The calculation for grid one is:
(3 x 7) – (1 x 9)
21 – 9 = 12
In grid two it will follow the same rule. I shall call the answer A.
A = N (N²- N + 1) – (1 x N²)
I can now use this formula to work out the
Answer with a 7 x 7 grid.
7 (7² - 7 + 1 – (1 x 7²)
7 (49 – 6) – 49
(7 x 43) – 49
301 – 49 = 252
If I then put this answer in the difference table:
The number in difference column 3 is the same as in the previous difference table I did. So the answer I got, 252 is correct.
I will just check the formula with the 5 x 5 grid:
N (N²- N + 1) – (1 x N²)
5(25- 5 + 1) – (1 x 25)
(5 x 21) – 25
105 – 25 =80: Which is correct.
However the starting number might not always be one. So I need to find an expression for it. If the bottom left square is ‘N² - N + 1’ then the top left square will follow the same rule and will be ‘N - N + 1’
So the new formula for the answer is:
N (N²- N + 1) – N²(N- N + 1)
If I now just check this with a 4 x 4 grid:
4 (16 – 4 + 1 – 16(4 – 4 + 1)
(4 x 13) – (16 x 1)
52 – 16 = 36: Which is correct.
Therefore the formula for number grids with equal side, with numbers increasing by one is:
A = N (N²- N + 1) – N²(N- N + 1)
This is a bit complicated so if I multiply out the brackets and simplify, the formula will be:
A = N (1 + N² – N) – N²(1 + N – N)
= N + N³ - N² – N²
= N³ - N² + N – N²
A = N³ - 2N² + N
I now have a formula, which I can use to work out the answers for number grids that are too large to draw.
Now I will investigate square number grids, where the numbers increase by one but where the starting number is not 1.
I have chosen 7 as the starting number in this case.
2 x 2 Grid
I have done another grid just to check.
3 x 3 Grid
The formula will not be the same for these. There will be a different formula for each kind. So the easiest way is to substitute the numbers in the grid for 1,2,3,4 and so on. Then you can use the formula for that type of grid to work out the answer. However this will only be easy for smaller number grids.
Sequences
Now I have tried grids with number sequences. First where the numbers increase by 2.
2 x 2 Grid
I have checked this by doing another 2 x 2 grid but with different numbers.
2 x 2 Grid
I have now done a 3 x 3 grid just to check this rule.
3 x 3 Grid
Now I have decided to try number grids where the numbers increase by 3, to see what happens.
2 x 2 Grid
I have realised that when the increase in the numbers is 1 the answer is 1 times as much, (1²).
When the increase in numbers is 2 the answer is 4 times as much, (2²)
When the increase in numbers is 3 the answer is 9 times as much, (3²)
So the increase squared is how many times more the answer is, than the answer in a grid where the numbers increase by one.
So presumably the answer for a grid where the numbers increase by 4 would be 16 (4²) times as much.
So if I check it:
2 x 2 Grid
I have made a table with this data. And from it you can predict the answer for any square grid and for different increase in the numbers.
Therefore the formula for number grids where:
The increase in numbers is 2 = 4(N³ - 2N² + N)
The increase in numbers is 3 = 9(N³ - 2N² + N)
The increase in numbers is 4 = 16(N³ - 2N² + N)
The increase in numbers is 5 = 25(N³ - 2N² + N)
So the overall formula is N³ - 2N² + N multiplied by the increase in the numbers squared.
A = Increase² (N³ - 2N² + N)
Rectangles
Now I will investigate rectangular number grids. I have chose to keep the width (W) the same but vary the depth (D) of the grid.
2 x 3 Grid
2 x 4 Grid
2 x 5 Grid
2 x 6 Grid
Now I have put this data in a table to look for patterns.
All the numbers in difference column 1 are the same so a formula for these rectangular grids may involve 2.
If I substitute the numbers in the 2 x 3 grid for algebraic expression I may be able to find out a formula.
I have used this to predict the answer to a 3 x 4 Grid.
4(3²) + 3 – 9 – 12
= 4 x 9 + 3 – 21
= 4 x 9 – 18
= 36 – 18
= 18
I have checked this by drawing the grid:
Therefore the formula for rectangular grids with consecutive numbers is:
A = D (W²) + W – W² – WD
I have now investigated rectangular number grids where the numbers increase by 2.
2 x 3 Grid
I have done a 2 x 3 grid where the numbers increase by 4 to check this.
Therefore a formula for these grids would be:
A = The increase² (D (W²) + W – W² – WD)
I have now investigated whether the formula above works with square number grids.
In a 2 x 2 grid with consecutive numbers the answer is 2.
The formula however will not be exactly the same. Instead of having D it will be replaced by W, because it is a square grid.
It will be:
W (W²) + W – W² - W²
= W³ + W – 2W²
= 2³ + 2 – 2 x 2²
= 8 + 2 – 8
= 2 : Which is correct.
I have checked this with a 4 x 4 grid. The answer should be 36.
W³ + W – 2W²
= 4³ + 4 – 2 x 4²
= 64 + 4 – 32
= 68 – 32
= 36: Which is correct
Therefore this formula does work.
