Number Grid Investigation
Number Grid Investigation
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0
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00
This is a simple 10 x 10 number grid.
I am now going to investigate 2x2 squares on this 10x10 number grid.
A
B
C
D
A, B, C and D all represent numbers in a 2x2 square, I am going to multiple the 2 diagonal numbers together and find the difference of their products.
I will pick a 2x2 number square at random and investigate it:
23 24 23x34= 782 Difference = 10
33 34 24x33= 792
The difference between the two products is 10 I will now choose another 2x 2
Square to see if this pattern follows.
47 48 47x58= 2726 Difference = 10
57 58 57x48= 2736
The difference in both these 2x2 squares is 10 a pattern seems to be emerging, I will now try another 2x2 square to convince me this pattern works throughout the number grid.
86 87 86x97= 8342 Difference = 10
96 97 87x96= 8352
I am now convinced that the difference between the two diagonal products in a 2x2 number square is 10.
I will now prove my results using algebra:
n
n+1
n+10
n+11
n x n+1 = n2 + 11n Difference = 10
n+1 x n+10 = n2 + 11n + 10
All I have done is substituted the first number of my 2x2 grid in the top left corner with the letter n, I have used the same process as before and I have found that the difference is always 10.
I will now go on to investigate 3 x 3 number squares, I predict that the difference will be 20 because in the in a 2 x 2 its 10 so I think it will go up another 10 for the 3 x 3 square.
For the 3 x 3 grid I will use the same process as before...using the 2 diagonal corners to find 2 products then calculating the difference between them.
2
3
4
22
23
24
32
33
34
2 x 34 = 408 Difference = 40
4 x 32 = 448
The difference is 40 I will now check another 3 x 3 grid to see if the pattern continues:
27
28
29
37
38
39
47
48
49
27 x 49= 1323 Difference = 40
29 x 47= 1363
A pattern seems to be forming the difference is 40 on both of these 3x3 squares; however I will try another square to convince me.
4
5
6
4
5
6
24
25
26
4 x 26= 104 Difference = 40
6 x 24= 144
After seeing sufficient evidence that the difference in a 3 x 3 number square is 40, I will now prove this by algebra.
n
n+1
n+2
n+10
n+11
n+12
n+20
n+21
n+22
n x n+22 = n2 + 22n Difference = 40
n+2 x n+20 = n2 + 22n + 40
I have done exactly the same process as before, by substituting in the letter n to prove that the differences between the 2 pairs of diagonal products have a difference of 40.
I will now go on to investigate 4 x 4 squares on the number grid; I will again be multiplying the 2 pairs of diagonal corners together then finding the difference between them. I predict the difference will be 70, I think this because there was 30 added to the 2 x 2 difference to get the 3 x 3 difference so I feel another 30 will be added to get the 4 x 4 difference.
3
4
5
6
3
4
5
6
23
24
25
26
33
34
35
36
3 x 36 = 108 Difference = 90
6 x 33 = 198
My prediction was incorrect the difference is 90 in this 4 x 4 square; I will now do this with a different set of numbers.
32
33
34
35
42
43
44
45
52
53
54
55
62
63
64
65
32 x 65 = 2080 Difference = 90
35 x 62 = 2170
The difference is again 90 I will now try one more 4 x 4 square to convince me the difference is always 90.
65
66
67
68
75
76
77
78
85
86
87
88
95
96
97
98
65 x 98 = 6370 Difference = 90
68 x 95 = 6460
As you can see the difference is 90, I am now convinced that the difference is always 90 in a 4 x 4 square on this number grid. I will prove this using algebra:
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
n x n+33 = n2 + 33n Difference 90
n+3 x n+30 = n2 + 33n + 90
This proves that any 4 x 4 square will have a difference of 90.
I will now go on to investigate 5 x 5 squares on the grid, using the same process I used for the 4 x 4, 3 x 3 etc. I predict that the difference will be 160 because so far it has been all the square numbers multiplied by 10 in ascending order.
2
3
4
5
1
2
3
4
5
21
22
23
24
25
31
32
33
34
35
41
42
43
44
45
x 45 = 45 Difference=160
5 x 41 = 205
My prediction was correct, the difference in this 5 x 5 square is 160 I will now try a different set of numbers in a 5 x 5 square.
