Assignment Part 1
Below is a 10x10 number grid:
The total on the numbers coloured in blue = 90 (i.e. 1+11+12+21+22+23)
Therefore the stair total in this 3-step stair = 90
Part 1 Objective:
For other 3-step stairs, investigate the relationship between the stair total and the position of the stair shape on the grid.
To start the investigation a 10x10-numbered grid square is used as illustrated below in table 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
36c
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 fc;
86
87
88
89.
90
91:
92
93
94
95.
96
97:
98:
99:
00
1
2!
21
22
23
From the complete 10 x 10 numbered grid square, we use part of it to carry out our initial investigation, for example the grid box on the right shows a slice of the 10 x 10 gird square, i.e. 6 boxes representing the numbers 1,11,12,21,22,23 (The 3-step stair)
From this basic numbered square that looks like stairs or steps we can start to establish if there is a pattern. If a pattern is found then we can use an algebra equation to represent this pattern and use the equation for a 10 x 10 numbered Grid Square.
By using the 3-step stair example we know there are [6] squares and lets assume in a 3-step stair the bottom grid box is equal to [x], therefore in our 3-step stair x = 21
Using the values in algebra the formula(s) would look like this:
The 1st square = 1 then the formula is x - 20 = 1
The 2nd square = 11 then the formula is x - 10 = 11
The 3rd square = 12 then the formula is x - 9 = 12
The 4th square = 21 it is simply just x = 21
The 5th square = 22 then the formula is x + 1 = 22
The 6th square = 23 then the formula is x + 2 = 23
The above algebra equations are shown below in our 3-step stair:
x - 20
x - 10
x - 11
X
x + 1
x + 2c
By adding the values from the equation (-20 - 10 - 9 + 1 + 2) = [-36]
Thus we can use the algebra equation 6x - 36 = [the total value of squares in a 10x10 square]
Using the above logic and method we can use it in other grids, such as an 11x11 and a 12x12 numbered grid square. Using the same 3-step stair approach we can use the theory for the 11x11 and 12x12-numbered square to find a pattern.
Below are the results from the such an exercise on a 10x10, 11x11 and 12x12 numbered grid box
Table 4: 10 x 10 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
55c
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.
84c
85;
86
87
88
89
90
91
92
93
94:
95
96
97
98
99
00
Table 5: Using the algebra equation and adding the values we get -36
x - 20
x - 10
x - 11
X
x + 1:
x + 2.
Table 6: 11 x 11 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.
76c
77
78
79
80
81
82
83
84
85
86
87c
88
89
90.
91
92
93
94
95.
96
97
98
99
00
01
02
03
04.
05:
06
07
08
09
10.
Table 7:
x - 32
x - 11
x - 10
X
x + 1
X + 2
Using the algebra equation and adding the values we get -40
Table 8: 12 x 12 Grid
2
3
4
5
6
7.
8
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
Table 9:
x - 24
x - 12
x - 11
x.
x + 1
x + 2
Using the algebra equation and adding the values we get -44
If we closely examine the results from the 3-step stairs from the 3 numbered grid squares, i.e. 10x10, 11x11 and 12x12 we can see there is a constant number that is consistent, which is [4]
We can draw up a table by using this constant number as illustrated below in table 10:
Table 10:
From the table on the left and using our results from the investigation we can see a pattern emerging. Every time the size of the grid box increases by 1, the value in the algebra formula increases by 4. For example for a 10x10 numbered grid using a 3-step stair the formula is 6x-36 where x is the number in the bottom left hand square, then we increase the grid size by 1, 11x11 and using the same 3-stepped stair approach the formula is 6x-40, etc. We can clearly see the constant number [4] is consistent every time the grid size increases by [1].
For any 3-step numbered gird box as shown in table 2 or in the six examples below, and [x] as the value from the bottom left hand square and the algebra theory used to calculate the equation the total value of the squares can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 gird squares. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 6x-36 or 6x-44 we can prove the results for any 3-step grid. The follow exercises in table 11 shows the expected results:
Table 11
33.
