= n²+11nw-11n
Stage B: Bottom left number x Top right number
= (n+10w-10)(n+w-1)
= n²+nw-n+10nw+10w²-10w-10n-10w+10
= n²+11nw-11n-20w+10w²+10
Stage B – Stage A: = [ n²+11nw-11n-20w+10w²+10] - (n²+11nw-11n)
= 10w²-20w+10
The result of these calculations means that for any size square box, the difference should be easily calculated using the formula. For example, if one of the previously tested 4x4 boxes is examined:
This means that if the width (w) of 4 is inserted into the formula, the difference of 90 should be returned.
Difference (d) = 10 x 4² - 20 x 4 + 10
Difference (d) = 160 - 80 +10
Difference (d) = 90
The above calculation has proved that the formula does in fact work and returns a valid result. It will therefore now be used in order to predict another difference of a box not yet tested (of w=6).
Difference (d) = 10 x 6² - 20 x 6 + 10
Difference (d) = 360 - 120 +10
Difference (d) = 250
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.
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.
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.
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
The above grid simplifies to form:
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.
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.
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.
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
The above grid simplifies to form:
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.
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.
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.
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
The above grid simplifies to form:
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
Overall, constant difference of 2 x w grids: 10w-10
3 x w
Overall, constant difference of 3 x w grids: 20w-20
4 x 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:
Which simplifies easily into:
Stage A: Top left number x Bottom right number
= n(n+10h+w-11)
= n2+10hn+nw-11n
Stage B: Bottom left number x Top right number
= (n+w-1)(n+10h-10)
= n2+10hn-10n+nw+10hw-10w-n+10h+10
= n2+10hn-11n+nw+10hw-10w+10h+10
Stage B – Stage A: (n2+10hn-11n+nw+10hw-10w+10h+10)-(n2+10hn+nw-11n)
= 10hw-10w-10h+10
When finding the general formula for any number (n) and any width (w), both answers begin with the equation n2+10hn-11n+nw, which signifies that they can be manipulated easily. Because the second answer has +10hw-10w+10h+10 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 10hw-10w+10h+10 will always be present.
Algebraic Investigation 3: Boxes on a ‘g’ x ‘g’ Grid, with Increments of ‘s’
The second investigation comprises of two main parts
Part A: Changing Grid Size, ‘g’
After completing all the stages previously detailed, it is now a logical decision to consider and investigate changing the actual dimensions of the overall grid.
Part B: Changing Increment Size, ‘s’
Another part of the investigation will be to analyse the overall difference formula in the event that a ‘step-size’ larger of 1 is used. In the diagram below, s is representing the jump (increment) size.
Part A: Changing Grid Size: 5 x 5 Grid
Experimental number box:
Algebraic box:
Which (when brackets are multiplied out) simplifies to:
Stage A: Top left number x Bottom right number
= n(n+5h+w-6)
= n2+nw+5hn-6n
Stage B: Bottom left number x Top right number
= (n+5h-5)(n+w-1)
= n2+nw-n+5hn+5hw-5h-5n-5w+5
= n2+nw+5hn-6n+5hw-5h-5w+5
Stage B – Stage A: (n2+nw+5hn-6n+5hw-5h-5w+5)-(n2+nw+5hn-6n)
= 5hw-5h-5w+5
When finding the general formula for any number (n), any height (h), and any width (w) both answers begin with the equation n2+nw+5hn-6n, which signifies that they can be manipulated easily. Because the second answer has +5hw-5h-5w+5 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 5hw-5h-5w+5 will always be present.
Testing:
Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on the 5x5 grid in question. As the number box clearly demonstrates, a difference of 5 should be present. This means the algebraic equation should be:
Difference: (5x2x2)-(5x2)-(5x2)+5 = 5
Therefore, the formula gained has been proven to be correct and valid for calculating the difference.