22
23
24
25
26
32
33
34
35
36
42
43
44
45
46
52
53
54
55
56
62
63
64
65
66
22 x 66 = 1452 Difference= 160
26 x 62 = 1612
I am convinced that the difference between the two pairs of diagonal corners have a difference of 160 when multiplied together in a 5 x 5 square. I will now prove this using algebra.
n
n+1
n+2
n+3
n+4
n+10
n+11
n+12
n+13
n+14
n+20
n+21
n+22
n+23
n+24
n+30
n+31
n+32
n+33
n+34
n+40
n+41
n+42
n+43
n+44
n x n+44 = n2+44n Difference
n+4 x n+40= n2+44n+160 = 160
This proves that any 5 x 5 square will have a difference of 160 as I predicted.
I am now going to investigate 6 x 6 squares on the number grid; I will use the same process as before to work out the difference. I predict that the difference will be 250, again because the results have shown that it has been the square numbers multiplied by 10. The next square number is 25 so the difference should be 250.
4
5
6
7
8
9
24
25
26
27
28
29
34
35
37
37
38
39
44
45
47
47
48
49
54
55
56
57
58
59
64
65
66
67
68
69
4 x 69= 966 Difference =
19 x 64= 1216 250
The difference is 250 as I predicted, I will now do the same using a different set of numbers.
2
3
4
5
6
1
2
3
4
5
6
21
22
23
24
25
26
31
32
33
32
35
36
41
42
43
44
45
46
51
52
53
54
55
56
x 56 = 56 Difference =
6 x 51 = 306 250
Again the difference is 250 I am now convinced this works for every 6 x 6 square on the grid, I will now prove this by algebra.
n
n+1
n+2
n+3
n+4
n+5
n+10
n+11
n+12
n+13
n+14
n+15
n+20
n+21
n+22
n+23
n+24
n+25
n+30
n+31
n+32
n+33
n+34
n+35
n+40
n+41
n+42
n+43
n+44
n+45
n+50
n+51
n+52
n+53
n+54
n+55
n x n+55 = n2+55n Difference =
n+5 x n+50 =n2+55n+250 250
This proves that the difference in a 6 x 6 square ...
This is a preview of the whole essay
n
n+1
n+2
n+3
n+4
n+5
n+10
n+11
n+12
n+13
n+14
n+15
n+20
n+21
n+22
n+23
n+24
n+25
n+30
n+31
n+32
n+33
n+34
n+35
n+40
n+41
n+42
n+43
n+44
n+45
n+50
n+51
n+52
n+53
n+54
n+55
n x n+55 = n2+55n Difference =
n+5 x n+50 =n2+55n+250 250
This proves that the difference in a 6 x 6 square on the grid has a difference of 250.
2 x 2
0
3 x 3
40
4 x 4
90
5 x 5
60
6 x 6
250
I am now going to work out a formula which will allow me to see what the difference is for any square on the number grid.
Size of Number Square Difference
This table shows the number squares with their differences.
It is clear from this table that there is a pattern in the results, the differences are all the square numbers multiplied by 10. It is also apparent that the 3 x 3 difference is equal to 2 x 2 multiplied by 10. So the difference is always 1 number square size smaller multiplied by 10. For example to work out the 8 x 8 difference I would say 7 x 7 = 49 multiplied by 10 = 490. So 490 is the difference for the 8 x8 number square. The table below shows how I get the answer to any size number square on the grid, after multiplying the opposite corners and finding the difference between the two products.
n
n + (a-1)
n + 10(a-1)
n + (a-1)
+ 10(a-1)
n x [n +(a-1)+10(a-1)] = n2+n(a-1)+10n(a-1)
[n+(a-1)] x [n +10(a-1)]= n2+10n+10(a-1)2
The difference between these is 10(a-1)2
0(a-1)2. a (length) represents the number square size you want to find...I will substitute in a number square size I have proven to convince me this formula works. 10 (4x4 - 1)...10 (3x3) = 90, this is correct as I proved earlier that the difference for a 4 x4 square is 90. The formula which works for any size square on the 10 x 10 number grid is: 10 ( a - 1 )2
I am now going to investigate rectangles on the 100 grid.