FORMULA = 6x - 36
43c
44
: ( 6 x 53) - 36 = 282
53
54
55
2: 33+43+44+53+54+55=282
77
FORMULA = 6x - 36
87
88
: (6 x 97) - 36 = 546
97
98
99
2: 77+87+88+97+98+00=546
3
FORMULA = 6x - 40!
24
25
: (6 x 35) - 36 = 170c
35c
36
37.
2: 13+24+25+35+36+37=170
51
FORMULA = 6x - 40
62
63
: (6 x 73) - 40 = 398
73
74!
75
2: 51+62+63+73+74+75=398
21
FORMULA = 6x - 44
33
34
: (6 x 45) - 44 = 226
45
46
47
2: 21+33+34+45+46+47=226
53
FORMULA = 6x - 44c
65
66
: (6 x 77) - 44 = 418
77
78
79
2: 53+65+66+77+78+79=418
From the above 6 examples, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving the equation to be correct.
From the above exercises and the results we can state the hypothesis as:
. Each time the step stair increases so does the amount of squares within the grid, which is called the formula of triangular numbers.
2. Also as the size of the grid increase by x number of squares the value of the squares total will also change again known as the triangular numbers.
We continue with the exercise and so far we have our algebra equation is 6 x - 36 for a 3-step grid in a 10x10 boxed grid
6 x - 40 for an 11x11 boxed grid and 6 x - 44 for a 12x12 boxed grid.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 36,40 & 44 and show them as shown below:
We know that the above values increase by the constant number [4] and also that in any 3-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor)
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 3-step grids (4) the calculations are 9, 10 & 11 i.e:
9 x 4 = 36 10 x 4 = 40 11 x 4 = 44
We can now use 4 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:
6 x - 4n The next step is to test the equation, we will use 15 as the grid size.
n = 15 and our 3-step grid numbers are 1,16,17,31,32,33 as shown below and x = 31:
Table 12:
6
7;
31
32
33;
Using the equation 6 x - 4(n) we can use the above values to see if the equation works
(6 x 31) - (4 x 15) = 126
To check the equation is correct we added all the numbers:
+16+17+31+32+33 = 130 Therefore 126 <> 130 and our equation is incorrect.
The value from the equation is higher that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 6 x - 4 (n - 1)
We can again test this using the above 3-step grid.
(6 x 31) - (4 x (15 - 1)
86 - 56 = 130
30 = 130 therefore the result from our equation is the same as it is by ...
This is a preview of the whole essay
+16+17+31+32+33 = 130 Therefore 126 <> 130 and our equation is incorrect.
The value from the equation is higher that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 6 x - 4 (n - 1)
We can again test this using the above 3-step grid.
(6 x 31) - (4 x (15 - 1)
86 - 56 = 130
30 = 130 therefore the result from our equation is the same as it is by adding up the numbers.
To prove our algebra equation is correct we test it on two other 3-step grids, 32 & 45 as shown below:
Table 13 a 32x32 and 45x45 grid:
33
34.
46
47
65
66.
67
91
92;
93
Using our equation we use the [x] and [n] values we see the results below:
(6 x 65) - (4 x (32 - 1) = 266 and also 1+33+34+65+66+67 = 266
Add for the 45x45 grid box the result is:
(6 x 91) - (4 x (45 - 1) = 370 and also 1+46+47+91+92+93 = 370
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case for a 3-step stair in any grid size:
THE GENERAL FORMULA IS:
Assignment Part 2
Part 2 Objective
Investigate further the relationship between the stair totals and other step stairs on other number grids.
Using our algebra equations and our proven theory for 4-step, 5-step and 6-step stairs for any numbered grid size. using the same approach and theory as we used in the 3-step exercise we can apply the same to the 4-step stair to find the general formula Durkheim obfuscated jagga's structuration .
Four step stairs:
0 x 10 Gridc
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
59c
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
x-30
x-20.
x-19
x-10
x-9
x-8.
Xc
X+1
X+2
X+3
For example if x = 31 then the square above would be x - 10 = 11 (31 - 10 = 21) and so on, Using the algebra equation and adding the values we get is -90
1 x 11 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
01
02
03
04
05
06.
