+11n+10

Stage B – Stage A:        (n2+11n+10)-(n2+11n) = 10

When finding the general formula for any number (n), both answers begin with the equation n2+11n, which signifies that they can be manipulated easily. Because the second answer has +10 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 10 will always be present.

2x3 Rectangle

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 6 7 8 16 17 18
 n n+1 n+2 n+10 n+11 n+12

Stage A:        Top left number x Bottom right number =        n(n+12) =        n2+12n

Stage B:        Bottom left number x Top right number =         (n+10)(n+2)=        n2+2n+10n+20

=        n2+12n+20

Stage B – Stage A:        (n2+12n+20)-(n2+12n) = 20

When finding the general formula for any number (n), both answers begin with the equation n2+12n, which signifies that they can be manipulated easily. Because the second answer has +20 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 20 will always be present.

2x4 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 81 82 83 84 91 92 93 94
 n n+1 n+2 n+3 n+10 n+11 n+12 n+13

Stage A:        Top left number x Bottom right number =        n(n+13) =        n2+13n

Stage B:        Bottom left number x Top right number =         (n+10)(n+3)=        n2+3n+10n+30

=        n2+13n+30

Stage B – Stage A:        (n2+13n+30)-(n2+13n) = 30

When finding the general formula for any number (n), both answers begin with the equation n2+13n, which signifies that they can be manipulated easily. Because the second answer has +30 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 30 will always be present.

Any rectangular box with width ‘w’, and a height of 2

It is possible to ascertain from the above examples that each box follows a certain trend (each progressive increase in width shows an overall increase in the difference by +10). It is possible to use the 2x2, 3 and 4 rectangles above to find the overall, constant formula for any 2xw rectangle on a 10x10 grid, by implementing the formulae below to find a general calculation and grid rectangle.

It is additionally possible to see that the numbers that are added to n (mainly in the corners of the grids) follow certain, and constant sets of rules, which demonstrates confirmation of a pattern.

The bottom right number in the rectangle is directly linked to the top right and bottom left numbers. It is the sum of these that equal the bottom right, or:

Formula 1:                Bottom Right (BR)  =  Top Right (TR)  +  Bottom Left (BL)

As also shown by the summary boxes and examples above, the formula for the top right number remains constant, and is linked with the width, w, of the box in the following way:

Formula 2:                Top Right (TR)  =  n+ Width (w)  -  1

It is also evident from the examples calculated that the bottom left number is also linked with the height, w, of the box using a formula that remains constant:

Formula 3:                Bottom Left (BL)  = n+ (Height (h)  -  1)  x  10

 n ~ n+w-1 n+10 ~ (n+10)+(w-1)

The above grid simplifies to form:

 n ~ n+w-1 n+10 ~ n+9+w

Stage A:        Top left number x Bottom right number         =        n(n+9+w)

=        n2+9n+nw

Stage B:        Bottom left number x Top right number         =         (n+10)(n+w-1)

=        n2+nw-n+10n+10w-10

=         n2+nw+9n+10w-10

Stage B – Stage A:        (n2+nw+9n+10w-10)-(n2+9n+nw)         =         10w-10

When finding the general formula for any number (n), both answers begin with the equation n2+nw+9n, which signifies that they can be manipulated easily. Because the second answer has +10w-10 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 10w-10 will always be present.

3 x Width Rectangles

3x2 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 75 76 85 86 95 96
 n n+1 n+10 n+11 n+20 n+21

Stage A:        Top left number x Bottom right number =        n(n+21) =        n2+21n

Stage B:        Bottom left number x Top right number =         (n+20)(n+1)=        n2+1n+20n+20

=        n2+21n+20

Stage B – Stage A:        (n2+21n+20)-(n2+21n) = 20

When finding the general formula for any number (n), both answers begin with the equation n2+21n, which signifies that they can be manipulated easily. Because the second answer has +20 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 20 will always be present.

3x3 Rectangle

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 21 22 23 31 32 33 41 42 43
 n n+1 n+2 n+10 n+11 n+12 n+20 n+21 n+22

Stage A:        Top left number x Bottom right number =        n(n+22) =        n2+22n

Stage B:        Bottom left number x Top right number =         (n+20)(n+2)=        n2+2n+20n+40

=        n2+22n+40

Stage B – Stage A:        (n2+22n+40)-(n2+22n) = 40

When finding the general formula for any number (n), both answers begin with the equation n2+22n, which signifies that they can be manipulated easily. Because the second answer has +40 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 40 will always be present.