(L x W)
For the rectangles I will be doing the same as what I did with the squares, multiplying the two pairs of diagonal corners and finding the difference between the two.
I will begin by investigating a 2 x 3 rectangle on the grid.
5
6
5
6
25
26
5 x 26 = 130 Difference= 20
6 x 25 = 150
I will now do the same for another set of numbers on the grid to see if the pattern continues.
23
24
33
34
43
44
23 x 44= 1012 Difference= 20
24 x 43= 1032
I am now convinced that the difference on a 2 x 3 rectangle is 20; I will now prove this by using algebra.
n
n+1
n+10
n+11
n+20
n+21
n x n+21= n2+21n Difference = 20
n+1 x n+20 = n2+21n+20
This proves that the difference of the 2 pairs of diagonally opposite after being multiplied have a difference of 20 in a 2 x 3 rectangle.
I will now go on to investigate 2 x 4 rectangles on the number grid. I predict that the difference in the 2 x 4 rectangle will be 40, I think this because the 2 x 3 rectangle had a difference of 20...I think that the next rectangle size
9
0
9
20
29
30
39
40
9 x 40= 360 Difference = 30
0 x 39= 390
I will now prove that the difference in a 2 x 4 rectangle is 30 using algebra.
n
n+1
n+10
n+11
n+20
n+21
n+30
n+31
n x n+31 = n2+31n Difference=30
n+1 x n+30 = n2+31n+30
This proves that the difference between the two diagonally opposite products is 30.
I will now go on to investigate 2 x 5 rectangles on the number grid. I predict that the difference for the 2 x 5 rectangle will be 40 because from the 2 x 3 to 2 x 4 the difference has increased by 10...I think it will increase another 10 to get 40.
7
8
27
28
37
38
47
48
57
58
7 x 58= 986 Difference = 40
8 x 57= 1026
As I predicted the difference was 40, I will go on and prove this by using algebra.
n
n+1
n+10
n+11
n+20
n+21
n+30
n+31
n+40
n+41
n+1 x n+40 = n2+41n+40 Difference = 40
n x n+41 = n2+41n
This formula proves that the difference in a 2 x 5 rectangle is 40.
I will now investigate 3 x 2 rectangles on the 10 x 10 grid; I predict that the difference will be 20 because I feel it will produce the same answer as the 2 x 3 grid. As it doesn't really matter which is the length or the width.
26
27
28
36
37
38
26 x 38 = 988 Difference = 20
28 x 36 = 1008
I will now prove that the difference in a 3 x 2 rectangle is 20, I will also be proving that it doesn't matter which is length or the width...2 x 3 and 3 x 2 will always be the same.
n
n+1
n+2
n+10
n+11
n+12
n x n+12= n2+12n Difference = 20
n+2 x n+10= n2 +12n+20
I have proven all the things I said I would prove in the above.
I will now investigate 3 x 4 rectangles on the 10 x 10 number grid. I predict that the 3 x 4 rectangle will have a difference of 60, I think this because the 2 x 3 had a difference of 20 then the 3 x 3 had a difference of 40...(going up in 20's) I believe that it will go up 20 again to 60 for the next grid size up, (3 x 4)
45
46
47
55
56
57
65
66
67
75
76
77
45 x 77= 3465 Difference= 60
47 x 75= 3525
As I predicted the difference was 60, I will now go on to investigate the 3 x 5 rectangles I predict that the difference will be 80.
I will go on to prove this by algebra:
n
n+1
n+2
n+10
n+11
n+12
n+20
n+21
n+22
n+30
n+31
n+32
n+40
n+41
n+42
n x n+42= n2+42n Difference= 80
n+2 x n+40= n2+42n+80
I have now established that the difference for a 3 x 5 rectangle is 80.
I will now investigate 4 x 5 rectangles on the 10 x 10 grid, I predict that the difference will be 120, because of the pattern of that has occurred which is going up in multiples of 30. 2 x 5= 50, 3x5= 80 therefore the 4 x5 must equal 120. I will now prove this by using algebra:
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
n+40
n+41
n+42
n+43
n x n+43= n2+43n Difference= 120
n+3 x n+40= n2+43n+120
I have now proved that the difference in a 4 x5 square on a 10 x 10 grid is 120.
