Samantha Whittaker
Number 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
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
Pick out 2x2 squares
Multiply the diagonals
Find the difference
34
35
44
45
INVESTIGATE
27
28
37
38
82
83
92
93
68
69
78
79
9
0
9
20
First number
Second number
Third number
Fourth number
stx4th
2ndx3rd
Difference
34
35
44
45
530
540
0
27
28
37
38
026
036
0
82
83
92
93
7626
7636
0
68
69
78
79
5372
5382
0
9
0
9
20
80
90
0
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 2x2 grid from a 10x10 grid, times the diagonals, the difference between the products of the diagonals is always 10.
36
37
46
47
If this rule is correct, then by using this grid:
I predict that the difference between the
Product of 36 and 47 compared with that of
37 and 46 will equal 10.
36x47=1692
37x46=1702
Difference=10
My prediction was correct as the difference between 36x47 (1692), and 37x46 (1702) was 10.
To prove that this theory will work for any 2x2 grid from a 10x10 number square, I am going to express it as an algebraic equation.
X
X+1
X+10
X+11
(X)(X+11)=(X+1)(X+10)
X²+11X=X²+11X+10
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 10 that is underlined in the above equation, proves that the latter part of the equation will always be 10 more than the first half of the equation.
So, by using a 2x2 grid and multiplying the diagonals, the difference will always be 10.
I am now going to investigate whether there is any relationship between the products of the diagonals of a 3x3 grid.
42
43
44
52
53
54
62
63
64
78
79
80
88
89
90
98
99
00
6
7
8
6
7
8
26
27
28
8
9
20
28
29
30
38
39
40
2
3
4
2
3
4
22
23
24
First number
Third number
Seventh number
Ninth number
stx9th
3rdx7th
Difference
42
44
62
64
2688
2728
40
78
80
98
00
7800
7840
40
6
8
26
28
68
208
40
8
20
38
40
720
760
40
2
4
22
24
48
88
40
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 3x3 grid from a 10x10 grid, times the diagonals, the difference between the products of the diagonals is always 40.
31
32
33
41
42
43
51
52
53
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 31 and 53 compared with that of
33 and 51 will equal 40.
31x53=1643
33x51=1683
Difference=40
My prediction was correct as the difference between 31x53 (1643), and 33x51 (1683) was 40.
To prove that this theory will work for any 3x3 grid from a 10x10 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+10
X+11
X+12
X+20
X+21
X+22
(X)(X+22)=(X+2)(X+20)
X²+22X=X²+22X+40
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 40 that is underlined in the above equation, proves that the latter part of the equation will always be 40 more than the first half of the equation.
I am now going to extend my investigations further and see whether there is any relationship between the products of the diagonals of a 4x4 grid.
24
25
26
27
34
35
36
37
44
45
46
47
54
55
56
57
62
63
64
65
72
73
74
75
82
83
84
85
92
93
94
95
64
65
66
67
74
75
76
77
84
85
86
87
94
95
96
97
3
4
5
6
3
4
5
6
23
24
25
26
33
34
35
36
27
28
29
30
37
38
39
40
47
48
49
50
57
58
59
60
First number
Fourth number
Thirteenth number
Sixteenth number
stx16th
4thx13th
Difference
24
27
54
57
368
458
90
62
65
92
95
5890
5980
90
64
67
94
97
6208
6298
90
3
6
33
36
08
98
90
27
30
57
60
620
710
90
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 4x4 grid from a 10x10 grid, times the diagonals, the difference between the products of the diagonals is always 90.
43
44
45
46
53
54
55
56
63
64
65
66
73
74
75
76
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 43 and 76 compared with that of
46 and 73 will equal 90.
43x76=3568
46x73=3358
Difference=90
My prediction was correct as the difference between 43x76 (3568), and 46x73 (3358) was 90.
To prove that this theory will work for any 4x4 grid from a 10x10 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+3
X+10
X+11
X+12
X+13
X+20
X+21
X+22
X+23
X+30
X+31
X+32
X+33
(X)(X+33)=(X+3)(X+30)
X²+33X=X²+33X+90
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 90 that is underlined in the above equation, proves that the latter part of the equation will always be 90 more than the first half of the equation.