07
08
09
10.
x-33
x-22
x-21
x-11
x-10 cd.
x-9
X
X+1
X+2
X+3
For example if x = 34 then the square above would be x - 11 = 12 (34 - 11 = 23) and so on, Using the algebra equation and adding the values we get -100
2 x 12 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
95c
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
x-36
x-24
x-23
x-12!
x-11
x-10
x
X+1
X+2
X+3
For example if x = 37 then the square above would be x - 12 = 25 (37 - 12 = 25) and so on Using the algebra equation and adding the values we get -110
If we closely examine the results from the 4-step stairs from the 3 numbered Grid Square, i.e. 10x10, 11x11 and 12x12 we can see that the constant value [10]
We can then complete the rest of the table by using these numbers square as shown below:
From the table below and using our results from the investigation we can see a pattern emerging. Every time the size of the grid square increases, the value in the algebra formula increases by 10. For example for a 10x10 numbered grid using a 4-step stair the formula is 10x-90 then we increase the grid size by 1, 11x11 and using the same 4-stepped stair approach the formula is 10x-100, etc. We can clearly see the constant number [10] is consistent every time the grid size increases by [1].
For any 4-step numbered grid box as shown below, using the bottom left grid box's value as x and the algebra theory used to calculate the equation the value of the box can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 grid boxes. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 10x-90 or 10x-110 we can prove the results for any 4-step grid:
21
Formula: 10x- 90
31
32
: 10 x 51 - 90= 420
41
42
43
2: 21+31+41+51+32+42+52+43+53+54= 420
51.
52
53
54
37
Formula: 10x- 90
47
48
: 10 x 67 - 90= 580
57
58
59
2: 37+47+57+67+48+58+68+59+69+70= 580c
67c
68.
69
70
24
Formula: 10x- 100
36
: 10 x - 100= 470
46
47
48
2: 24+35+46+57+36+47+58+48+59+60= 470
57
58
59
60
51
Formula: 10x- 100
62
63:
: 10 x - 100= 740
73
74
75
2: 51+62+73+84+63+74+85+75+86+87= 740
84
85;
86
87
30
Formula: 10x- 110
42.
43
: 10 x - 110= 740
54
55
56
2: 30+42+54+66+43+55+67+56+68+69= 740
66
67
68
69
From the above 5 examples of such exercise, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving our equation to be correct.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 90,100 & 100 and show them as shown below: Foucault oppressed jagga's realism hypothesis.
We know that the above values increase by the constant number [10] and also that in any 4-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor) Marx denied jagga's structuration theory.
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 4-step grids (10) the calculations are 9, 10 & 11 i.e:
9 x 10 = 90 10 x 10 = 100 11 x 10 =110
We can now use 10 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:cocg cgr secgcgw orcg cgk incg focg cg;
0 x - 10(n) and the next step is to test the equation; we will use 15 as the grid size.
n = 15 and our 4-step grid numbers are 1,16,17,31,32,33,46,47,48,49 as shown below and x = 46:
6
7
31
32
33
46
47
48
49
Using the equation 10 x - 10(n) we can use the above values to see if the equation works
(10 x 46) - (4 x 15) = 400 To check the equation is correct we added all the numbers:
+16+17+31+32+33+46+47+48+49 = 320 Therefore 400 <> 320 and our equation is incorrect.
The value from the equation is lower that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 10 x - 10 (n - 1) we can again test this using the above 3-step grid.
(10 x 46) - (10 x (15 - 1) 460 - 140 = 320
320 = 320 therefore the result from our equation is the same as it is by adding up the numbers.
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case:
THE GENERAL FORMULA IS:
Five step stairs: 10 x 10 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:
x-40
x-30
x-29
x-20
x-19.
x-18
x-10
x-9
x-8
x-7
Xc
X+1
X
X+3
X+4
Using the algebra equation and adding the values we get -180
1 x 11 Grid
2!
3
4.