X
2
3
4
5
2
0
20
30
40
3
20
40
60
80
4
30
60
90
20
5
40
80
20
60
This table simply shows the results of all the squares and rectangles...the blue numbers across the top 2,3,4 etc can be either the width or length the same goes for the blue numbers along the left hand side, as it doesn't matter which is length or width as I proved earlier. There are significant patterns throughout these results. In the 3 column across the top and on the left hand side the numbers go up in multiples of 20, this the 2 times table multiplied by 10. For the 4 column it's the 3 times table multiplied by 10...as you can see to get the pattern you must take 1 away from the width or length you will be working with then multiple them by the other side of the rectangle or square you want to work with to find the difference. I will now be working out a formula to prove this pattern.
n
n + (w-1)
n + 10(L-1)
N +10(L-1) +
(w-1)
n x [n+10(L-1)+(w-1)] = n2+ 10n (L-1)+n(w-1)
[n + (w-1)] x [n + 10(L-1)] = n2 + 10n(l-1) (w-1) + 10(w-1) (L-1)
Difference= 10(W-1) (L-1)
The 'W' represents the length of the rectangle whereas the 'L' represents the width. The final formula which enable me to work out the difference for any rectangle on a 10 x 10 grid is 10(L-1) (W-1)
I am now going to change the size of my 10 x 10 grid, I will change It to an 8 x 8 grid.
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
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
58
60
61
62
63
64
This is a simple 8 x 8 number grid...I will be investigating it further.
I will now investigate 2 x 2 number squares on the 8 x 8 grid. I will be multiplying the two diagonally opposite corners and finding the difference, like I did for the 10 x 10 grid.
23
24
31
32
23 x 32= 736 Difference=8
24 x 31= 744
I the difference is 8, I will now do another example to inform me of whether this works for every place on the 8 x 8 number grid.
2
9
0
x 10 = 10 Difference = 8
2 x 9 = 18
The difference was once again 8, I will now go on and prove this by using algebra.
n
n+1
n+8
n+9
n x n+9 = n2+9n Difference = 8
n+1 x n+8 = n2+9n+8
I have now proven that the difference in a 2 x 2 square on an 8 x 8 grid is 8. I will now investigate 3 x 3 squares on an 8 x 8 grid.
22
23
24
30
31
32
38
39
40
22 x 40 = 880 Difference= 32
24 x 38 = 912
I know that will be same for all the 3 x 3 squares on the number grid so I will now prove this using algebra.
n
n+1
n+2
n+8
n+9
n+10
n+16
n+17
n+18
n x n+18 =n2+18n Difference= 32
n+2xn+16= n2+18n+32
I have now proved that the difference for all 3 x3 squares on an 8 x 8 grid is 32.
I will now continue the sequence of squares on the 8 x 8 grid and investigate 4 x 4 squares within the gird.
I predict that the difference on the 4 x4 grid will be 56 because from the 2x2 to the 3x3 it increased by 24 so I think it may increase by another 24.
5
6
7
8
3
4
5
6
21
22
23
24
29
30
31
32
5 x 32 = 160 Difference= 72
8 x 29 = 232
I will now prove that this example shows the correct difference by using algebra.
n
n+1
n+2
n+3
n+8
n+9
n+10
+11
n+16
n+17
n+18
n+19
n+24
n+25
n+26
n+27
n x n+27= n2+27n Difference= 72
n+3 x n+24= n2+27n+72
I will now investigate 5 x5 number squares on the 8 x8 grid, I predict that the difference will be 128 because I have realised that to get the difference you use the square size previous multiplied by 8. e.g 4 x 4 x 8 = 128
2
3
4
5
9
0
1
2
3
7
8
9
20
21
25
26
27
28
29
33
34
35
36
37
x 37 = 37 Difference = 128
5 x 33 = 165
I will now back up this example by proof, using algebra.
n
n+1
n+2
n+3
n+4
n+8
n+9
n+10
n+11
n+12
n+16
n+17
n+18
n+19
n+20
n+24
n+25
n+26
n+27
n+28
n+32
n+33
n+34
n+35
n+36
n x n+36n = n2+36 Difference = 128
n+4 x n+32 = n2+36n+128
After finishing up to the 5 x5 squares on the 8 x 8 I will now work out a formula which enables me to work out any number square on the 8 x 8 grid.