3x4 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 37 38 39 40 47 48 49 50 57 58 59 60
 n n+1 n+2 n+3 n+10 n+11 n+12 n+13 n+20 n+21 n+22 n+23

Stage A:        Top left number x Bottom right number =        n(n+23) =        n2+23n

Stage B:        Bottom left number x Top right number =         (n+20)(n+3)=        n2+3n+20n+60

=        n2+23n+60

Stage B – Stage A:        (n2+23n+60)-(n2+23n) = 60

When finding the general formula for any number (n), both answers begin with the equation n2+23n, which signifies that they can be manipulated easily. Because the second answer has +60 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 60 will always be present.

Any rectangular box with width ‘w’, and a height of 3

It is possible to ascertain from the above examples that each box follows a certain trend (each progressive increase in width shows an overall increase in the difference by +20). It is possible to use the 3x2, 3 and 4 rectangles above to find the overall, constant formula for any 3xw rectangle on a 10x10 grid, by implementing the formulae below to find a general calculation and grid rectangle.

It is additionally possible to see that the numbers that are added to n (mainly in the corners of the grids) follow certain, and constant sets of rules, which demonstrates confirmation of a pattern.

The bottom right number in the rectangle is directly linked to the top right and bottom left numbers. It is the sum of these that equal the bottom right, or:

Formula 1:                Bottom Right (BR)  =  Top Right (TR)  +  Bottom Left (BL)

As also shown by the summary boxes and examples above, the formula for the top right number remains constant, and is linked with the width, w, of the box in the following way:

Formula 2:                Top Right (TR)  =  n+ Width (w)  -  1

It is also evident from the examples calculated that the bottom left number is also linked with the height, w, of the box using a formula that remains constant:

Formula 3:                Bottom Left (BL)  =  n+ (Height (h)  -  1)  x  10

 n ~ n+w-1 ~ ~ ~ n+20 ~ n+20+w-1

The above grid simplifies to form:

 n ~ n+w-1 ~ ~ ~ n+20 ~ n+19+w

Stage A:        Top left number x Bottom right number         =        n(n+19+w)

=        n2+19n+nw

Stage B:        Bottom left number x Top right number         =         (n+20)(n+w-1)

=        n2+nw-n+20n+20w-20

=         n2+nw+19n+20w-20

Stage B – Stage A:        (n2+nw+19n+20w-20)-(n2+19n+nw)         =         20w-20

When finding the general formula for any number (n), both answers begin with the equation n2+nw+19n, which signifies that they can be manipulated easily. Because the second answer has +20w-20 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 20w-20 will always be present.

4 x Width Rectangles

4x2 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 2 3 12 13 22 23 32 33
 n n+1 n+10 n+11 n+20 n+21 n+30 n+31

Stage A:        Top left number x Bottom right number =        n(n+31) =        n2+31n

Stage B:        Bottom left number x Top right number =         (n+30)(n+1)=        n2+1n+30n+30

=        n2+31n+30

Stage B – Stage A:        (n2+31n+30)-(n2+31n) = 30

When finding the general formula for any number (n), both answers begin with the equation n2+31n, which signifies that they can be manipulated easily. Because the second answer has +30 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 30 will always be present.

4x3 Rectangle

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 17 18 19 27 28 29 37 38 39 47 48 49
 n n+1 n+2 n+10 n+11 n+12 n+20 n+21 n+22 n+30 n+31 n+32

Stage A:        Top left number x Bottom right number =        n(n+32) =        n2+32n

Stage B:        Bottom left number x Top right number =         (n+30)(n+2)=        n2+2n+30n+60

=        n2+32n+60

Stage B – Stage A:        (n2+32n+60)-(n2+32n) = 60

When finding the general formula for any number (n), both answers begin with the equation n2+32n, which signifies that they can be manipulated easily. Because the second answer has +60 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 60 will always be present.

4x4 Rectangles

Firstly, a rectangle with applicable numbers from the grid will be selected as a baseline model for testing.

 32 33 34 35 42 43 44 45 52 53 54 55 62 63 64 65
 n n+1 n+2 n+3 n+10 n+11 n+12 n+13 n+20 n+21 n+22 n+23 n+30 n+31 n+32 n+33

Stage A:        Top left number x Bottom right number =        n(n+33) =        n2+33n

Stage B:        Bottom left number x Top right number =         (n+30)(n+3)=        n2+3n+30n+90

=        n2+33n+90

Stage B – Stage A:        (n2+33n+90)-(n2+33n) = 90

When finding the general formula for any number (n), both answers begin with the equation n2+33n, which signifies that they can be manipulated easily. Because the second answer has +90 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 90 will always be present.

