For example,
For the 2x2 square.
n = any number
(n+1)(n+10) - n(n+11)
n²+10n+1n +10-n² -11n
n²+11n+10 - n²-11n = 10
This proves that for any number in a 2x2 square the grid difference will always be 10.
I would now like to look at the difference in a 3x3 square on a 10x10 grid.
My prediction is that once I find out the distance for the first 3x3 square then the rest will be the same I will give 3 examples.
a)
If I multiply the top right number with the bottom left number and do the same with the top left and bottom right number I will be able to find the product.
1x23= 23
3x21= 63
To work out the product I will subtract 23 from 63.
63 – 23 = 40
The difference is 40.
b)
74 x 96 = 7104
76 x 94 = 7144
7144 – 7104 = 40
The difference is 40.
c)
30 x 52 = 1560
32 x 50 = 1600
1600 – 1560 = 40
The difference is again 40.
All the examples differences were 40, with a 3x3 square in a 10x10 grid the difference is 40.
I will record the results on a table.
I will again use algebra to prove this.
For a 3x3 square:
n = any number
(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.
This formula could be used for any 3x3 square in a 10x10 grid.
I will now try with a 4x4 square in a 10 x 10 grid and my prediction is the same as above, the difference will be the same for all 3 4x4 squares.
a)
I will multiply the top right number and the bottom left and the top left and bottom right.
1x34 = 34
4 x31 =124
To find the difference I will subtract the numbers.
124 – 34 = 90
The difference is 90.
b)
42 x 75 = 3150
45 x 72 = 3240
3240 – 3150 = 90
The difference is 90.
c)
24 x 57 = 1368
27 x 54 = 1458
1458 – 1368 = 90
The difference is again 90.
The table below shows this.
I will again prove this with algebraically.
(n + 3)(n + 30) – n(n+33)
n² +30n+3n+90-n² -33n
n²+33n+90 - n²- 33n = 90
My prediction was right any number in a 4x4 square within a 10x10 grid has a difference of 90.
I drew a table to show my results so far.
From the product differences I noticed a pattern emerging so I took the noughts away from the numbers, I would be left with square numbers i.e. 1, 4, and 9.
The next square number would be 16 so if I added the naught I would get 160, so I predict that a 5x5 square will have the difference of 160.
I will now test my prediction.
a)
.
1 x 45 = 45
5 x 41 = 205
205 – 45 = 160
The difference is 160 as predicted.
b)
50 x 94 = 4700
54 x 90 = 4860
4860 – 4700 = 160
My prediction was right I will now prove this will algebra.
(n + 4)(n + 40) – n(n + 44)
n² +40n+4n+160-n² -44n
n² + 44n+160 - n² - 44n = 160
With this you can see the difference for any number in a 5x5 square on a 10x10 grid is 160, which proves my prediction.
Using the same method I predict that the 6x6 square on a 10x10 grid difference will be 250, because if I took the nought away it would be the next square number in the sequence, 25.
1, 4, 9, 16, and 25
I will test my theory using the same method I have used throughout.
a)
1 x 56 = 56
6 x 51 = 306
306 – 56 = 250
My prediction was correct as the difference is 250, I will do another example to make sure.
b)
45 x 100 = 4500
50 x 95 = 4750
4750 – 4500 = 250
Again the difference is 250 so my prediction was correct I will now use algebra to prove this.
For a 6x6 square:
(n +5)(n + 50) – n(n +55)
n² + 50n +5n +50- n² – 55n
n² +55n+50 – n² – 55n = 250
This is the formula for any square on a 10x10 grid:
10(n – 1)(n - 1)
n stands for the height and the width of the window.
This can be simplified to:
10(n – 1)²
I then decided to see if I can find any other pattern so I looked at the difference of my results.
2 x 2 3 x 3 4 x 4 5 x 5 6x6
10 40 90 160 250
\ / \ / \ / \ /
30 50 70 90
I then noticed a pattern so I found the difference of the new results.
30 50 70 90
\ / \ / \ /
20 20 20
The 3rd difference is always 20.
Conclusion
For any square in a 10x10 grid I have found out 3 things.
- If you take the nought of the difference you get square numbers.
-
The 3rd difference is always 20.
- The formula for any square is 10(n – 1)²
I would now like to see what changing the square to a rectangle would do to the product and find the difference and a formula.
I will first try a 2x3 rectangle in a 10 x 10 grid, I will use the same method of multiplying the top right number and the bottom left number and multiply the top left number by the bottom right and then find the difference by taking away the products.
a)
1 x 13 = 13
3 x 11 = 33
33 – 13 = 20
The difference is 20.
I will try one more time.
b)
77 x 89 = 6853
79 x 87 = 6873
87 – 79 = 20
I will show this algebraically.
n(n + 12) – (n + 2)(n + 10)
(n² + 12n) – (n² + 12n + 20) = 20
This proves the difference is 20.
I will know look at a 4x2 rectangle on a 10x10 grid and find the difference.
a)
44 x 57 = 2508
47 x 54 = 2538
2538 – 2508 = 30
The difference is 30.
b)
87 x 100 = 8700
90 x 97 = 8730
8730 – 8700 = 30
The difference is 30 again.
I will now prove this algebraically.
n(n + 31) = (n2) + (31n)
n + 1 x n + 30 = n2 + 30n + 1n + 30
n2 + 30n + 1n + 30 – n2 + 31n = 30
I will draw a table to show my results.
From my results I can see an increase of 10 each time, so the 2nd difference is 10, I can now predict that a 2x5 rectangle will have a difference of 40, I will test my prediction.
a)
1 x 15 = 15
5 x 11 = 55
55 – 15 = 40
The difference is 40.
b)
65 x 79 = 5135
69 x 75 = 5175
5175 – 5135 = 40
My prediction was correct the difference was 40, I will prove this algebraically.
n(n + 14) – (n + 4)(n + 10) =
(n² + 14n) – (n² +14n + 40) = 40.
This proves the difference is 40 and the number has gone up by 10 again so you can then predict 2x6 difference of 60 and 2x7 difference of 70.
I am now going to find a formula.
I tried
(n – 1) X 10
n =the width of the box.
7 x 2
7 – 1 = 6
6 x 10 = 60
I predict using my formula that the difference of this 7 x 2 square will be 60.
22 x 38 = 836
28 x 32 = 896
896 – 836 = 60
My prediction was correct.
I could use this formula to work out any size of rectangle.
I.e. 18 X 2 = 170
Conclusion
I found 3 things out about rectangles in a 10x10 grid.
- The rectangles I used differences increased by 10 every time.
-
The 2nd difference is always 10.
- The formula for any rectangle is (n – 1) X 10
Evaluation
By doing lots of experiments and recording my results I found different patterns in a square and a rectangle by finding the difference of the corner numbers, I also used algebra to prove the differences. I found 2 formulas which can be used in any size square and any size rectangle in a 10x10 grid.
I could have expanded my coursework if I had more time and looked at different grid sizes and maybe different shapes like triangles or t shape patterns.
This coursework has helped me to understand number patterns and has developed my ability to introduce algebra when trying to spot patterns and I am happy with my results.