Square size
Difference
2 x 2
8
3 x 3
32
4 x 4
72
5 x 5
28
I have noticed that the differences are all square numbers multiplied by 8, just like I found on the 10 x 10 grid. It is also noticeable that the rule doesn't change for the 8 x 8 grid as to find a difference you use the square size smaller than what you want to find and multiply it by 8. e.g to get the 4 x 4 difference I would say 3 x 3 x 8 = 72. The table below shows how I get the answer to any size number square on the 8 x 8 grid, after multiplying the opposite corners and finding the difference between the two products.
n
n + (a-1)
n + 8(a-1)
n + (a-1)
+ 8(a-1)
n x [n +(a-1)+8(a-1)] = n2+(a-1)+8n(a-1)
n+(a-1) x [n +8(a-1)]= n2+8n+8(a-1)2
The difference between these is 8(a-1)2
As you can see the table is almost indistinguishable from the 10 x 10 algebra grid I completed earlier. The only thing that has been altered is the number 10 which has been to changed to an 8 I suspect that this is because the grid has been changed to an 8 x 8. In the table a (length) represents the number square size you want to find...I will substitute in a square size I have proven to convince me this formula works. 8 (4x4 - 1)...8 (3x3) = , this is correct as I proved earlier that the difference for a 4 x4 square is 90. The formula which works for any size square on the 10 x 10 number grid is: 10 ( a - 1 )2
I will now start to investigate rectangles on the 8 x 8 grid, to start off the rectangles I will be investigating 2 x 3 number rectangles on the grid.
5
6
3
4
21
22
5 x 22 = 110 difference = 16
6 x 21 = 126
I will now do another example of a 2 x 3 square to convince me that this difference is the throughout the number 8 x 8 number grid.
37
38
45
46
53
54
37 x 54 = 1998 difference = 16
38 x 53 = 2014
I am now convinced the difference is 16 on a 2 x 3 rectangle, I will now proceed in proving this by algebra.
n
n+1
n+8
n+9
n+16
n+17
n x n+17 = n2+17n difference = 16
n+1 x n+16= n2+17n+16
I have proven that the difference between the products of the 2 diagonally opposite corners is 16.
I will now advance the investigation of rectangles on the 8 x 8 grid by exploring 2 x 4 rectangles.
25
26
33
34
41
42
49
50
25 x 50 = 1250 difference= 24
26 x 49 = 1274
I will now prove that the difference on this 2 x 4 rectangle is 24.
n
n+1
n+8
n+9
n+16
n+17
n+24
n+25
n x n+25 = n2+25n difference = 24
n+1 x n+24 = n2+25n+24
To further my investigation I will now be investigating 2 x 5 rectangles on the 8 x 8 grid. I predict that the 2 x 5 rectangle on the 8 x 8 grid will have a difference of 32 I think this because from the 2 x 3 to the 2 x 4 it has gone up 8 I think it will continue to go up 8 for the 2 x 5
21
22
29
30
37
38
45
46
53
54
21 x 54 = 1134 Difference= 32
22 x 53 = 1166
As I predicted the difference was 32 I will now prove this by using algebra.
n
n+1
n+8
n+9
n+16
n+17
n+24
n+25
n+32
n+33
n x n+33 = n2+33n Difference = 32
n x n+32 = n2 + 33n + 32
I have now proven that the difference in a 2 x 5 on an 8 x 8 grid is always 32, anywhere on the grid.
I will now go on to investigate 3 x 4 rectangles on the number grid. I predict that the difference will be 64...I think this because the 3 x 3 difference was 32 I think it might increase by another 32 for the next rectangle size up.