I will now investigate further by taking a 2x2 grid from a 5x5 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
2
6
7
3
4
8
9
9
20
24
25
4
5
9
0
6
7
21
22
First number
Second number
Third number
Fourth number
stx4th
2ndx3rd
Difference
2
6
7
7
2
5
3
4
8
9
247
252
5
9
20
24
25
475
480
5
4
5
9
0
40
45
5
6
7
21
22
352
357
5
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 2x2 grid from a 5x5 grid, times the diagonals, the difference between the products of the diagonals is always 5.
3
4
8
9
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 3 and 9 compared with that of
4 and 8 will equal 5.
3x9= 27
4x8= 32
Difference= 32-27=5
My prediction was correct as the difference between 3x9 (27), and 4x8 (32) was 5.
To prove that this theory will work for any 2x2 grid from a 5x5 number square, I am going to express it as an algebraic equation.
X
X+1
X+5
X+6
(X)(X+6)= (X+1)(X+5)
X² +6X= X²+6X+5
The equation is set out with ...
This is a preview of the whole essay
I predict that the difference between the
product of 3 and 9 compared with that of
4 and 8 will equal 5.
3x9= 27
4x8= 32
Difference= 32-27=5
My prediction was correct as the difference between 3x9 (27), and 4x8 (32) was 5.
To prove that this theory will work for any 2x2 grid from a 5x5 number square, I am going to express it as an algebraic equation.
X
X+1
X+5
X+6
(X)(X+6)= (X+1)(X+5)
X² +6X= X²+6X+5
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 5 that is underlined in the above equation, proves that the latter part of the equation will always be 5 more than the first half of the equation.
I will now investigate further by taking a 3x3 grid from a 5x5 grid.
8
9
0
3
4
5
8
9
20
1
2
3
6
7
8
21
22
23
2
3
6
7
8
1
2
3
3
4
5
8
9
20
23
24
25
7
8
9
2
3
4
7
8
9
First number
Third number
Seventh number
Ninth number
stx9th
3rdx7th
Difference
8
0
8
20
60
80
20
1
3
21
23
253
273
20
3
1
3
3
33
20
3
5
23
25
325
345
20
7
9
27
29
33
53
20
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 3x3 grid from a 5x5 grid, times the diagonals, the difference between the products of the diagonals is always 20.
2
3
4
7
8
9
22
23
24
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 12 and 24 compared with that of
4 and 22 will equal 20.
2x24=288
4x22=308
Difference= 308-288=20
My prediction was correct as the difference between 12x24 (288), and 4x8 (308) was 20.
To prove that this theory will work for any 3x3 grid from a 5x5 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+5
X+6
X+7
X+10
X+11
X+12
(X)(X+12)=(X+2)(X+10)
X²+12X= X²+12X+20
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 20 that is underlined in the above equation, proves that the latter part of the equation will always be 20 more than the first half of the equation.
I shall investigate further by using a 2x2 grid from a 7x7 to start off with.
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
6
7
23
24
26
27
33
34
37
38
44
45
6
7
3
4
8
9
25
26
First number
Second number
Third number
Fourth number
stx4th
2ndx3rd
Difference
6
7
24
25
384
393
7
26
27
33
34
884
891
7
37
38
44
45
665
672
7
6
7
3
4
84
91
7
8
9
25
26
468
475
7
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 2x2 grid from a 7x7 grid, times the diagonals, the difference between the products of the diagonals is always 7.
If this rule is correct, then by using this grid:
41
42
48
49
I predict that the difference between the
product of 41 and 49 compared with that of
42 and 48 will equal 7.
41x49=2009
42x48=2016
Difference= 2016-2009=7
My prediction was correct as the difference between 41x49 (2009), and 42x48 (2016) was 7.
To prove that this theory will work for any 2x2 grid from a 7x7 number square, I am going to express it as an algebraic equation.
X
X+1
X+7
X+8
(X)(X+8)=(X+1)(X+7)
X²+8X= X²+8X+7
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 7 that is underlined in the above equation, proves that the latter part of the equation will always be 7 more than the first half of the equation.
I will now extract 3x3 grids from the 7x7 number square.
3
4
5
0
1
2
7
8
9
8
9
20
25
26
27
32
33
34
22
23
24
29
30
31
36
37
38
33
34
35
40
41
42
47
48
49
2
3
8
9
0
5
6
7
First number
Third number
Seventh number
Ninth number
stx9th
3rdx7th
Difference
3
5
7
9
57
85
28
8
20
32
34
612
640
28
22
24
36
38
836
864
28
33
35
47
49
617
645
28
3
5
7
7
45
28
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 3x3 grid from a 7x7 grid, times the diagonals, the difference between the products of the diagonals is always 28.
