31 32 33 23x41 = 943 This calculation was worked out by: 943-903 = 40
41 42 43
From the calculations of the 3x3 square grid the difference was 40.
The algebraic formula for the 3x3 square grid is the following:
(n + 20) (n + 2) – n(n + 22)
n² +2n+20n+40-n² -22n
n²+22n+40 - n²- 22n = 40
From this algebraic calculation it shows that the answer is 40, which is the difference of the 3x3 square grid.
This formula could be used for any 3x3 square grid in a 10x10 grid.
After calculating the 2x2 and 3x3 square grids, I will now calculate the 4x4 square grid and calculate the difference with the same rule in the 10x10 master grid.
This is the calculation for the 4x4 square grid. I will do two calculations to prove the differences of the products are correct.
7 8 9 10 7x40 = 280 The difference for this 4x4 square grid is 90.
17 18 19 20 10x37 = 370 This calculation was worked out by: 370-280 = 90
27 28 29 30
37 38 39 40
I will do another 4x4 square grid to prove that this is correct.
45 46 47 48 45x78 = 3510 The difference for this 4x4 square grid is 90.
55 56 57 58 48x75 = 3600 This calculation was worked out by: 3600-3510 = 90
65 66 67 68
75 76 77 78
From the calculations of the 4x4 square grids above the difference of the products is 90.
The algebraic calculation for the 4x4 square grid is the following:
(n + 3)(n + 30) – n(n+33)
n² +30n+3n+90-n² -33n
n²+33n+90 - n²- 33n = 90
From this calculation it shows that the answer is the correct, hence the 4x4 square grid product difference is 90.
This algebraic formula proves that it can be used for any 4x4 square grid in a 10x10 master grid.
I will now calculate the products of the 5x5 square grid to find the difference.
35 36 37 38 39 35x79 = 2765 The difference for this 5x5 square grid is 160.
45 46 47 48 49 39x75 = 2925 This calculation was worked out by:
55 56 57 58 59 2925-2765 = 160
65 66 67 68 69
75 76 77 78 79
55 56 57 58 59
65 66 67 68 69 55x99 = 5445 The difference for this 5x5 square grid is 160.
75 56 77 78 79 59x95 = 5605 This calculation was worked out by:
85 86 87 88 89 5605-5445 = 160
95 96 97 98 99
From the above two calculation for the 5x5 square grid the difference is 160.
The algebraic calculation for the 5x5 square grid is the following:
(n + 4)(n + 40) – n(n + 44)
n² +40n+4n+160-n² -44n
n² + 44n+160 - n² - 44n =160
From this calculation it shows that the answer is the correct, hence the 5x5 square grid product difference is 160.
This algebraic formula proves that it can be used for any 5x5 square grid in a 10x10 master grid.
By doing the 2x2, 3x3, 4x4 and the 5x5 square grids my hypothesis is proved correct, as when I calculated the top left and bottom right numbers the calculation was less than the top right and bottom left numbers.
I will show the different sized square grids and their product differences in a table.
Table of results
The above results were obtained from the calculations and the rule mentioned in the above investigation.
From the product differences I noticed that if I took the noughts away from the numbers then I would be left with square numbers i.e. 1, 4, 9, 16.
From the square numbers I predict that the 6x6 number grid product difference will be 250, as if I took the nought away it would the next square number in the sequence.
I will now work out the difference of each product to find a pattern of the product differences of the five square grids in the 10x10 master grid. This is achieved by the following calculations, through subtracting the 3x3 product difference which is 40 with the 2x2 product difference which is 10, then the 4x4 product difference which is, 90 from the 3x3 i.e. 40 and the 5x5 product difference i.e.160 from the 4x4 i.e. 90.
This method is shown by:
40 (3x3) – 10 (2x2) = 30 The difference of each of these sequences is 20.
90 (4x4) – 40 (3x3) = 50 I calculated this by: 50-30 = 20.
160 (5x5) – 90 (4x4) = 70 70-50 = 20
From this above calculation I can then predict that the 6x6 square grid will also have the difference of 20. Also I can predict from the above calculations that the product difference of the 6x6 square grid is going to be 250. This product difference was worked out by: 160 (5x5) + 90 (4x4) = 250. Therefore the 6x6 square grid will have a product difference of 250.
This follows on from the above calculations where if I subtracted 250 (6x6) -160 (5x5) = 90. I will now subtract 90 – 70 which equals 20. From this the product difference for the 6x6 square grid is correct.
The above algebraic calculation methods can be used to work out any size square grid in a 10x10 grid.
I will now investigate when I do a 2x2 square grid in a 6x6 master grid using the same rule, when calculating a square grid in a 10x10 master grid.
I will calculate the 2x2 square grid on a 6x6 master grid.
2 3 2x9 = 18 This is a difference of 6.
8 9 3x8 = 24 I worked this out by: 24-18 = 6
From this calculation it has been found that the difference of 2x2 square size grid in the master grid 6x6 is 6.
From this calculation I can now show it in algebraic terms.
(n +1)(n +6) – n (n +7)
n² +6n+1n+7-n² -7n
n² + 7n + 6 - n²-7n = 6
I will now do a 3x3 square grid in the master grid of 6x6 to calculate the product difference.
3 4 5 3x17 = 51 The difference is 24.
9 10 11 5x15 = 75 This was worked out by: 75-51 = 24
15 16 17
The product difference of this 3x3 square grid is 24.
I can use this calculation to show it in algebraic terms.
(n +20)(n +2) – n (n +22)
n² +2n+20n+20-n² -22n
n² + 22n + 24 - n²-22n = 24
I will now calculate a 4x4 square grid on a 6x6 master grid to find the product difference.