5c
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.
x-44
x-33c!
x-32
x-22
x-21
x-20
x-11
x-10
x-9
x-8
X
X+1
X+2
X+3
X+4;
Using the algebra equation and adding the values we get -200
2 x 12 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
52c
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67c
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
13c
14
15
16
17
18
19
20
x-48
x-36!
x-35.
x-24
x-23
x-22
x-12
x-11
x-10
x-9
X
X+1
X+2
X+3
X+4
Using the algebra equation and adding the values we get -220
If we closely examine the results from the 5-step stairs from the 3 numbered grid boxes, i.e. 10x10, 11x11 and 12x12 we can see there is a constant number that is consistent, which is [20]
We can then complete the rest of the table by using this constant number as shown below:
From the table below and using our results from the investigation we can see a pattern emerging. Every time the size of the grid square increases, the value in the algebra formula increases by 20. For example for a 10x10 numbered grid using a 5-step stair the formula is 15x-180 then we increase the grid size by 1, 11x11 and using the same 5-stepped stair approach the formula is 15x-200, etc. We can clearly see the constant number [20] is consistent every time the grid size increases by [1].
For any 5-step numbered grid box as illustrated, using the bottom left grid box's value as x and the algebra theory used to calculate the equation the value of the box can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 grid boxes. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 15x-180 or 15x-220 we can prove the results for any 5-step grid:
1
Formula: 15x-180
21
22
: 15x51-180= 585
31
32
33
2: 21+31+41+51+32+42+52+43+53+54+11+22+33+44+55= 585
41
42
43
44
51
52
53
54
55c
3
Formula: 15x-180
23
24
: 15x53-180= 615
33
34
35
2: 13+23+33+43+53+24+34+44+54+35+45+55+46+56+57= 615.
43
44
45
46
53
54
55
56
57
3
Formula: 15x-200
24
25
: 15x57-200= 655.
35:
36
37
2: 24+35+46+57+36+47+58+48+59+60+13+25+37+49+61= 655
46
47
48
49
57
58
59
60
61
2
Formula: 15x-200
23
24
: 15x56-200= 640
34
35
36
2: 12+23+34+45+56+24+35+46+57+36+47+58+48+59+60= 640
45
46
47
48
56
57
58
59.
60
8.
Formula: 15x-220
30
31
: 15x66-220= 770
42
43
44.
2: 30+42+54+66+43+55+67+56+68+69+18+31+44+57+70= 770
54.
55
56
57.
66
67
68
69.
70
Formula: 15x-220.
3
4
: 15x49-220= 515
25
26
27
2: 1+13+25+37+49+14+26+38+50+27+39+51+40+52+53= 515
37.
38!
39:
40:
49!
50
51.
52
53
From the above 5 examples of such exercise, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving our equation to be correct.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
Figure 1
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 280,200 & 220 and show them as shown below:
We know that the above values increase by the constant number [20] and also that in any 5-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor)
Figure 2
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 5-step grids (20) the calculations are 9, 10 & 11 i.e:
9 x 20 = 180 10 x 20 = 200 11 x 20 =220
We can now use 20 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:
5 x - 20(n) The next step is to test the equation; we will use 15 as the grid size.
n = 15 and our 5-step grid numbers are 1,16,17,31,32,33,46,47,48,49,61,62,63,64,65 as shown below and x =61:
.
6
7.
31
32
33
46
47
48
49
61
62
63
64
65
Using the equation 15 x - 20(n) we can use the above values to see if the equation works
(15 x 61) - (20 x 15) = 615 To check the equation is correct we added all the numbers:
+16+17+31+32+33+46+47+48+49+61+62+63+64+65 = 635
Therefore 615 <> 635 and our equation is incorrect.
The value from the equation is lower that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 15 x - 20 (n - 1) We can again test this using the above 5-step grid.c
(15 x 61) - (20 x (15 - 1) 915 - 280 = 635
635 = 635 therefore the result from our equation is the same as it is by adding up the numbers.
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case:
Figure 3
THE GENERAL FORMULA IS:
Six step stairs:
0 x 10 Grid
c
2
3
4
5
6
7
8c
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 ed:
52
53
54.
55
56
57.
58
59
60
61
62
63
64
65
66
67
68
69c
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
X-50
X-40.
X-39
X-30
X-29
X-28
X-20.
X-19
X-18
X-17.
X-10
X-9
X-8
X-7
X-6
Xc
X+1
X+2
X+3
X+4
X+5
Using the algebra equation and adding the values we get -315
2
3
4
5
6c
7
8
9.