If this rule is correct, then by using this grid:
9
20
21
26
27
28
33
34
35
I predict that the difference between the
product of 19 and 35 compared with that of
21 and 33 will equal 28.
9x35=665
21x33=693
Difference=663-665=28
My prediction was correct as the difference between 19x35 (665), and 21x33 (693) was 28.
To prove that this theory will work for any 3x3 grid from a 7x7 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+7
X+8
X+9
X+14
X+15
X+16
(X)(X+16)=(X+2)(X+14)
X²+16X= X²+16X+28
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 28 that is underlined in the above equation, proves that the latter part of the equation will always be 28 more than the first half of the equation.
Finally, from a 7x7 number square, I will extract 4x4 grids.
2
3
4
5
9
0
1
2
6
7
8
9
23
24
25
26
1
2
3
4
8
9
20
21
25
26
27
28
32
33
34
35
25
26
27
28
32
33
34
35
39
40
41
42
46
47
48
49
22
23
24
25
29
30
31
32
36
37
38
39
43
44
45
46
8
9
0
1
5
6
7
8
22
23
24
25
29
30
31
32
First number
Fourth number
Thirteenth number
Sixteenth number
stx16th
4thx13th
Difference
2
5
23
26
52
15
63
1
4
32
35
385
448
63
25
28
46
49
225
288
63
22
25
43
46
012
075
63
8
1
29
32
256
319
63
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 4x4 grid from a 7x7 grid, times the diagonals, the difference between the products of the diagonals is always 63.
9
0
1
2
6
7
8
9
23
24
25
26
30
31
32
33
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 9 and 33 compared with that of
2 and 30 will equal 63.
9x33=297
2x30=360
Difference=360-297=63
My prediction was correct as the difference between 9x33 (297), and 12x30 (360) was 63.
To prove that this theory will work for any 4x4 grid from a 7x7 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+3
X+7
X+8
X+9
X+10
X+14
X+15
X+16
X+17
X+21
X+22
X+23
X+24
(X)(X+24)=(X+3)(X+21)
X²+24X= X²+24X+63
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 63 that is underlined in the above equation, proves that the latter part of the equation will always be 63 more than the first half of the equation.
I am now going to re-investigate a 10x10 grid, yet this time extracting rectangles from the grid. I shall start by looking at grids of 2x3.
23
24
33
34
43
44
65
66
75
76
85
86
8
9
8
9
28
29
79
80
89
90
99
00
56
57
66
67
76
77
First number
Second number
Fifth number
Sixth Number
stx6th
2ndx5th
Difference
23
24
43
44
012
032
20
65
66
85
86
5590
5610
20
8
9
28
29
232
252
20
79
80
99
00
7900
7920
20
56
57
76
77
4312
4332
20
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 2x3 grid from a 10x10 grid, times the diagonals, the difference between the products of the diagonals is always 20.
9
0
9
20
29
30
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 9 and 30 compared with that of
0 and 29 will equal 20.
9x30=270
0x29=290
Difference= 290-270=20
My prediction was correct as the difference between 9x30 (270), and 10x29 (290) was 20.
To prove that this theory will work for any 2x3 grid from a 10x10 number square, I am going to express it as an algebraic equation.
X
X+1
X+10
X+11
X+20
X+21
(X)(X+21)=(X+1)(X+20)
X²+21X=X²+21X+20
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 20 that is underlined in the above equation, proves that the latter part of the equation will always be 20 more than the first half of the equation.
I shall now investigate grids of 5x3 from a 10x10 number square.
2
3
4
5
1
2
3
4
5
21
22
23
24
25
42
43
44
45
46
52
53
54
55
56
62
63
64
65
66
76
77
78
79
80
86
87
88
89
90
96
97
98
99
00
6
7
8
9
0
6
7
8
9
20
26
27
28
29
30
35
36
37
38
39
45
46
47
48
49
55
56
57
58
59
First number
Fifth number
Eleventh number
Fifteenth number
stx15th
5thx11th
Difference
5
21
25
25
05
80
42
46
62
66
2772
2852
80
76
80
96
00
7600
7680
80
6
0
26
30
80
260
80
35
39
55
59
2065
2145
80
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 5x3 grid from a 10x10 grid, times the diagonals, the difference between the products of the diagonals is always 80.