Any rectangular box with width ‘w’, and a height of 3

It is possible to ascertain from the above examples that each box follows a certain trend (each progressive increase in width shows an overall increase in the difference by +30). It is possible to use the 4x2, 3 and 4 rectangles above to find the overall, constant formula for any 4xw rectangle on a 10x10 grid, by implementing the formulae below to find a general calculation and grid rectangle.

It is additionally possible to see that the numbers that are added to n (mainly in the corners of the grids) follow certain, and constant sets of rules, which demonstrates confirmation of a pattern.

The bottom right number in the rectangle is directly linked to the top right and bottom left numbers. It is the sum of these that equal the bottom right, or:

Formula 1:                Bottom Right (BR)  =  Top Right (TR)  +  Bottom Left (BL)

As also shown by the summary boxes and examples above, the formula for the top right number remains constant, and is linked with the width, w, of the box in the following way:

Formula 2:                Top Right (TR)  =  Width (w)  -  1

It is also evident from the examples calculated that the bottom left number is also linked with the height, w, of the box using a formula that remains constant:

Formula 3:                Bottom Left (BL)  =  n+ (Height (h)  -  1)  x  10

 n ~ n+w-1 ~ ~ ~ ~ ~ ~ n+30 n+30+w-1

The above grid simplifies to form:

 n ~ n+w-1 ~ ~ ~ ~ ~ ~ n+30 n+29+w

Stage A:        Top left number x Bottom right number         =        n(n+29+w)

=        n2+29n+nw

Stage B:        Bottom left number x Top right number         =         (n+30)(n+w-1)

=        n2+nw-n+30n+30w-30

=         n2+nw+29n+30w-30

Stage B – Stage A:        (n2+nw+29n+30w-30)-(n2+29n+nw)         =         30w-30

When finding the general formula for any number (n), both answers begin with the equation n2+nw+29n, which signifies that they can be manipulated easily. Because the second answer has +30w-30 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 30w-30 will always be present.

Part B:Changing the Height, ‘h’

2 x w

 n ~ n+w-1 n+10 ~ n+9+w

Overall, constant difference of 2 x w grids:        10w-10

3 x w

 n ~ n+w-1 ~ ~ ~ n+20 ~ n+19+w

Overall, constant difference of 3 x w grids:        20w-20

4 x w

 n ~ n+w-1 ~ ~ ~ ~ ~ ~ n+30 n+29+w

Overall, constant difference of 4 x w grids:        30w-30

From the trend clearly displayed in both the algebraic summary boxes and the overall difference equation above, it is possible to create a sample algebra box for a grid size of any h x w:

 n ~ n+w-1 ~ ~ ~ n+10(h-1) ~ n+10h-10+w-1

Which simplifies easily into:

 n ~ n+w-1 ~ ~ ~ n+10h-10 ~ n+10h+w-11

Conclusion

To obtain the overall algebraic box, for any increment size, it is necessary to refer back to the previously stated formulae for calculating the individual terms that should be present in the corners. It is therefore possible to use the grids to find the overall, constant formula for any rectangle on any size grid, with any increment size, by implementing the formulae below to find a general calculation and grid rectangle.

It is additionally possible to see that the numbers that are added to n (mainly in the corners of the grids) follow certain, and constant sets of rules, which demonstrates confirmation of a pattern.

The bottom right number in the rectangle is directly linked to the top right and bottom left numbers. It is the sum of these that equal the bottom right, or:

Formula 1:                Bottom Right (BR)  =  Top Right (TR)  +  Bottom Left (BL)

As also shown by the summary boxes and examples above, the formula for the top right number remains constant, and is linked with the width, w, of the box in the following way:

Formula 2:                Top Right (TR)  =  n+ Increment size (s) x (Width -1)

It is also evident from the examples calculated that the bottom left number is also linked with the height, w, of the box using a formula that remains constant:

Formula 3:                Bottom Left (BL)  =  n+ Increment size x Gridsize x (height -1)

Using these rules, it is possible to establish an algebraic box that could be used to calculate the difference for any hxw box on any gxg grid.

 n ~ n+s(w-1) ~ ~ ~ n+gs(h-1) ~ n+s(w-1)+gs(h-1)

Stage A:        Top left number x Bottom right number

= n² +sn (w-1)+gns (h-1)

Stage B:        Bottom left number x Top right number

= n²+ns (w-1)+gns (h-1)+gs² (h-1)(w-1)

Stage B – Stage A:

= gs² h-1)(w-1)

= s²ghw-s²gh-s²gw+s²g

= s²g(hw-h-w+1)

= s²g(h-1)(w-1)

Charankamal Singh Theora

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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