0
1.
2
3
4.
5
6.
7
8.
9
20
21
22
23 cc;
24
25
26
27
28
29.
30
31
32
33
34
35
36
37
38c
39
40:
41
42
43
44
45
46
47
48
49
50
5.
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
87r
88!
89
90
91
92
93
94
95.
96
97
98
99.
00
01
02
03
04
05
06
07
08
09
10
X-55
X-44
X-43
X-33
X-32
X-31:
X-22
X-21
X-20
X-19c
X-11
X-10
X-9
X-8.
X-7
Xc
X+1
X+2
X+3
X+4
X+5
Using the algebra equation and adding the values we get -350
2 x 12 Grid
2
3
4
5
6
7:
8c.
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
34c:
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
76c.
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
X-60
X-48
X-47c
X-36
X-35
X-34c
X-24
X-23
X-22.
X-21
X-12
X-11
X-10
X-9
X-8
X
X+1
X+2
X+3
X+4
X+5
Using the algebra equation and adding the values we get -385
If we closely examine the results from the 6-step stairs from the 3 numbered grid boxes, i.e. 10x10, 11x11 and 12x12 we can see there is a constant number that is consistent, which is [35]
We can then complete the rest of the table by using this incremental number as shown below:
From the table below and using our results from the investigation we can see a pattern emerging. Every time the size of the grid square increases, the value in the algebra formula increases by 35. For example for a 10x10 numbered grid using a 6-step stair the formula is 21x-315 then we increase the grid size by 1, 11x11 and using the same 6-stepped stair approach the formula is 21x-350, etc. We can clearly see the constant number [35] is consistent every time the grid size increases by [1].
For any 6-step numbered grid box as shown below, using the bottom left grid box's value as x and the algebra theory used to calculate the equation the value of the box can be found. This theory has been put to the test using a 10x10, 11x11 and 12x12 grid boxes. The results are conclusive and consistent, proving the theory to be accurate and reliable.
Using any of the algebra equation from the above table, e.g. 21x-315 or 21x-385 we can prove the results for any 6-step grid:
Formula: 21x-315
1!
2
: 21x51-315= 756
21
22
23
+11+21+31+41+51+12+22+32+42+52+23+33+43+
2: 53+34+44+54+45+55+56= 756
31
32
33
34
41
42.
43
44
45
51
52c
53
54
55
56
3
Formula: 21x-315
23
24
: 21x63-315= 1008
33
34
35
2: 13+23+33+43+53+63+24+34+44+54+64+35+45+.
55+65+46+56+66+57+67+68= 1008
43
44
45
46
53
54
55
56
57.
63
64
65
66
67
68
24
Formula: 21x-350
35
36
: 21x79-350= 1309
46
47
48
2: 24+35+46+57+68+79+36+47+58+69+70+81+80+
48+59+60+71+82+72+83+84= 1309
57
58
59
60
68
69.
70
71
72
79
80
81
82
83
84
37.
Formula: 21x-350
48
49
: 21x92-350= 1582
59
60
61
2: 37+48+59+70+81+92+49+60+71+82+93+61+72+
83+94+73+84+95+85+96+97= 1582
70
71
72
73
81
82
83
84
85.
92
93
94
95
96
97
Formula: 21x-385
3
4
: 21x61-385= 896
25
26
27
2: 1+13+25+14+26+27+37+38+39+40+49+50+51+
52+53+61+62+63+64+65+66= 896
37.
38
39
40
49
50
51
52
53
61
62
63
64;
65
66.
40
Formula: 21x-385
52
53
: 21x100-385= 1715.
64
65
66
2: 40+52+64+76+88+100+53+65+77+89+101+66+
78+90+102+79+91+103+92+104+105= 1715
76c
77
78
79
88
89
90
91
92
00
01
02c
03
04
05
From the above 5 examples of such exercise, it clearly indicates that the algebra equation works and the theory is accurate.
In this exercise the method of adding the numbers was used to ensure that the algebra equation gives the correct results and in all cases it did, proving our equation to be correct.
We continue to proceed with the exercise to complete our formula (The General Formula). In the algebra equation we need to include the value of the grid size, i.e. is it a 4x4 [4] or 16x16 [16] represented as algebra to give us the total of the numbers added together.