54
55
56
57
58
64
65
66
67
68
74
75
76
77
78
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 54 and 78 compared with that of
58 and 74 will equal 80.
54x78=4212
58x74=4292
Difference= 4292-4212=80
My prediction was correct as the difference between 54x78 (4212), and 58x74 (4292) was 80.
To prove that this theory will work for any 5x3 grid from a 10x10 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+3
X+4
X+10
X+11
X+12
X+13
X+14
X+20
X+21
X+22
X+23
X+24
(X)(X+24)=(X+4)(X+20)
X²+24X=X²+24X+80
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 80 that is underlined in the above equation, proves that the latter part of the equation will always be 80 more than the first half of the equation.
I shall now investigate grids of 3x5 from a 10x10 number square.
47
48
49
57
58
59
67
68
69
77
78
79
87
88
89
2
3
4
2
3
4
22
23
24
32
33
34
42
43
44
58
59
60
68
69
70
78
79
80
88
89
90
98
99
00
23
24
25
33
34
35
43
44
45
53
54
55
63
64
65
28
29
30
38
39
40
48
49
50
58
59
60
68
69
70
First number
Third number
Thirteenth number
Seventeenth number
stx17th
3rdx13th
Difference
47
49
87
89
4183
4263
80
2
4
42
44
88
68
80
58
60
98
00
5800
5880
80
23
25
63
65
495
575
80
28
30
68
70
960
2040
80
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 3x5 grid from a 10x10 grid, times the diagonals, the difference between the products of the diagonals is always 80.
54
55
56
64
65
66
74
75
76
84
85
86
94
95
96
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 54 and 96 compared with that of
56 and 94 will equal 80.
54x96=5184
56x94=5264
Difference=5264-5184=80
My prediction was correct as the difference between 54x96 (5184), and 56x94 (5264) was 80.
To prove that this theory will work for any 3x5 grid from a 10x10 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+10
X+11
X+12
X+20
X+21
X+22
X+30
X+31
X+32
X+40
X+41
X+42
(X)(X+42)=(X+2)(X+40)
X²+42X=X²+42X+80
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 80 that is underlined in the above equation, proves that the latter part of the equation will always be 80 more than the first half of the equation.
I shall now investigate grids of 4x3 from a 5x5 number square.
1
2
3
4
6
7
8
9
21
22
23
24
7
8
9
0
2
3
4
5
7
8
9
20
2
3
4
5
7
8
9
0
2
3
4
5
2
3
4
5
7
8
9
20
22
23
24
25
2
3
4
6
7
8
9
1
2
3
4
First number
Fourth Number
Ninth number
Twelfth number
stx12th
4thx9th
Difference
1
4
21
24
264
294
30
7
0
7
20
40
70
30
2
5
2
5
30
60
30
2
5
22
25
300
330
30
4
1
4
4
44
30
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 4x3 grid from a 5x5 grid, times the diagonals, the difference between the products of the diagonals is always 30.
6
7
8
9
1
2
3
4
6
7
8
9
If this rule is correct, then by using this grid: I predict that the difference between the
product of 6 and 19 compared with that of
9 and 16 will equal 30.
6x19=114
9x16=144
Difference=144-114=30
My prediction was correct as the difference between 6x19 (114), and 9x16 (144) was 30.
To prove that this theory will work for any 4x3 grid from a 5x5 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+3
X+5
X+6
X+7
X+8
X+10
X+11
X+12
X+13
(X)(X+13)=(X+3)(X+10)
X²+13X=X²+13X+30
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 30 that is underlined in the above equation, proves that the latter part of the equation will always be 30 more than the first half of the equation.
I will now investigate grids of 5x2 from a 5x5 number square.
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
20
6
7
8
9
0
1
2
3
4
5
First number
Fifth number
Sixth number
Tenth number
stx10th
5thx6th
Difference
5
6
0
0
30
20
1
5
6
20
220
240
20
6
0
1
5
90
10
20
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 5x2 grid from a 5x5 grid, times the diagonals, the difference between the products of the diagonals is always 20.
If this rule is correct, then by using this grid:
6
7
8
9
20
21
22
23
24
25
I predict that the difference between the
product of 16 and 25 compared with that of
20 and 21 will equal 20.
6x25=400
20x21=420
Difference=420-400=20
My prediction was correct as the difference between 6x25 (400), and 20x21 (420) was 20.