32
33
34
40
41
42
48
49
50
56
57
58
32 x 58 = 1856 Difference = 48
34 x 56 = 1904
My prediction was wrong the difference in this example is 48; I will now prove that the difference is 48 by means of algebra.
n
n+1
n+2
n+8
n+9
n+10
n+16
n+17
n+18
n+24
n+25
n+26
n x n+26 = n2 + 26n Difference = 48
n+2 x n+24 = n2 + 26n+48
I have now proven that the difference in a 3 x4 rectangle within an 8 x 8 grid is 48. I am now going to investigate 3 x 5 rectangles on the 8 x 8 grid. I predict that the difference will be 64. I think this because in the 3 x 2, 3 x 3, 3 x 4 the differences have gone up in multiples of 16. The 3 x 4 difference was 48 so if I add another 16 on to that which equals 64 that should come to the difference for this 3 x 5 rectangle.
1
2
3
9
20
21
27
28
29
35
36
37
43
44
45
1 x 45 = 495 difference = 64
3 x 43 = 559
I will now go on to prove that the difference is defiantly 64. I will do this by the means of algebra.
n
n+1
n+2
n+8
n+9
n+10
n+16
n+17
n+18
n+24
n+25
n+26
n+32
n+33
n+34
n x n+34= n2 + 34n difference= 64
n+2 x n+32 = n2 + 34n+64
I have now proven that the difference of the diagonally opposite products in a 3 x 5 number rectangle on an 8 x 8 grid is 64.
I will now go on to investigate 4 x 5 rectangles on the 8 x 8 grid, I will first of all do an example of the 4 x 5 grid then work out the difference.
3
4
5
6
1
2
3
4
9
20
21
22
27
28
29
30
35
36
37
38
3 x 38 = 114 difference= 96
6 x 35 = 210
I will now go on and prove the difference is 96 by using algebra.
n
n+1
n+2
n+3
n+8
n+9
n+10
n+11
n+16
n+17
n+18
n+19
n+24
n+25
n+26
n+27
n+32
n+33
n+34
n+35
n x n+35 = n2+35n difference= 96
n+3 x n+32 =n2+35n+96
I have now proved that the difference in a 4 x 5 square on an 8 x 8 grid is 92.
I will now go on to investigate the patterns in the results I am finding within the rectangles on the 8 x 8 grid, then hopefully come up with a formula which will prove every rectangle on the 8 x 8 grid.
X
2
3
4
5
2
8
6
24
32
3
6
32
48
64
4
24
48
72
96
5
32
64
92
28
This table simply shows the results of all the squares and rectangles...the blue numbers across the top 2,3,4 etc can be either the width or length the same goes for the blue numbers along the left hand side, as it doesn't matter which is length or width as I proved earlier. There are significant patterns throughout these results. In the 3x column across the top and on the left hand side the numbers go up in multiples of 16's, this the 2 times table multiplied by 8. For the 4x column it's the 3 times table multiplied by 8...as you can see to get the pattern you must take 1 away from the width and length...e.g. 3 x 4 take away 1 from length and width would equal 2 x 3, then to find the actual difference you would multiply that by 8. (2 x 3 x 8 = 48)
I am now going to work out the formula below, the algebra will be very similar to the 10 x 10 rectangles formula but all the 10's will change to 8's because it is on an 8 x 8 grid instead of a 10 x 10.
n
n + (w-1)
n + 8(L-1)
N + 8(L-1)
(w-1)
n x [n+8(L-1) (w-1)] = n2+ 8n (L-1)(w-1)
n + (w-1) x [n + 8(L-1)] = n2 + 8n(l-1) (w-1) + 8(w-1) (L-1)
Difference= 8(W-1) (L-1)
The 'W' represents the length of the rectangle
Whereas the 'L' represents the width. The final formula which enables me to work out the difference for any rectangle on a 8 x 8 grid is 8(L-1) (W-1)
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
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
00
01
02
03
04
05
06
07
08
09
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
To further my investigation I will change the grid size again, this time to a
3 x 13 grid which is this time larger than the 10 x 10 rather than smaller like the 8 x8 grid which I have already investigated.
58
59
71
72
58x72 = 4176 difference = 13
59x71 = 4189
The difference of the 2x2 square on the 13x13 grid is 13. I will do one more example to convince me that this difference will continue throughout the number grid.