To start the exercise we need to establish the highest common factor, by using our values above, i.e. 315,350 & 385 and show them as shown below:
We know that the above values increase by the constant number [35] and also that in any 6-step grid square if the grid size increases by 1 then the constant number is added to the value (The highest Common Factor)
In the above illustration the grid size is increasing by 1, i.e. from 10 to 11 and then to 12, using the highest common factor for 6-step grids (35) the calculations are 9, 10 & 11 i.e:
9 x 35 = 315 10 x 35 = 350 11 x 35 =385
We can now use 35 as the constant number and [n] as the grid size in our algebra equation. Therefore the algebra equation starts to look like this:
21 x - 35(n) the next step is to test the equation; we will use 15 as the grid size. n = 15 and our 5-step grid numbers are 1,16,17,31,32,33,46,47,48,49,61,62,63,64,65,76,77,78,79,80,81 as shown below and x =76:
6
7
31;
32
33.
46
47
48
49.
61
62
63
64
65
76
77
78c.
79
80
81c
Using the equation 21 x - 35(n) we can use the above values to see if the equation works
(21 x 76) - (35 x 15) = 1071 To check the equation is correct we added all the numbers:
+16+17+31+32+33+46+47+48+49+61+62+63+64+65+76+77+78+79+80+81 = 1106 Therefore 1071 <> 1106 and our equation is incorrect.
The value from the equation is lower that the correct result, therefore we need to reduce n, and we can start this by 1 i.e. (n - 1).
For example 21 x - 35 (n - 1) We can again test this using the above 6-step grid.
(21 x 76) - (35 x (15 - 1) 1596 - 490 = 1106
106 = 1106 therefore the result from our equation is the same as it is by adding up the numbers.
From the steps we have taken we have formulated an algebra equation which when tested, gives us positive results in every case:
THE GENERAL FORMULA IS:
Conclusion
From the investigation and the exercises conducted in this assignment we can conclude that by using the general formula to calculate the total of the grid numbers:
· The step stairs, regardless to whether it is a 3-stepped, 4-stepped, or 7-stepped etc or where its position is in a numbered grid, i.e. 10x10, or 21x21 etc the general formula will always give the total of the gird numbers in that step stair.
· The investigations carried out in part 1 and part 2 show clearly illustrates the common relationship when using the general formula
The Universal Formula
The aim of a Universal Formula is to calculate the total value of the numbers in a step stair.
In the hypothesis we stated the notion of Triangular Numbers, whereby when there is an increase in the number of steps in a numbered grid the total of the squares increase.
Below is a representation of triangular numbers:
Figure 6
P = 1.
3n
P = 2.
6n
P = 3:
0n.
P = 4
5n
P = 5
21n
From the triangular numbers diagram the first set of numbers are referred to as the "a" numbers (1,4,10,20,35,56), 2nd set as the "c" numbers (1st difference - 3,6,10,15,21,28), 3rd set as the "d" numbers (2nd difference - 3,4,5,6,7) and finally the last set "e" numbers (common difference all the 1's). p is referred to as the step index.
Therefore if p = 5 then there will be 21 squares (1st difference) in the step stair, making it a 6-step stair as shown on the right:
53:
69
70
85
86
87
01
02
03
04
17
18
19
20
21
33
34.
35.