6 x 6 Grid
Experimental number box:
Algebraic box:
Which (when brackets are multiplied out) simplifies to:
Stage A: Top left number x Bottom right number
= n(n+6h+w-7)
= n2+nw+6hn-7n
Stage B: Bottom left number x Top right number
= (n+6h-6)(n+w-1)
= n2+nw-n+6hn+6hw-6h-6n-6w+6
= n2+nw+6hn-7n+6hw-6h-6w+6
Stage B – Stage A: (n2+nw+6hn-7n+6hw-6h-6w+6)-(n2+nw+6hn-7n)
= 6hw-6h-6w+6
When finding the general formula for any number (n) any height (h), and any width (w) both answers begin with the equation n2+nw+6hn-7n, which signifies that they can be manipulated easily. Because the second answer has +6hw-6h-6w+6 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 6hw-6h-6w+6 will always be present.
Testing:
Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on the 6x6 grid in question. As the number box clearly demonstrates, a difference of 6 should be present. This means the algebraic equation should be:
Difference: (6x2x2)-(6x2)-(6x2)+6 = 6
Therefore, the formula gained has been proven to be correct and valid for calculating the difference.
7 x 7 Grid
Experimental number box:
Algebraic box:
Which (when brackets are multiplied out) simplifies to:
Stage A: Top left number x Bottom right number
= n(n+7h+w-8)
= n2+nw+7hn-8n
Stage B: Bottom left number x Top right number
= (n+7h-7)(n+w-1)
= n2+nw-n+7hn+7hw-7h-7n-7w+7
= n2+nw+7hn-8n+7hw-7h-7w+7
Stage B – Stage A: (n2+nw+7hn-8n+7hw-7h-7w+7)-(n2+nw+7hn-8n)
= 7hw-7h-7w+7
When finding the general formula for any number (n) any height (h), and any width (w) both answers begin with the equation n2+nw+7hn-8n, which signifies that they can be manipulated easily. Because the second answer has +7hw-7h-7w+7 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 7hw-7h-7w+7 will always be present.
Testing:
Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on the 7x7 grid in question. As the number box clearly demonstrates, a difference of 7 should be present. This means the algebraic equation should be:
Difference: (7x2x2)-(7x2)-(7x2)+7 = 7
Therefore, the formula gained has been proven to be correct and valid for calculating the difference.
Summary Boxes and Grid Size of ‘g’
5x5 Grid
Algebraic Summary Box:
Overall difference calculation: 5hw-5h-5w+5
6x6 Grid
Algebraic Summary Box:
Overall difference calculation: 6hw-6h-6w+6
7x7 Grid
Algebraic Summary Box:
Overall difference calculation: 7hw-7h-7w+7
gxg Grid
As can be observed from the examples above, the algebraic boxed and the overall difference formulae show a distinct trend. In order to prove that this trend is exhibited in all cases, however, it is necessary to conduct some further algebraic research.
To obtain the overall algebraic box, for any gxg grid, 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, 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
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.
Which can simplify into:
Stage A: Top left number x Bottom right number
= n(n+gh+w-g-1)
= n2+ngh+nw-ng-n
Stage B: Bottom left number x Top right number
= (n+gh-g)(n+w-1)
= n2+ngh-ng+nw+ghw-gw-n-gh+g
= n2+ngh+nw-ng-n+ghw-gw-gh+g
Stage B – Stage A: (n2+ngh+nw-ng-n+ghw-gw-gh+g)-( n2+ngh+nw-ng-n)
= ghw-gw-gh+g
When finding the general formula for any number (n) any height (h), any width (w), and on any gxg grid size, both answers begin with the equation n2+ngh+nw-ng-n, which signifies that they can be manipulated easily. Because the second answer has +ghw-gw-gh+g at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of ghw-gw-gh+g will always be present.
Part B: Changing Increment Size (s) Increment Size: 2
Experimental number box:
Algebraic box:
Simplifies to:
Stage A: Top left number x Bottom right number
= n(n+2w+2gh-2g-2)
= n2+2nw+2ghn-2gn-2n
Stage B: Bottom left number x Top right number
= (n+2gh-2g)(n+2w-2)
= n2+2nw-2n+2ghn+4ghw-4gh-2gn-4gw+4g
= n2+2nw+2ghn-2gn-2n+4ghw-4gh-4gw+4g
Stage B – Stage A: (n2+2nw+2ghn-2gn-2n+4ghw-4gh-4gw+4g)-(n2+2nw+2ghn-2gn-2n)
= 4ghw-4gh-4gw+4g
When finding the general formula for any number (n) any height (h), any width (w), and with any gird size, both answers begin with the equation n2+2nw+2ghn-2gn-2n, which signifies that they can be manipulated easily. Because the second answer has +4ghw-4gh-4gw+4g at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 4ghw-4gh-4gw+4g will always be present.
