 |
 |
 |
 |
 |
CourseworkHelp
:Pattens in Squares
The aim of the investigation is to find differences of small n x n squares in 10 x 10 square and then to see if there is any rule or pattern which connects the size of square chosen and the difference.
In order to find the difference of n x n square first step is to take nxn square and then multiply the corners diagonally. For example take 2x2 square and multiply its corners diagonally.
6 7 6 x 16 = 96
15 16 7 x 15 = 105
After doing that minus the small answer from the bigger answer. This gives the difference of 2x2 square.
105 – 96 = 9
After doing this I will check my answer using the nxn formula
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
Now I will start the investigation
There are a large number of possible starting points in 10x10 square but I will start with the smallest square possible and make my way up so that the investigation does not get too complicated.
2x2 square
n n+1 n(n+11) - [(n+1) (n+10)]
1 2
n2 +11n - n2 +10n + n+10
11 12
n+10 n+11 n2 +11n – n2 +11n +10
11 x 2 = 22 10
'font-size:14.0pt; '>12 x 1 = 12
'font-size:14.0pt; '>22 – 12 = 10
In both ways the difference is 10
3x3 square
n n+2 n(n+22) – [(n+2) (n+20)]
1 2 3
11 12 13 n2+22n - n2+20n +2n +40
21 22 23
n+20 n+22 n2 +22n - n2 +22n +40
21 x 3 = 63 40
23 x 1 = 23
63 – 23 = 40
In both ways the difference is 40
4x4 square
n n+2 n(n+33) - [(n+3) (n+30)]
1 2 3 4
11 12 13 14 n2+33n - n2+30n +3n +90
21 22 23 24
31 32 33 34 n2+33n - n2+33n + 90
n+30 n+33 10
31 x 4 = 124 In both ways the difference is 90
34 x 1 = 31
124 – 34 = 90
5x5 square
n n+4 n(n+44) - [(n+4) (n+40)]
1 2 3 4 5
11 12 13 14 15 n2+44n - n2+40n + 4n +160
21 22 23 24 25
31 32 33 34 35 n2+44n - n2+44n +160
41 42 43 44 45
n+40 n+44 160
41 x 5 = 205
45 x 1 = 45
205 – 45 = 160
In both ways the difference is 160
6x6 square
n n+5 n(n+55) - [(n+5) (n+50)
1 2 3 4 5 6
11 12 13 14 15 16 n2+55n - n2+50n + 5n + 250
21 22 23 24 25 26
31 32 33 34 35 36 n2+55n - n2+55n + 250
41 42 43 44 45 46
51 52 53 54 55 56 250
n+50 n+55
51 x 6 = 306
56 x 1 = 56
306 – 56 = 250
In both ways the difference is 250
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
In a 5x5 square the differences of smaller nxn squares
If you look closely at the results in the table , you will notice that if you multiply nxn number not square and then multiply the answer by 5, this gives you the answer of the next nxn square. For example if you multiply 2 by 2 and then multiply the answer by 5, the answer of this is 20 and 20 is the difference of 3x3 square. So the formula is :-
(n x n) x5 = the difference of next nxn square
'font-size:14.0pt; '>Same thing happens with all the nxn squares.so if I minus 1 from both n’s I think this would give the proper formula for nxn squares. Therefore the new formula is :-
(n-1) (n-1) x5 = the difference of nxn square
lets try the new formula with 4x4 square and see if the answer is 45.
(4-1) (4-1) x5
(16-4-4+1) x5
9 x 5 = 45
yes the new formula is right and now I can predict the differences of any nxn square which is within larger 5x5 square.
In a 6x6 square the differences of smaller nxn squares
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
If you look closely at the results in the table , you will notice that if you multiply nxn number not square and then multiply the answer by 6, this gives you the answer of the next nxn square. For example if you multiply 2 by 2 and then multiply the answer by 6, the answer of this is 24 and 24 is the difference of 3x3 square. So the formula is :-
(n x n) x6 = the difference of next nxn square
'font-size:14.0pt; '>Same thing happens with all the nxn squares.so if I minus 1 from both n’s I think this would give the proper formula for nxn squares. Therefore the new formula is :-
(n-1) (n-1) x6 = the difference of nxn square
lets try the new formula with 4x4 square and see if the answer is 54.
(4-1) (4-1) x6
(16-4-4+1) x6
9 x 6 = 54
yes the new formula is right and now I can predict the differences of any nxn square which is within larger 6x6 square.