36
37
38
Example using a 16x16 grid
A mathematical formula has already been written, referred to as the "Universal Formula" as shown below:
[1/2 p2 + (3p/2) + 1] X (x) - [1 + (p - 1)/6 (p2 + 4p + 6)] X (n - 1)
Where p is the number in a triangular grid, x = the value of the bottom left grid in the step stair and n = the size of the grid
Taking our 16x16 grid example we can calculate the total number of the value for the step stair:
P = 5 x = 133 n = 16
[[1/2 (5 x 5) + ((3 x 5)/2) + 1] X (x)] = 21x -
[1 + ((5 - 1) ((5 x 5) + (4 x 5) + 6))/6 x (n - 1) = 35 (n - 1)
= 21x - 35 (n - 1) = (21 x 133) - (35 x 15)
= 2793 - 525 = 2268
To test the answer we add the numbers:
53+69+70+85+86+87+101+102+103+104+117+118+119+120+121+133+134+135+136+137+138 = 2268
This demonstrates that the universal formula works for any step stair, any numbered grid size and combination of numbers:
We can do one more test using the same step stair, but starting the step at 104 as shown on the side
Therefore: p = 2 because there are 6 squares i.e. 6n and x = 136 and n = 16
Using the universal formula we can calculate:
[[1/2 (2 x 2) + ((3 x 2)/2) + 1] X (x)] = 6x - [1 + ((2 - 1) ((2 x 2) + (4 x 2) + 6))/6 x (n-1)=4(n - 1)
= 6x - 4 (n - 1) = (6 x 136) - (4 x 15)
= 816 - 60 = 756 Add the numbers: 104+120+121+136+137+138 = 756
The formula shows that no matter what the size of the step stair (i.e. number of squares), the position of the step stair (e.g. 1,11,12,21,22,23 or 81,82,83,91,92,93) and the grid size (e.g. 10x10, 25x25) the result will always be the total of the numbers in the step stair.
This relationship is known as the "triangular numbers"
Aim
The aim of this coursework is to find relationships and patterns in the total of
all the numbers in 'Number Stairs' such as the one below. For example, the total
of the number stair shaded in black is:
25+26+27+35+36+45 = 194
I have to investigate any relationships that might occur if this stair was in a
different place on the grid.
Part One
For other 3-stepped stairs, investigate the relationship between the stair total
and the position of the stair shape on the grid.
By looking at this grid, the stair total is:
25 + (25+1) + (25+2) + (25+3) + (25+10) + (25+11) + (25+20)
If we call 25, the number in the corner of the stair, 'n' then we get:
n + (n + 1) + (n + 2) + (n + 3) + (n + 10) + (n + 11) + (n + 20) =
6n + 1 + 2 + 3 + 10 + 11 + 20 =
6n + 44
This formula should work with every number stair that can fit onto a 10 by 10
grid. I can say the total for square n (Tn) is, "n + n + 1 + n + 2 + n + 3 + n +
0 + n + 11 + n + 20", because where ever n is on the grid, the number of the
square:
· One place right of it will be n +1
· Two places right of it will be n +2
· One place above it will be n +10
· One place above it then one place to the right will be n + 10 + 1 = n + 11
· Two places above it will be n + 20
So to get the total you multiply 'n' by six then add 1, 2, 10, 11 and 20 which
gives you a general formula of 6n + 44.
I can test my theory by making doing the following:
First I need to make a table giving a sample of results found by adding each
stair up, square by square:
Stair Number (n) 1 2 3 4 5 6 7 8
Stair Total (Tn) 50 56 62 68 74 80 86 92
I then need find the difference between the totals:
Total 50 56 62 68 74 80 86 92
First difference 6 6 6 6 6 6 6
Because there is an instant pattern in the first difference - the totals go up
in 6, I need find out what 6n would come to:
Stair Number (n) 1 2 3 4 5 6 7 8
6n 6 12 18 24 30 36 42 48
6n doesn't give me the correct answer so I add 44 to 6n and get the Stair Total:
Stair Number (n) 1 2 3 4 5 6 7 8
Stair Total (Tn) 50 56 62 68 74 80 86 92
6n 6 12 18 24 30 36 42 48
+ 44 44 44 44 44 44 44 44 44
6n + 44 50 56 62 68 74 80 86 92
Testing the Formula
To ensure the formula works in every case, I tested it in eight situations where
'n' was different each time.
Using the formula 6n + 44
If n = 1 then
· 6 x 1 + 44 = 6 + 44 = 50
If n = 12 then
· 6 x 12 + 44 = 72 + 44 = 116
If n = 23 then
· 6 x 23 + 44 = 138 + 44 = 182
If n = 34 then
· 6 x 45 + 44 = 270 + 44 = 314
If n = 56 then
· 6 x 56 + 44 = 336 + 44 = 380
If n = 67 then
· 6 x 67 + 44 = 402 + 44 = 446
If n = 78 then
· 6 x 78 + 44 = 468 + 44 = 512
Stair Number (n) 1 12 23 34 45 56 67 78
Stair Total (Tn) 50 116 182 248 314 380 446 512
6n + 44 50 116 182 248 314 380 446 512
All the results using the formula are correct, so I can come to the conclusion
that the formula for the total of a stair:
· On a 10 by 10 grid
· Which travels downwards from left to right
· With a height of 3 squares
Is:
6n + 44
Part Two (Extension)
Investigate further the relationship between the stair total and other step
stairs on the number grids.