2 3 4 5 2x20 = 40 The difference is 45.
7 8 9 10 5x17 = 85 This was worked out by: 85-40 = 45
12 13 14 15
17 18 19 20
I will now do the algebraic calculations for the 4x4 square grid.
(n + 30)(n +3 )-n (n + 33)
n² +3n+30n+30-n² -33n
n² + 33n + 45- n²-33n = 45
I will now test the 5x5 square grid in a 6x6 master grid.
2 3 4 5 6 2x21 = 42 The difference is 60.
7 8 9 10 11 6x17 = 102 I worked this out by: 102-42 = 60
12 13 14 15 16
17 18 19 20 21
The algebraic calculation for the 5x5 square grid is the following:
(n + 40)(n + 4) –n (n + 44)
n² +4n+40n+40-n² -44n
n² + 44n + 60 - n²- 44n = 60
From the algebra I have used for the master grids 10x10 and 6x6 the difference for a 10x10 grid is 10(n-1), and the 6x6 grid difference is 6(n-1).
Table of Results
From this table I obtained the results by calculating different sized square grids in a 6x6 master grid. As you can see the 2x2, 3x3 and 5x5 product differences all multiply by 6 i.e. 1x6=6 (2x2 grid), 4x6=24 (3x3), 10x6=60 (5x5) except the 4x4 grid as when multiplied by 6 the answer is 7.5.
I will now investigate even further and do calculations for rectangles supposed to squares using the same rule. This will then show the relations between the product differences of the squares and rectangles.
Prediction
I predict that as the rectangle sizes goes up i.e. 2x5, 3x5 and 4x5 the product difference will increase. This would be because the rectangle grid sizes are increasing and therefore would have a higher number in the bottom left and right corner of the rectangle to multiply by when calculating i.e. in a 2x5 grid I will draw it like this:
1 2 3 4 5
11 12 13 14 15
The calculation will be 1x15=15 and 5x11 =55.
This will compare to a 4x5 grid like: 1 2 3 4 5
11 12 13 14 15
21 22 23 24 25
31 32 33 34 35
And the calculation for this would be 1x35= 35 and 5x31= 155.
This justifies my prediction by showing workings of my theory by the size grids that I will be using. The following calculations will be able to justify and possibly prove my prediction correct or incorrect.
I will investigate the 2x5 rectangle on a 10x10 grid.
I will present this in a table showing the algebraic formula for the 2x5 rectangle.
1 2 3 4 5 1x15 = 15 The difference is 40.
11 12 13 14 15 5x11 = 55 I worked this out by: 55-15 = 40
I will now do the algebra for the 2x5 rectangle.
(n + 10) (n + 2) – n (n + 12)
n²+2n+10n+20-n² -12
n² +12n + 40- n²-12n = 40
From calculating the algebraic formula for the rectangle 2x5 in a 10x10 master grid there is a difference of 40.
I will do the algebra to calculate the difference for a 3x5 rectangle in a 10x10 master grid.
1 2 3 4 5 1x25 = 25 The difference is 80
11 12 13 14 15 5x21 = 105 I worked this out by: 105-25 = 80
21 22 23 24 25
I will now calculate the algebra for the 3x5 rectangle.
(n + 20)(n + 3) – n(n + 23)
n²+3n+20n+80-n² -23
n² + 23n + 80- n²-23n = 80
From calculating the difference in the 3x5 rectangle there is a difference of 80.
I will now investigate for a 4x5 rectangle in a 10x10 grid.
1 2 3 4 5 1x35 = 35 The difference is 120
11 12 13 14 15 5x31 = 155 I worked this out by: 155-35 = 120
21 22 23 24 25
31 32 33 34 35
I will now do the algebra for the 4x5 rectangle.
(n + 30)(n +4) –n(n+ 34)
n² +4n+30n+120-n² -34
n² + 34n + 120- n²-34n = 120
This shows that the difference of the 4x5 rectangle in the 10x10 master grid is 120.
The table below shows the product difference of different sized rectangular grids in a 10x10 master grid.
I will now compare the product difference with the different sized square grids in the 10x10 master grid investigation.
I will only compare three sets of results for the square product difference. This is because I only investigated three rectangles so this will provide for a fair comparison.
The two tables above show my results/findings from the investigations of different sized grids in a 10x10 master grid. The common findings from the tables show that as the size of the rectangle and square increases the product difference for them both increases.
Conclusion
By doing the investigations I found out from the number grid investigation that my hypothesis was correct as when I calculated the top left and bottom right number the product was less than the top right and bottom left number. This was proven correct throughout the calculations in the investigation.
I decided to carry out two examples for each individual size grid in a 10x10 master gird, as I wanted to ensure that I was conducting the calculations correctly. It therefore gave accurate and reliable results. Although I did not do two examples for the individual sized grids in the 6x6 master grid or for the rectangle size grids. This was because my method procedure was correct throughout the calculations for the squares in the 10x10 master grids so I did not feel it was necessary.
From the results/findings in the 6x6 master grid table it shows that the product differences of the 2x2, 3x3 and 5x5 square grids are multiples of six but not the 4x4 square grid product difference. The calculation for the 4x4 product difference is correct and as noted doesn’t follow the same pattern as the other square grid sizes.
The rectangles product differences increases by 40 each time as the grid size goes up, which proves my prediction correct. I stated in my prediction that as ‘the rectangle sizes goes up i.e. 2x5, 3x5 and 4x5 the product difference will increase. This would be because the rectangle grid sizes are increasing and therefore would have a higher number in the bottom left and right corner of the rectangle to multiply by when calculating’ From the prediction I showed an examples of an 2x5 and a 4x5 grid to justify my prediction.