In a 7x7 square the differences of smaller nxn squares
|
|
|
|
|
|
|
|
|
|
|
|
|
4x7 |
|
|
|
|
9x7 |
|
|
|
|
16x7 |
|
|
|
|
25x7 |
|
|
|
|
36x7 |
|
|
|
|
? |
|
If you look closely at the results in the table , you will notice that if you multiply nxn number not square and then multiply the answer by 7, this gives you the answer of the next nxn square. For example if you multiply 2 by 2 and then multiply the answer by 7, the answer of this is 28 and 28 is the difference of 3x3 square. So the formula is :-
(n x n) x7 = the difference of next nxn square
'font-size:14.0pt; '>Same thing happens with all the nxn squares.so if I minus 1 from both n’s I think this would give the proper formula for nxn squares. Therefore the new formula is :-
(n-1) (n-1) x7 = the difference of nxn square
lets try the new formula with 4x4 square and see if the answer is 63.
(4-1) (4-1) x7
(16-4-4+1) x7
9 x 7 = 63
yes the new formula is right and now I can predict the differences of any nxn square which is within larger 7X7 square.
In a 8X8 square the differences of smaller nxn squares
|
|
|
BREAK IT DOWN |
BREAK IT DOWN MORE |
|
|
2 X 2 |
8 |
|||
|
3 X 3 |
32 |
|
2 x 2 x 8 |
|
|
4 X 4 |
72 |
9 X 8 |
3 x 3 x 8 |
|
|
5 X 5 |
128 |
16 X 8 |
4 x 4 x 8 |
|
|
6 X 6 |
200 |
25 X 8 |
5 x 5 x 8 |
|
|
7 X 7 |
288 |
36 X 8 |
6 X 6 X 8 |
|
|
8 X 8 |
512 |
49 X 8 |
|
|
|
N X N |
? |
? |
? |
If you look closely at the results in the table , you will notice that if you multiply nxn number not square and then multiply the answer by 8, this gives you the answer of the next nxn square. For example if you multiply 2 by 2 and then multiply the answer by 8, the answer of this is 32 and 32 is the difference of 3x3 square. So the formula is :-
(n x n) x8 = the difference of next nxn square
'font-size:14.0pt; '>Same thing happens with all the nxn squares.so if I minus 1 from both n’s I think this would give the proper formula for nxn squares. Therefore the new formula is :-
(n-1) (n-1) x8 = the difference of nxn square
lets try the new formula with 4x4 square and see if the answer is 72.
(4-1) (4-1) x8
(16-4-4+1) x8
9 x 8 = 72
yes the new formula is right and now I can predict the differences of any nxn square which is within larger 8X8 square.
In a 9X9 square the differences of smaller nxn squares
|
N X N |
DIFFERENCE |
BREAK IT DOWN |
BREAK IT DOWN MORE |
|
|
2 X 2 |
9 |
|||
|
3 X 3 |
36 |
4 x 9 |
2 x 2 x 9 |
|
|
4 X 4 |
81 |
9 x 9 |
3 x 3 x 9 |
|
|
5 X 5 |
144 |
16 x 9 |
4 x 4 x 9 |
|
|
6 X 6 |
225 |
25 x 9 |
5 x 5 x 9 |
|
|
7 X 7 |
324 |
36 x 9 |
6 X 6 X 9 |
|
|
8 X 8 |
441 |
49 x 9 |
7 X 7 X 9 |
|
|
9 X 9 |
576 |
64 x 9 |
8 X 8 X 9 |
|
|
N X N |
? |
? |
? |
If you look closely at the results in the table , you will notice that if you multiply nxn number not square and then multiply the answer by 9, this gives you the answer of the next nxn square. For example if you multiply 2 by 2 and then multiply the answer by 9, the answer of this is 36 and 36 is the difference of 3x3 square. So the formula is :-
(n x n) x9 = the difference of next nxn square
'font-size:14.0pt; '>Same thing happens with all the nxn squares.so if I minus 1 from both n’s I think this would give the proper formula for nxn squares. Therefore the new formula is :-
(n-1) (n-1) x9 = the difference of nxn square
lets try the new formula with 4x4 square and see if the answer is 81.
(4-1) (4-1) x9
(16-4-4+1) x9
9 x 9 = 81
yes the new formula is right and now I can predict the differences of any nxn square which is within larger 8X8 square.
Conclusion
After doing all this work I have found out that to get the difference of any nxn square you just have to change the inside numbers of brackets to nxn square you want to find out and the outside numbers to nxn square you want to find out your difference from. So if you want to find out the difference of smaller 5x5 square from 20x20 square, the formula changes to the following :- Inside numbers outside number
(5 - 1) (5 - 1) x20
Therefore if you want to find out the difference of 10x10 square in 20x20 square the formula will be :-
Inside number outside number
(10 - 1) (10 - 1) x20
If you still can't find any of the coursework that you are looking for, click here to look at over 30,000 GCSE, A-Level and University Level essays on Coursework.Info.