To prove that this theory will work for any 5x2 grid from a 5x5 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+3
X+4
X+5
X+6
X+7
X+8
X+9
(X)(X+9)=(X+4)(X+5)
X²+9X=X²+9X+20
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 20 that is underlined in the above equation, proves that the latter part of the equation will always be 20 more than the first half of the equation.
I will now investigate grids of 3x4 from a 5x5 number square.
2
3
6
7
8
1
2
3
6
7
8
7
8
9
2
3
4
7
8
9
22
23
24
3
4
5
8
9
0
3
4
5
8
9
20
6
7
8
1
2
3
6
7
8
21
22
23
8
9
0
3
4
5
8
9
20
23
24
25
First number
Third number
Tenth number
Twelfth number
stx12th
3rdx10th
Difference
3
6
8
8
48
30
7
9
22
24
68
98
30
3
5
8
20
60
90
30
6
8
21
23
38
68
30
8
0
23
25
200
230
30
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 3x4 grid from a 5x5 grid, times the diagonals, the difference between the products of the diagonals is always 30.
2
3
4
7
8
9
2
3
4
7
8
9
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 2 and 19 compared with that of
4 and 17 will equal 20.
2x19=38
4x17=68
Difference=68-38=30
My prediction was correct as the difference between 2x19 (38), and 4x17 (68) was 30.
To prove that this theory will work for any 3x4 grid from a 5x5 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+5
X+6
X+7
X+10
X+11
X+12
X+15
X+16
X+17
(X)(X+17)=(X+2)(X+15)
X²+17X=X²+17X+30
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 30 that is underlined in the above equation, proves that the latter part of the equation will always be 30 more than the first half of the equation.
Finally, I will take rectangles from a 7x7 grid, starting with a rectangle of 2x5.
5
6
2
3
9
20
26
27
33
34
7
8
24
25
31
32
38
39
45
46
2
8
9
5
6
22
23
29
30
20
21
27
28
34
35
41
42
48
49
6
7
3
4
20
21
27
28
34
35
First number
Second number
Ninth number
Tenth number
stx10th
2ndx9th
Difference
5
6
33
34
70
98
28
7
8
45
46
782
810
28
2
29
30
30
58
28
20
21
48
49
980
008
28
6
7
34
35
210
238
28
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 2x5 grid from a 7x7 grid, times the diagonals, the difference between the products of the diagonals is always 28.
3
4
0
1
7
8
24
25
31
32
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 3 and 32 compared with that of
4 and 31 will equal 28
.
3x32=96
4x31=124
Difference=124-96=28
My prediction was correct as the difference between 3x32 (96), and 4x31 (124) was 28.
To prove that this theory will work for any 2x5 grid from a 7x7 number square, I am going to express it as an algebraic equation.
X
X+1
X+7
X+8
X+14
X+15
X+21
X+22
X+28
X+29
(X)(X+29)=(X+1)(X+28)
X²+29X=X²+29X+28
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 28 that is underlined in the above equation, proves that the latter part of the equation will always be 28 more than the first half of the equation.
I will now extract 6x3 grids from the same 7x7 grid.
2
3
4
5
6
8
9
0
1
2
3
5
6
7
8
9
20
5
6
7
8
9
20
22
23
24
25
26
27
29
30
31
32
33
34
29
30
31
32
33
34
36
37
38
39
40
41
43
44
45
46
47
48
23
24
25
26
27
28
30
31
32
33
34
35
37
38
39
40
41
42
2
3
4
5
6
7
9
0
1
2
3
4
6
7
8
9
20
21
First number
Sixth number
Thirteenth number
Sixteenth number
stx16th
6thx13th
Difference
6
5
20
20
90
70
5
20
29
34
510
580
70
9
24
43
48
392
462
70
23
28
37
42
966
036
70
2
7
6
21
42
12
70
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 6x3 grid from a 7x7 grid, times the diagonals, the difference between the products of the diagonals is always 70.
9
0
1
2
3
4
6
7
8
9
20
21
23
24
25
26
27
28
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 9 and 28 compared with that of
4 and 23 will equal 70.
9x28=252
4x23=322
Difference=322-252=70
My prediction was correct as the difference between 9x28 (252), and 14x23 (322) was 70.
To prove that this theory will work for any 6x2 grid from a 7x7 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+3
X+4
X+5
X+7
X+8
X+8
X+10
X+11
X+12
X+14
X+15
X+16
X+17
X+18
X+19
(X)(X+19)=(X+5)(X+14)
X²+19X=X²+19X+70
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 70 that is underlined in the above equation, proves that the latter part of the equation will always be 70 more than the first half of the equation.