Testing:
Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on a 5x5 grid, with an increment size (s) of 2. As the number box clearly demonstrates, a difference of 20 should be present. This means the algebraic equation should be:
Difference: (4x5x2x2)-(4x5x2)-(4x5x2)+(4x5)= 20
Therefore, the formula gained has been proven to be correct and valid for calculating the difference.
Increment Size: 3
Experimental number box:
Algebraic box:
Simplifies to:
Stage A: Top left number x Bottom right number
= n(n+3w+3gh-3g-3)
= n2+3nw+3ghn-3gn-3n
Stage B: Bottom left number x Top right number
= (n+3gh-3g)(n+3w-3)
= n2+3nw-3n+3ghn+9ghw-9gh-3gn-9gw+9g
= n2+3nw+3ghn-3gn-3n+9ghw-9gh-9gw+9g
Stage B – Stage A: (n2+3nw+3ghn-3gn-3n+9ghw-9gh-9gw+9g)-(n2+3nw+3ghn-3gn-3n)
= 9ghw-9gh-9gw+9g
When finding the general formula for any number (n) any height (h), any width (w), and with any gird size, both answers begin with the equation n2+3nw+3ghn-3gn-3n, which signifies that they can be manipulated easily. Because the second answer has +9ghw-9gh-9gw+9g at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 9ghw-9gh-9gw+9g will always be present.
Testing:
Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on a 5x5 grid, with an increment size (s) of 2. As the number box clearly demonstrates, a difference of 20 should be present. This means the algebraic equation should be:
Difference: (9x5x2x2)-(9x5x2)-(9x5x2)+(9x5)= 45
Therefore, the formula gained has been proven to be correct and valid for calculating the difference.
Increment Size: 4
Experimental number box:
Algebraic box:
Simplifies to:
Stage A: Top left number x Bottom right number
= n(n+4w+4gh-4g-4)
= n2+4nw+4ghn-4gn-4n
Stage B: Bottom left number x Top right number
= (n+4gh-4g)(n+4w-4)
= n2+4nw-4n+4ghn+16ghw-16gh-4gn-16gw+16g
= n2+4nw+4ghn-4gn-4n+16ghw-16gh-16gw+16g
Stage B – Stage A: (n2+4nw+4ghn-4gn-4n+16ghw-16gh-16gw+16g)-(n2+4nw+4ghn-4gn-4n)
= 16ghw-16gh-16gw+16g
When finding the general formula for any number (n) any height (h), any width (w), and with any gird size, both answers begin with the equation n2+4nw+4ghn-4gn-4n, which signifies that they can be manipulated easily. Because the second answer has +16ghw-16gh-16gw+16g at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 16ghw-16gh-16gw+16g will always be present.
Testing:
Using the experimental number box above, it is possible to prove that the formula works, and is correct. The box had a height and width of 2x2, and was based on a 5x5 grid, with an increment size (s) of 2. As the number box clearly demonstrates, a difference of 20 should be present. This means the algebraic equation should be:
Difference: (16x5x2x2)-(16x5x2)-(16x5x2)+(16x5)= 80
Therefore, the formula gained has been proven to be correct and valid for calculating the difference.
Summary Boxes and Increment Size of ‘s’
Increment Size (s): 2
Algebraic Summary Box:
Overall difference calculation: 4ghw-4gh-4gw+4g
Increment Size (s): 3
Algebraic Summary Box:
Overall difference calculation: 9ghw-9gh-9gw+9g
Increment Size (s): 4
Algebraic Summary Box:
Overall difference calculation: 16ghw-16gh-16gw+16g
Any Increment of ‘s’
As can be observed from the examples above, the algebraic boxed and the overall difference formulae show a distinct trend. In order to prove that this trend is exhibited in all cases, however, it is necessary to conduct some further algebraic research.
