Number Grid Investigation.
Number Grid Investigation
Introduction:
The coursework task is to investigate the patterns generated from using rules in a square grid. The grid provides a structured approach to learning number relationships.
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In accordance with the task there are four steps to follow:
. A box is drawn round four numbers [see grid above].
2. Find the product of the top left number and the bottom right number.
3. Do the same with the top right number and the bottom left number.
4. Calculate the difference between these numbers.
From the example in the above grid: [45 x 56 = 2520 - 46 x 55 = 2530] I find that the product difference for the diagonal square [2x2] is 10.
I am going to start by investigating as to whether or not the location of the 2x2 square on the grid is significant.
A. 81 x 92 = 7452 - 82 x 91 = 7462. Product difference is 10.
B. 9 x 20 = 180 - 10 x 19 = 190. Product difference is 10.
C. 12 x 23 = 276 - 13 x 22 = 286. Product difference is 10.
From the worked examples A, B & C I find that the product difference is 10. Taking these results into account, I predict for any 2x2 square the result will always be 10.
Below is a table of results for 2x2 squares that were randomly chosen from the 10x10 grid.
Table 1. 2x2 results.
st No. multiplication
2nd No. multiplication
Difference
5 x 16 = 80
6 x 15 = 90
0
7 x 28 = 476
8 x 27 = 486
0
32 x 43 = 1376
33 x 42 = 1386
0
68 x 79 = 5372
69 x 78 = 5382
0
The results from T.1 show that the difference is a constant 10, so my prediction was correct.
In Mathematics, Algebra is designed to help solve certain types of problems; letters can be used to represent values, which are usually unknown. I will now attempt to prove my results algebraically. Let `N` represent the number in the top left of the square.
N
N +1
N+10
N+11
This can be expressed into the equation:
( N + 1) ( N + 10) - N ( N + 11)
Multiply out the brackets:
N² + 10N + N + 10 - N² - 11N
Simplified to:
N² + 11N + 10 - N² - 11N
= 10.
By using `N` as the variable and constants together, the results for a 2x2 square prove to be 10. When subtracted the `N` cancels out leaving the number 10 which is actually the size of the number grid I am working with, and also equal to the constant.
I can see from the results that for every multiplication and subtraction carried out the difference will always be 10. Can this be true for any square size ? I shall carry this formula on, and investigate further and see if this works for different sized selection squares e.g. 3x3, 4x4 ect; again I will select random squares on the grid.
3 x3 Squares:
A. ( 3 x 21) - ( 1 x 23 ) = 40
B. ( 44 x 62) - ( 42 x 64 ) = 40
C. ( 8 x 26 ) - ( 6 x 28 ) = 40
D. ( 56 x 76 ) - ( 56 x 78 ) = 40
For the 3x3 squares the constant difference is 40. From the worked examples A, B & C all demonstrate that if you put a 3x3 square anywhere on a 10x10 grid, the product difference will equal 40. Therefore, I can predict for any 3x3 square the result will always be 40.
Now I will show the table for a 3x3 and the algebraic formula. Let `N` represent the top left number in the 3x3 square.
N
N+2
N+20
N+22
This can be expressed into the equation:
( N +2) ( N + 20) - N ( N +22)
Multiply out the brackets:
N² +20N +2N +40 - N² - 22N = 40
This also can be simplified, consequently you are left with 40, which is the constant difference for a 3x3 square.
4x4 Squares:
I will now investigate for the last time with square boxes using a larger 4x4 square. I will again place them randomly on the 10x10 grid.
A. ( 4 x 31) - ( 34 x 1) = 90
B. ( 9 x 36) - ( 6 x 39) = 90
C. ( 67 x 94) - ( 64 x 97) = 90
The difference for a 4x4 square is always going to be 90 no matter where you place the square on a 10x10 grid. I can now prove this by using algebra. Below is the table for a 4x4 and the algebraic formula, let `N` represent the top left number in the square.
N
N+3
N+30
...
This is a preview of the whole essay
A. ( 4 x 31) - ( 34 x 1) = 90
B. ( 9 x 36) - ( 6 x 39) = 90
C. ( 67 x 94) - ( 64 x 97) = 90
The difference for a 4x4 square is always going to be 90 no matter where you place the square on a 10x10 grid. I can now prove this by using algebra. Below is the table for a 4x4 and the algebraic formula, let `N` represent the top left number in the square.
N
N+3
N+30
N+33
This can be expressed into the equation:
( N +3) ( N+30) - N ( N+33)
Multiply out the brackets:
N² + 20N + 3N +90 - N² - 33N = 90
When simplified out this equation equates to 90, which is the constant of a 4x4 square.
To further this number grid investigation, I have explored into a varied selection of squares within the 10x10 grid. Table 2. represents the square size examples and the numerical differences between each of them.