3
4
6
7
3 x 17 = 51 difference = 13
4 x 16 = 64
I am now convinced that all 2x2 squares on a 13x13 grid will have the difference of 13. I will now prove this by algebra.
n
n+1
n+13
n+14
n x n+14 = n2+14n difference = 13
n+1 x n+13 = n2+14n+13
This has now proved that all 2x2 squares on a 13x13 grid will always have the difference of 13. I will now go on and investigate 3x3 squares.
29
30
31
42
43
44
55
56
57
29x57 = 1653 difference = 52
31x55 = 1705
The difference for the 3x3 squares on a 13 x 13 grid is 52, I am now convinced after doing 2 examples that the difference will account for every place on the 13 x 13 grid. I will now go on to prove this by the means of algebra.
n
n+2
n+26
n+28
n x n+28 = n2+28n difference= 52
n+2 x n+26 = n2+28n+52
I have now proved that all 3x3 squares on the 13x13 grid will always have the difference of 52. I will now investigate the 4x4 squares on the 13x13 grid and predict that the difference will be 208. I predict this because 13 was multiplied by 4 to get 52 so I shall multiply 52 by 4 to get the next difference.
21
22
23
24
34
35
36
37
47
48
49
50
60
61
62
63
21x63 = 1323 difference = 117
24x60 = 1440
My prediction was incorrect, the difference was in fact 117, I will now go on to prove this difference by using algebra as I know from previous squares that the difference will be the same throughout the grid.
n
n+3
n+39
n+42
n x n+42 = n2+42n difference = 117
n+3 x n+39 = n2+42n+117
I have now proven that all 4x4 squares on a 13x13 gird will always have the difference of 117. I will now further my investigation by exploring 5x5 squares on the grid.
82
83
84
85
86
95
96
97
98
99
08
09
10
11
12
21
22
23
24
25
34
35
36
37
38
82x138 = 11316 difference = 208
86x134 = 11524
The difference for the 5x5 squares on a 13x13 grid is 208. I will now prove this by using algebra.
n
n+4
n+52
n+56
n x n+56 = n2+56n difference= 208
n+4 x n+52 = n2+56n+208
I have now proven that 5x5 squares on a 13x13 grid will always have the difference of 208. I will now produce a table of results showing my differences, then I will find patterns and sequences within them.
Square Size
Difference
2x2
3
3x3
52
4x4
17
5x5
208
It is a apparent from the table that there are patterns in these results.
The table shows the pattern, which is that; the difference for any size square is the product of the previous square size multiplied by thirteen. For example;
3x3x13 = 117, and 117= the difference of the 4x4 square or 4x4x13=208= the difference of the 4 x 4 square. This pattern is highly similar to the pattern I found in both the 10 x 10 and 8 x 8 rectangles I done earlier. This means that the formula for the squares on a 13x13 grid will be the same as the 10x10 and the 8x8 but instead of multiplying it by 10 or 8 I will multiply it by 13 because that my grid is a 13 x 13 grid. The formula to prove this would consequently be:
a
n
n+ (a-1)
a
n+13(a-1)
N+(a-1)
+13(a-1)
n x [n+(a-1)+13(a-1)] = n +n(a-1)+13n(a-1)
[n+(a-1)] x [n+13(a-1)] = n2+13n (a-1)+
n(a-1)+10(a-1)2
Difference = 13(a-1)2
Formula = 13(a-1)2
As you can see the formula for the 13x13 squares is very similar to the formula of the 10x10 and 8x8 grids that I completed earlier the only change is the number which you multiply the formula by. The overall formula now, for all the squares on an 13x13 grid is 13(a-1)2.
I will now investigate rectangles on a 13x13 grid. I will start off by investigating 2x3 rectangles on the 13x13 grid.
61
62
63
74
75
76
61x76 = 4636 difference = 26
63x74 = 4662
The difference of the 2x3 rectangle is 26. I am convinced that it will stay the same for all the 2x3 rectangles on the 13x13 grid so I will now prove this by using algebra.
n
n+2
n+13
n+15
n x n+15 = n2+15n difference = 26
n+2 x n+13 = n2+15n+26
This now proves by formula that all 2x3 rectangles all have the same difference of 26. I will now proceed by investigating 2x4 rectangles on the 13x13 grid.