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.
Which through simple algebraic process can simplify into:
Stage A: Top left number x Bottom right number
= n(n+sw+ghs-gs-s)
= n2+nsw+ghns-gns-ns
Stage B: Bottom left number x Top right number
= (n+ghs-gs)(n+sw-s)
= n2+nsw-ns+ghns+ghs2w-ghs2-gns+gs2w+gs2
= n2+nsw+ghns-gns-ns+ghs2w-ghs2+gs2w+gs2
Stage B – Stage A: (n2+nsw+ghns-gns-ns+ghs2w-ghs2+gs2w+gs2)-(n2+nsw+ghns-gns-ns)
= ghs2w-ghs2+gs2w+gs2
= s2ghw-s2gh+s2gw+s2g
When finding the general formula for any number (n) any height (h), any width (w), and with any gird size, both answers begin with the equation n2+nsw+ghns-gns-ns, which signifies that they can be manipulated easily. Because the second answer has +ghs2w-ghs2+gs2w+gs2 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of s2ghw-s2gh+s2gw+s2g will always be present.
Difference Relationships in a 10x10x10 Cube
Top Face (TF)
Bottom Face (BF)
Difference in Cube:
Stage A: TF Top Left x BF Bottom Right
Stage B: TF Bottom Left x BF Top Right
Stage B - Stage A: Difference
Difference in Cube: Stage A: = 1 x 1000
= 1000
Stage B: = 91 x 910
= 82810
Difference in Cube: Stage B- Stage A: 81810
Algebra Faces
Due to the shape in question having equal height, width and breadth (depth), a universal algebraic term (e.g. ‘x’) could be used to replace the variables ‘h’, ‘w’ and ‘b’. Instead, in order to ensure the grid could be used to calculate a cube or cuboid of any height, width and depth, with any step size, the original terms are used.
n: The top left number of the front face grid. This will remain constant.
w: The width of the cube in question.
h: The height of the cube in question.
d: The depth of the cube in question.
g: The overall gridsize contained in the cube.
s: The increment (step) size between numbers in the grid.
Top Face (TF)
Bottom Face (BF)
Difference in Cube:
Stage A: (TF) Top left number x (BF) Bottom right number
= n(n+100d-100+sw+ghs-gs-s)
= n2+100dn+100n+nsw+ghns-gns-ns
Stage B: (TF) Bottom left number x (BF) Top right number
= (n+ghs-gs)( n+100d-100+sw-s)
= n2+100dn-100n+nsw-ns+ghsn+100dghs-100ghs+wghs2-ghs2-nsg-100dsg+100sg-gws2+gs2
= n2+100dn-100n+nsw+ghsn-nsg-ns+100dghs-100ghs+wghs2-ghs2-100dsg+100sg-gws2+gs2
Stage B – Stage A: (n2+100dn-100n+nsw+ghsn-nsg-ns+100dghs-100ghs+wghs2-ghs2-100dsg+100sg-gws2+gs2) - (n2+100dn+100n+nsw+ghns-gns-ns)
= 100dghs-100ghs-100dgs+100gs+ghws2-ghs2-gws2+gs2
When finding the general formula for 10x10x10 cube dimensions, both answers begin with the equation n2+100dn+100n+nsw+ghns-gns-ns, which signifies that they can be manipulated easily. Because the second answer has +100dghs-100ghs-100dgs+100gs+ghws2-ghs2-gws2+gs2 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 100dghs-100ghs-100dgs+100gs+ghws2-ghs2-gws2+gs2 will always be present.
Difference Relationships in a 5x5x5 Cube
Top Face (TF)
Bottom Face (BF)
Difference in Cube:
Stage A: TF Top Left x BF Bottom Right
Stage B: TF Bottom Left x BF Top Right
Stage B - Stage A: Difference
Difference in Cube: Stage A: = 1 x 125
= 125
Stage B: = 21 x 105
= 2205
Difference in Cube: Stage B- Stage A: 2080
Algebra Faces
Due to the shape in question having equal height, width and breadth (depth), a universal algebraic term (e.g. ‘x’) could be used to replace the variables ‘h’, ‘w’ and ‘b’. Instead, in order to ensure the grid could be used to calculate a cube or cuboid of any height, width and depth, with any step size, the original terms are used.
n: The top left number of the front face grid. This will remain constant.
w: The width of the cube in question.
h: The height of the cube in question.
d: The depth of the cube in question.
g: The overall gridsize contained in the cube.
s: The increment (step) size between numbers in the grid.