Table. 2.
Square selection size.
Product difference
Difference between each difference (p.d.)
Increase
2 x 2
0
3 x 3
40
30
20
4 x 4
90
50
20
5 x 5
60
70
20
6 x 6
250
90
20
7 x 7
360
10
20
M x M
M+p.d.+20
M+20
20
If I wanted to find out what the next difference would be, I would add to the difference previously. E.g. 40 is the product difference for a 3x3 square, if I wanted to find out what a 4x4 square difference would be, I would add 40 plus the p.d. which is 30 (difference between 2x2 & 3x3) and then add 20 to the result ( because this is the p.d. difference, which increases by 20 each time); the result gives me 90 - the correct answer for a 4x4 square.
3 x 3 = 40 4 x 4 = ? = 40 + 30 (p.d.) + 20 = 90.
The difference between each p.d. increases by 20 on the previous one each time, (see increase T.2.) Therefore, this can be shown as M+20, where M is the difference previously. E.g. If I wanted to find the difference for a 3x3 square and a 4x4 square, I would find the difference between a 2x2 square and a 3x3 square and call this M. subsequently the difference between a 3x3 and a 4x4 would M+20.
Sequences
Mathematics is full of patterns, this helps us to add, subtract, multiply etc. By putting the results in a table I have noticed that there is an obvious pattern. These patterns / sequences can be expressed in terms of the nth term of a sequence.
The sequence of numbers are: 10, 40, 90, 160 ( Product differences in T.2.)
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2nd
3rd
4th
(2x2)
(3x3)
(4x4)
(5x5)
0
40
90
60
30
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70
20
20
Nth term
Product difference
p.d
Increase
From my calculations I have worked out a formula for finding the nth term in this particular sequence of numbers.
Formula: 10 n ²
Now I have a general formula for the sequence I can work out ant term in the sequence.
* 4th term - 10 x 4² = 160 ( correct - see above)
* 6th term - 10 x 6² = 360 ( correct see T.2, 7x7)
* 8th term - 10 x 8² = 640 ( correct. 8th term is a 9x9 square,( 9x81) - (1x 89) =640)
Having checked a further series of results I conclude that the general formula for figuring out product differences in a 10x10 grid is 10n².
The general formula works if you know the term in the sequence, which gives you the square size. What if the square selection size is not known? An overall formula is needed linking the square selection sizes and grid sizes.
The formula would be: G ( s - 1) ²
G = is the grid size, S = is the square size.
Lets see if this works on the 10 x 10 grid with a 5 x5 square selection size:
0 ( which is grid size) ( 5( which is square size) - 1 ) ²: this works out as 10 ( 5 - 1) ² = 160, which is the correct result.
Once again having checked further sets of results, I conclude that the overall formula for finding the product difference from any sized square in a 10 x 10 grid is : 10 ( s - 1)²
Investigating further
Now I have worked algebraic and nth term formulas for different sized squares in the 10x10 grid, I now can investigate further to see if there are any differences or similarities in different sized grids, e.g. 8x8 & 9x9. I shall use the same rules and processes as before.
8 x8 Grid
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Recap method: The top left x the bottom right - the top right x the bottom left.
2 x2 results
A. ( 2x 9) - ( 1x 10) = 8
B. ( 29x 36) - ( 28x 37) = 8
C. ( 56x 63) - ( 55x 64) = 8
From the results A, B & C first observations enable me to predict that for any 2x2 square chosen randomly from the 8x8 grid will give a product difference of 8. I will now produce a second set of results to help my investigation.
Table.3. 2x2 results
st No. multiplication
2nd No. multiplication
Difference
34x43 =1462
35x42=1470
8
5x14 = 70
6x13 = 78
8
51x60 = 3060
52x59 = 3068
8
7x16 = 112
8x15 =120
8
Once again the results in T.3 prove my prediction was correct - the constant product difference is 8. Therefore, the equation can be shown algebraically, let `N` represent the number in the top left of the square.
N
N+1
N+8
N+9
This can be expressed into the equation:
( N+ 1) ( N+ 8) - N ( N + 9)
Multiply out the brackets:
N² + 8N + N +8 - N²- 9N
Simplified to:
N² + 9N + 8 -N² -9N = 8
To reiterate, when subtracted the 'N` cancels out leaving the number 8. Once again this is the size of the grid I am working with, and also it is the constant difference from a 2x2 square.