Finally from a 7x7 number square, I will take out grids of 5x2.
2
3
4
5
8
9
0
1
2
7
8
9
20
21
24
25
26
27
28
38
39
40
41
42
45
46
47
48
49
24
25
26
27
28
31
32
33
34
35
29
30
31
32
33
36
37
38
39
40
First number
Fifth number
Sixth number
Tenth number
stx10th
5thx6th
Difference
5
8
2
2
40
28
7
21
24
28
476
504
28
38
42
45
49
862
890
28
24
28
31
35
840
868
28
29
33
36
40
160
188
28
I have put my results into a table so that they are easier to analyse and compare.
What I have found is; when you take a 5x2 grid from a 7x7 grid, times the diagonals, the difference between the products of the diagonals is always 28.
9
0
1
2
3
6
7
8
9
20
If this rule is correct, then by using this grid:
I predict that the difference between the
product of 9 and 20 compared with that of
3 and 16 will equal 28.
9x20=180
3x16=208
Difference=208-180=28
My prediction was correct as the difference between 9x20 (180), and 13x16 (208) was 28.
To prove that this theory will work for any 5x2 grid from a 7x7 number square, I am going to express it as an algebraic equation.
X
X+1
X+2
X+3
X+4
X+7
X+8
X+8
X+10
X+11
(X)(X+11)=(X+4)(X+7)
X²+11X=X²+11X+28
The equation is set out with the first half being the top left multiplied by the bottom right and the second half being the top right multiplied by the bottom left.
The 28 that is underlined in the above equation, proves that the latter part of the equation will always be 28 more than the first half of the equation.
Now, using all the material I have collaborated in my previous investigations, I am going to devise and explain a formula or formulae that will aid when finding the nth term.
I shall start off by looking at squares taken from number grids.
Side of original grid
Side of extracted grid
Difference
0
2
0
0
3
40
0
4
90
5
2
5
5
3
20
7
2
7
7
3
28
7
4
63
Let A be the side of the original grid, B be the side of the extracted grid and C be the difference.
C= (B-1)² x A
This means that the difference between the products of each diagonal on a square taken from a number grid will be the side of the extracted grid minus one then multiplied by the side of the original number square.
CHECK
49
50
51
52
53
54
55
56
57
64
65
66
67
68
69
70
71
72
79
80
81
82
83
84
85
86
87
94
95
96
97
98
99
00
01
02
09
10
11
12
13
14
15
16
17
24
25
26
27
28
29
30
31
32
39
40
41
42
43
44
45
46
47
54
55
56
57
58
59
60
61
62
69
70
71
72
73
74
75
76
77
The above is a 9x9 grid taken from a 15x15 number square. According to my calculations, the difference between 49x177 and 57x169 should be 960.
49x177=8673
57x169=9633
Difference=9633-8673=960
My prediction was correct; therefore the formula I constructed will apply in all cases of squares taken from number squares.
I will now try and devise a formula to find out the difference when multiplying the diagonals of rectangles taken from a number square.
Side of original grid
Length of extracted grid
Width of extracted grid
Difference
0
2
3
20
0
5
3
80
0
2
5
80
5
4
3
30
5
5
2
20
5
3
4
30
7
2
5
28
7
6
3
70
7
5
2
28
Let A be the side of the original grid, B be the length of the extracted grid, C be the width of the extracted grid and D be the difference.
D= (B-1)(C-1) x A
This means that the difference between the products of each diagonal on a square taken from a number grid will be the length of the extracted grid minus one times the width of the extracted grid minus one then multiplied by the side of the original number square.
CHECK
92
93
94
95
96
97
98
99
00
01
02
03
04
05
07
08
09
10
11
12
13
14
15
16
17
18
19
20
22
23
24
25
26
27
28
29
30
31
32
33
34
35
37
38
39
40
41
42
43
44
45
46
47
48
49
50
52
53
54
55
56
57
58
59
60
61
62
63
64
65
67
68
69
70
71
72
73
74
75
76
77
78
79
80
The above is a 14x6 grid taken from a 15x15 number square. According to my calculations, the difference between 92x180 and 105x167 should be 975.
92x180=16560
05x167=17535
Difference=17535-16560=975
My prediction was correct; therefore the formula I constructed will apply in all cases of rectangles taken from number squares.