Top Face (TF)
Bottom Face (BF)
Difference in Cube:
Stage A: (TF) Top left number x (BF) Bottom right number
= n(n+25d-25+sw+ghs-gs-s)
= n2+25dn-25n+nsw+ghns-gns-ns
Stage B: (TF) Bottom left number x (BF) Top right number
= (n+ghs-gs)(n+25d-25+sw-s)
= n2+25dn-25n+nsw-ns+ghns+25dghs-25ghs+s2ghw-ghs2-gns-25dgs+25gs-s2gw+gs2
= n2+25dn-25n+nsw+ghns-gns-ns+25dghs-25ghs+s2ghw-ghs2-25dgs+25gs-s2gw+gs2
Stage B – Stage A: (n2+25dn-25n+nsw+ghns-gns-ns+25dghs-25ghs+s2ghw-ghs2-25dgs+25gs-s2gw+gs2) - (n2+25dn-25n+nsw+ghns-gns-ns)
= 25dghs-25ghs-25dgs+25gs+ghws2-ghs2-gws2+gs2
When finding the general formula for 5x5x5 cube dimensions, both answers begin with the equation n2+25dn-25n+nsw+ghns-gns-ns, which signifies that they can be manipulated easily. Because the second answer has +25dghs-25ghs-25dgs+25gs+ghws2-ghs2-gws2+gs2 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of 25dghs-25ghs-25dgs+25gs+ghws2-ghs2-gws2+gs2 will always be present.
Summary Boxes and Increment Size of ‘s’
Cube Size of 10x10x10
Top Face (TF)
Bottom Face (BF)
Overall Difference: 100dghs-100ghs-100dgs+100gs+ghws2-ghs2-gws2+gs2
Cube Size of 5x5x5
Top Face (TF)
Bottom Face (BF)
Overall Difference: 25dghs-25ghs-25dgs+25gs+ghws2-ghs2-gws2+gs2
Any Cube size of ‘c’
As can be observed from the examples above, the algebraic boxed and the overall difference formulae show a distinct trend. In order to prove that this trend is exhibited in all cases, however, it is necessary to conduct some further algebraic research.
To obtain the overall algebraic boxes, for any cube 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 cube/cuboid dimensions, by implementing the formulae below to find a general calculation and grid rectangles.
Top Face of Cube ‘c x c x c’
Bottom Face of Cube ‘c x c x c’
Difference in Cube:
Stage A: (TF) Top left number x (BF) Bottom right number
= n(n+c2d-c2+sw+ghs-gs-s)
= n2+c2dn-c2n+nsw+ghns-gns-ns
Stage B: (TF) Bottom left number x (BF) Top right number
= (n+ghs-gs)(n+c2d-c2+sw-s)
= n2+c2dn-c2n+nsw-ns+ghns+c2dghs-c2ghs+s2ghw-ghs2-gns-c2dgs+c2gs-s2gw+gs2
= n2+c2dn-c2n+nsw+ghns-gns-ns+c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2
Stage B – Stage A: (n2+c2dn-c2n+nsw+ghns-gns-ns+c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2) - (n2+c2dn-c2n+nsw+ghns-gns-ns)
= c2dghs- c2ghs- c2dgs+ c2gs+ghws2-ghs2-gws2+gs2
When finding the general formula for any cube dimensions, both answers begin with the equation n2+c2dn-c2n+nsw+ghns-gns-ns, which signifies that they can be manipulated easily. Because the second answer has + c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2 at the end, it demonstrates that no matter what number is chosen to begin with (n), a difference of c2dghs-c2ghs+s2ghw-ghs2-c2dgs+c2gs-s2gw+gs2 will always be present.
Using this formula, it is possible to calculate the difference for any cube size, with any dimentions, increment size and gridsize.