3x3 Results
A. ( 7x 21 ) - ( 5x 23 ) = 32
B. ( 20x 34) - ( 18x 36) = 32
C. ( 40x 54 ) - ( 38x 56) = 32
Algebraic table for 3x3:
N
N+2
N+16
N+18
Equation:
(N + 2) ( N + 16) - N ( N + 18)
N² +2N +16N +32 - N² + 18N
+ 18N + 32 - 18N = 32
4x4 Results
A. ( 4x 25 ) - ( 1x 28 ) = 72
B. ( 24x 45) - ( 21x 48 ) = 72
C. ( 36x 57 ) - ( 33x 60) = 72
Algebraic table for 4x4:
N
N+3
N+24
N+ 27
Equation:
( N + 3 ) ( N + 24) - N ( N + 27)
N² + 3N + 24N + 72 - N² + 27N
+27N + 72 - 27N = 72
Table 4. Results for 8x8 grid.
Square selection size
Product difference
Difference between each difference p.d.
Increase
2 x 2
8
3 x 3
32
24
6
4 x 4
72
40
6
5 x 5
28
56
6
6 x 6
200
72
6
7 x 7
288
88
6
By investigating further and looking at my results for T.4. I have established there is a formula for an nth term for any square selection size within an 8x8 grid.
st
2nd
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4th
(2x2)
(3x3)
(4x4)
(5x5)
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32
72
28
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6
Nth term
Product difference
p.d.
Increase
The formula for finding any term in any size from the 8x8 grid is 8n² E.g.
* 2nd term - 8 x 2² = 32 (correct, see T.4. 3x3)
* 4th term - 8 x 4² = 128 (correct, see T.4. 5x5)
* 7th term - 8 x 7²= 392 (correct, 7th term is an 8x8 square( 8x5 - 1x64 = 392)
I can conclude that the general formula for finding out product difference in an 8x8 grid is 8n²
Again from the results, I can see there is a pattern for each square size and grid size. This means I shall try my overall formula for a `s` sized square on a `G` sized grid.
Formula is : G ( s - 1)²
8 (grid size) ( 4 (square size) - 1) ²
8 (4 - 1) ² = 72
The overall formula proves to be correct, see T4, a 4 x 4 square has a product difference of 72.
.
9x9 Grid
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Recap method: The top left x the bottom right - the top right x the bottom left.
2x2 Results
A. ( 2x 10) - ( 1x 11) = 9
B. ( 25 x 33 ) - ( 24x 34 ) = 9
C. ( 48x 56) - ( 47x 57 ) = 9
Results A, B & C show a product difference of 9. After further calculations of randomly chosen 2x2 squares prove that there is a constant product difference of 9.
I now present the equation algebraically, let `N` represent the number in the top left square.
N
N+1
N+9
N+10
Equation:
( N + 1) ( N + 8) - N ( N + 9)
N² + N +9N + 9 - N² + 10N
+ 10N + 9 -10N = 9
3x3 Results
A. ( 9x 27) - ( 7x 27) = 36
B. ( 57x 73) - ( 55x 75) = 36
C. ( 63x 79) - ( 61x 81) = 36
Algebraic table for 3x3:
N
N+2
N+18
N+20
Equation:
( N + 2) ( N + 18) - N ( N + 20 )
N² + 2N + 18N + 36 - N² - 20N
+20N +36 -20N = 36
4x4 Results
A. ( 4x 28 ) - ( 1x 31) = 81
B. ( 9x 33 ) - ( 6x 36) = 81
C. ( 49x 73) - ( 46x 76) = 81
Algebraic table for 4x4:
N
N+3
N+27
N+30
Equation:
( N +3) ( N + 27) - N ( N + 30 )
N²+3N +27N +81 - N²- 30N
+30N +81 - 30N = 81
Table. 5. Results for 9x9 grid
Square selection size
Product difference
Difference between each difference p.d
Increase
2 x 2
9
3 x 3
36
27
8
4 x 4
81
45
8
5 x 5
44
63
8
6 x 6
225
81
8
7 x 7
324
99
8
Once more I find from my results in T.5. There is a pattern forming, so I can predict that there will be an nth term for finding any product difference within the sequence for the 9x9 grid.
st
2nd
3rd
4th
2x2
3x3
4x4
5x5
9
36
81
44
27
45
63
8
8
N th term
Product difference
p.d.
Increase
My prediction suggests the nth term formula is 9n². E.g.
* 3rd term - 9 x 3² = 81 (correct - see T.5. 4x4 )
* 5th term - 9 x 5² = 225 (correct - see T.5 6x6)
* 8th term - 9 x 8² = 576 (correct - 8th term is 9x9 square ( 9x73 - 1x 81 = 576)
My prediction was correct, there is a general formula for finding the nth term in a 9x9 grid - 9n².
From previous results and workings I can predict that the overall formula for finding any square selection from a 9 x9 grid will be G ( s - 1) ²
9 (grid size) ( 7 ( square size) - 1) ²
9 ( 7 - 1) ² = 324.
I have checked this result: 7x7 square within a 9x9 grid:
7 x 55 - 1 x 61 = 324.