I am going to investigate patterns and relationships in:
· The position of the stairs
· The height of the stairs
· The width of the grid
I will then put everything together and produce a universal formula.
Throughout this section, the symbols for the variable inputs will be as follows:
Tn = Total of stair
n = Stair number (the number in the bottom left had corner of the stair)
h = Number of squares high the stair is
w = Width of grid (number of squares)
Relationships between different stair heights on a 10 by 10 grid
To find a pattern, I kept 'n' constant (n = 1) and I changed the height, 'h'.
Height (h) 1 2 3 4 5 6
Total (T1) 1 14 50 120 235 406
Now I need to find the differences between these numbers:
Height (h) 1 2 3 4 5 6
Total (T1) 1 14 50 120 235 406
Difference 1 13 36 70 115 171
Difference 2 23 34 45 56
Difference 3 11 11 11
I repeated this with n = 33
Height (h) 1 2 3 4 5 6
Total (T33) 33 110 242 440 715 1078
Difference 1 77 132 198 275 363
Difference 2 55 66 77 88
Difference 3 11 11 11
The third difference for both of them is 11, which tells me the formula may have
something to do with 11 and h3. I made a table to see if there were any
patterns:
(n = 1)
Height (h) 1 2 3 4 5 6
+11h3 11 88 297 704 1375 2376
-11h -11 -22 -33 -44 -55 -66
Total 0 66 264 660 1320 2310
h3 1 8 27 64 125 216
Total (T1) 1 14 50 120 235 406
There were no obvious patterns so to help myself, I made this grid:
This pattern will apply anywhere on a 10 by 10 grid
If h = 1 then Tn =
n + 0
If h = 2 then Tn =
3n + 11
If h = 3 then Tn =
6n + 44
If h = 4 then Tn =
0n + 107
If h = 5 then Tn =
5n + 211
If h = 6 then Tn =
21n + 366
There is a pattern - the 'h' triangle number multiplies 'n' (so if h = 3, the
3rd triangle would multiply n). The formula for triangle numbers is:
h2 + h
2
So the first part of the formula will be:
n (h2 + h)
2
If I include this in a table, I may get some better results:
Height (h) 1 2 3 4 5 6
Stair Number (n) 1 1 1 1 1 1
Triangle Number of h 1 3 6 10 15 21
Triangle Number of h x n 1 3 6 10 15 21
1h3 - 11h 0 66 264 660 1320 2310
(11h3 - 11h)?6 0 11 44 110 220 385
(Triangle Number of h x n) + (11h3 - 11h)?6 1 14 50 120 235 406
Total (T1) 1 14 50 120 235 406
I included "11h3 - 11h" from my last table. I divided "11h3 - 11h" by 6. This is
because 6 was the number that got "11h3 - 11h" closest to the required total. I
noticed that if you add the "Triangle Number of h x n" and "(11h3 - 11h)?6"
together you get the Total. So my formula for any height Number stair on a 10 by
0 grid is:
n (h2 + h) + (11h3 - 11h)
2 6
Testing the formula
Tn = Total of stair
n = Stair number (the number in the bottom left had corner of the stair)
h = Number of squares high the stair is
w = Width of grid (number of squares)
Test One - Square 25, Height 3
Total = 25 + 26 + 27 + 35 + 36 + 45 = 194
T25 = 25 x ((32 + 3) ? 2) + ((11 x 33 - 11 x 3) ? 6)
T25 = 25 x (11 ? 3) + (297 - 33) ? 6
T25 = 25 x 6 + 44
T25 = 194
Test Two - Square 68, Height 3
Total = 68 + 69 + 70 + 78 + 79 + 88 = 194