29
30
31
32
42
43
44
45
29x45 = 1305 Difference = 39
32x42 = 1344
The difference of the 2x4 rectangle is 39. I will now prove by using algebra that all 2x4 rectangles have a difference of 39.
n
n+3
n+13
n+16
n x n+16 = n2+16n Difference= 39
n+3 x n+13 = n2+16n+39
I have now proven that all 2x4 rectangles on the 13x13 grid have the difference of 39.
I will now investigate this further by looking at the 2x5 rectangles. I predict that the difference for these will be 52. I think this because I think the pattern of differences is the multiples of 13 so I have to add 13 to the previous difference to obtain 52.
12
13
14
15
16
25
26
27
28
29
12x129 = 14448 difference = 52
16x125 = 14500
As I predicted the difference was 52. I will now show this using algebra and prove that all 2x5 rectangles will have the difference of 52.
n
n+4
n+13
n+17
n x n+17 = n2+17n difference=52
n+4 x N+13 = n2+17n+52
This now proves that all 2x5 rectangles have the same difference of 52. I will now go on and investigate 3x4 as the information before this has already been proved. I will predict that the 3x4 rectangle will have the difference of 78.
43
44
45
46
56
57
58
59
69
70
71
72
43x72 = 3096 Difference = 78
46x69 = 3174
My prediction was correct and the difference is 78. I can now use all this information to predict correctly the rest of the differences for the rectangles on a 13x13 grid. I will now prove by algebra that all 3x4 rectangles have the difference of 78.
n
n+3
n+26
n+29
n x n+29 = n2+29n difference= 78
n+3 x n+26 = n2+29n+78
I have now proved by algebra that all 3x4 rectangles will always have the difference of 78. I can now use all the information gathered to predict with a lot of confidence the other differences for rectangles on a 13x13 grid. I will now draw a table to prove how I know.
X
2
3
4
5
2
3
26
39
52
3
26
52
78
04
4
39
78
17
56
5
52
04
56
208
I have completed the table of differences by predicting the rest of the results however I have proved in recent rectangles that the difference goes up in a pattern. The pattern on which I followed to complete the table was that firstly for the 2x rectangles go up in multiples of 13. Then the 3x rectangles go up in multiples of 26 so it had doubled. The patterns were always correct when I predicted them so I went ahead and tripled and then multiplied by 4 the 13 to complete the table of differences.
W
n
n+(W-1)
L
n+13(L-1)
N+13(L-1)
(W-1)
n x [n+13(L-1)(W-1)] = n2+13n(L-1)+n(W-1)
[n+(W-1)] x [n+13(L-1)] = n2+13n(L-1)+n(W-1)+
3(L-1)(W-1)
Difference = 13(L-1)(W-1)
3 x 13 grid formula= 13(L-1)(W-1)
This is the overall formula that proves the difference of any sized rectangle on a 13x13 grid.
After doing these investigations into number squares and rectangles on different size grids. I can now come up with a formula that shows how to work out the difference on any size number square or rectangle on any size grid. My formula is n(L-1)(W-1). n represents the size of the number grid, so if the number grid was 15 x 15 n would be 15. L and W represent length and width respectively. I will now do an example to prove that this formula works.
I will try this formula by trying to solve the difference of a 3 x 4 rectangle on a 6 x 6 number grid. I will now transfer this rectangle in the formula,
6 (4-1)(3-1)
6 x 3 x 2 = 36. The difference is 36 I will now prove this first by using numbers and then by using algebra.
2
3
4
5
8
9
0
1
4
5
6
7
2 x 17 = 34 difference= 36
5 x 14 = 70
n
n+1
n+2
n+3
n+6
n+7
n+8
n+9
n+12
n+13
n+14
n+15
n x n+15 = n2+15n difference= 36
n+3 x n+12 = n2+15n+36
The difference was 36, just as I expected after putting in the rectangle size and grid size into my formula, n(L-1)(W-1). This formula works for any square or rectangle on any size grid.
![number grid investigation]](https://mbt-essays-prod-public.s3.eu-west-1.amazonaws.com/810836/listing/810836_1.jpg)