My prediction was correct; the overall formula for finding any `s` square sized selection from a 9x9 grid is : 9 ( s-1)²
Summary
With the information I have gathered; tables of results, algebra and sequences have provided me with a broad range of results, enabling me to have a more definitive view of the initial investigation. From the worked examples I have observed that:
0x10 grid:
* The smallest square selection size 2x2 gives the result 10, which is the grid size.
* The product difference and ` difference between each differences` are all multiples of 10.
* The increase between product difference and p.d is 20 ( 2x10).
* The formula for the nth term is 10n² .
8x 8 Grid:
* The smallest square selection size 2x2 gives the result 8,which is the grid size.
* The product difference and ` difference between each difference` are all multiples of 8.
* The increase between product difference and p.d is 16 (2x8).
* The formula for the nth term is 8n².
9x 9 Grid:
* The smallest square selection size 2x2 gives the result 9, which is the grid size.
* The product difference an `difference between each difference` are all multiples of 9.
* The increase between product difference and p.d. is 18 (2x9).
* The formula for the nth term is 9n².
All the grid selection sizes have an overall formula for finding the product difference; from any size grid, any sized square selection the constant formula is: G (s -1)²
Taking all these observations into account, could these findings be true for any grid size? Would a 5 x5 grid produce results that are all multiples of 5, and linked to the number 5? I am going to predict this.
5x5 Grid Results
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2 x 2 =
5
3 x 3 =
20
4 x 4 =
45
5 x 5 =
80
Nth term
5n²
My prediction was right. Therefore, for any size grid such as, 12 x12 will have a formula and results that link and incorporate the number 12.
My findings show also by using algebra in the equations, I can construct an algebraic formula. By using `N` in the top left of the algebraic table, the equation, when fully worked out will always give the correct constant answer in any square selection size in any size grid.
Extension
As a continuation of this investigation I have chosen to examine rectangles in place of square selection sizes. I will use the same rule as before and also the 10x10 grid. I will try three different sizes of rectangles, each will be 2 numbers in width, but I will increase each length by 1 number e.g. 2x3, 2x4.
0x10 Grid
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I will show only my results, not all the working out,
2 x 3 Results
4 x 12 - 2 x 14 = 20
2 x 4 Results
50 x 57 - 47 x 60 =30
2x 5 Results
85 x 91 - 81 x 95 = 40
By looking at the results and carrying out more results, I can see that the product difference is equal to 10 times the nth term plus 10 (difference between each difference). I will now change the length of the rectangle to `D`, giving me an algebraic formula for finding a 2x ? rectangle.
2 x D = 10 n + 10
I now have a formula for a 2 x D rectangle.
E.g. 2 x 6 - 4th term in sequence - 10 x 4 +10 = 50
6 x 11 - 1 x 16 = 50.
My formula of 10n + 10 is correct.
I will now enlarge the rectangle and also try three different sizes of rectangles. Each will be 3 numbers in width, but I will increase the length e.g 3x3, 3x4.
3 x 3 Results
3 x 21 - 1 x 23 = 40
3 x 4 Results
33 x 51 - 31 x 53 = 60
3 x 5 Results
8 x 24 - 4 x 28 = 80
This set of results show that the difference is equal to 20 times the nth term plus 20 (difference between each difference). I will now change the length of the rectangle to `D`, this can be shown algebraically.
3 x D = 20n + 20
I now have a formula for a 3 x D rectangle.
E.g. 3 x 7 - 5th term in the sequence - 20 x 5 + 20 = 120.
7 x 21 - 1 x 27 = 120.
My formula of 20n + 20 is correct.
From this I can begin to see the product differences are increasing by 10`s, and also the formulas are going up by 10`s. I can predict that the formula for finding a 4 x? rectangle will be 30 n + 30, but I will have to test this to see if this is true.
A. 4x 3 = 60
B. 4 x 4 = 90
C. 4 x 5 = 120
My prediction was correct. You need to multiply the nth term by 30 this time, and then add 30 which is the difference between each difference. I can now construct a formula for a 4 x D rectangle: 4 x D = 30 n + 30.
E.g. 4 x 6 - 4th term in the sequence - 30 x 4 + 30 = 150
6 x 31 - 1 x 36 = 150
M y formula of 30n + 30 is correct.
These workings out and formulas give me a rule for any sized rectangle, providing I know the width. Possibly there could be a rule for find any width size rectangle with an unknown sized length?
Conclusion
In this project I have found that number grids are an extremely powerful tool for a wide range of maths concepts such as, number patterns, problem solving and investigation. I have successfully predicted what can come next in square selection sized differences, nth terms and algebraic equations that simplify expressions.
There are many different rules formulas within grid sizes and square or rectangle selections. If I were to extend this project I would possibly change the original rules with regards to diagonal multiplication and subtraction, to see if there are any different patterns, or I would change the grid sizes and rectangle sizes or perhaps try different shapes.
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