Even though the formulas I got before I simplified them look different, when simplified they are the same:
N (1 + N² – N) – N²(1 + N – N)
= N³ – 2N² + N
W (W²) + W – W² - W²
= W³ - 2W² + W
Decimals
Now I have decided to investigate decimals in number grids.
I have realised that the answers to these decimal number grids are a hundredth of the grids with a number increase of 1.
The answer to a 2 x 2 grid numbered 1,2,3,4 = 2
2/100 = 0.02: the answer to the 2 x 2 grid numbered 0.1,0.2,0.3,0.4.
Therefore a formula for these grids would be:
A = N³ - 2N² + N
100
Now I will try decimal number grids where the numbers increase by more than 0.1:
I predict that the answer will be 0.04 times as much because 0.2 is the increase and 0.2² = 0.04. This was the rule for the other grids. I will check my theory:
The answer was in fact 4 times that of the previous answer:
0.02 x 4 = 0.08
I will check this with another grid:
2 x 2 grid where increase in numbers is 0.3.
(0.4 x 0.7) – (0.1 x 1)
0.28 – 0.1 = 0.18
Answer = 0.18
I have realised that the answers are a hundredth.
Therefore a formula for these grids would be:
N³ - 2N² + N
100
So if a decimal grid’s numbers increase by 2, then the formula for the answer will be:
N³ - 2N² + N x 4
100
Now I will investigate number grids with decimal and whole numbers:
For all the grids so far I have had the numbers going horizontally from left to right.
But now I will have the numbers arranged differently:
The numbers are arranged horizontally from right to left.
I will just check this with another grid.
So a formula for these grids would be:
A = - (N³ - 2N² + N)
Grids with numbers that have powers
I have now investigated number grids with square numbers.
2 x 2 grid
3 x 3 grid
4 x 4 grid
5 x 5 grid
(5² x 21²) – (1² x 25²)
(25 x 441) – (1 x 625)
11025 – 625 = 10400
Answer = 10400
7 x 7 grid
(7² x 43²) – (1² x 49²)
(49 x 1849) – (1 x 2401)
90601 – 2401 = 88200
Answer = 88200
I have now put these results in a table.
All the numbers in difference column 6 are the same so the formula will involve something to the power of six.
I have now substituted algebraic expressions in the number grid.
Therefore the formula will be:
N²(N² - N + 1)² – N4
= N²(N² - N + 1) (N² - N + 1) – N4
= N² (N4 – N³ + N² - N³ + N² - N + N² - N + 1) – N4
I have checked this for a 2 x 2 grid:
4 (16 – 8 + 4 – 8 + 4 – 2 + 4 – 2 + 1) – 16
= (4 x 9) – 16
= 36 – 16
= 20
Which is the correct answer.
I have checked it for 3 x 3 aswell:
9(81 – 27 + 9 – 27 + 9 – 3 + 9 – 3 + 1) - 81
9 x 441 – 81 = 360 Which is correct
If I then simplify the formula:
=N² (N4 + N² + N² + N² - N³ - N³ - N - N + 1) - N4
= N6 + N4 + N4 + N4 – N5 – N5 – N³ – N³ + N² – N4
= N6 + 2N4 – 2N5 – 2N³ + N²
I have checked this formula with a 3 x 3 grid:
= 729 + 162 – 486 – 54 + 9
= 900 – 540
= 360: Which is the correct answer.
Therefore a formula for the answer in number grids with square numbers is:
A = N6 + 2N4 – 2N5 – 2N³ + N²
The formula does involve a term to the power of six as I predicted.
I have now investigated number grids with cubed numbers:
I have substituted the numbers in the grid for algebraic expressions to discover a formula.
Therefore the formula will be:
N³(N³ - N + 1)³ - N6
N³(N² - N + 1) (N² - N + 1) (N² - N + 1) - N6
I have checked this formula for a 2 x 2 grid:
2³(2² - 2 + 1) (2² - 2 + 1) (2² - 2 + 1) - 26
= 8 x 3 x 3 x 3 – 64
= 216 – 64
= 152: Which is correct
This formula is rather complicated and hard to simplify. So I must find a way.
The original formula I got for square grids with consecutive numbers was:
N³ - 2N² + N
So if I factorise this I get:
= N (N² - 2N + 1)
= N (N – 1) (N – 1)
= N (N – 1)²
Therefore presumably a formula for cubed number grids would be:
A = N³ (N³ - 1)² - N6
I will check to see if the formula works for a 3 x 3 grid:
N³(N³ - N + 1)³ - N6
= 3³ (3² - 3 + 1)³ - 36
= 27 x 7³ - 36
= 9261 – 729
= 8532: which is correct.
Because the formula for cubed number grids is N³(N³ - N + 1)³ - N6, I predict the formula for a grid with numbers to the power of four will be:
A = N4 (N² - N + 1) 4 – N8
Proof: I will check this for a 2 x 2 with numbers to the power of four.
=24 (2² - 2 + 1) 4 – 28
=16 x 3 x 3 x 3 x 3 - 256
= 1296 – 256
=1040
I will now do the calculation to check this:
(24 x 34) – (14 x 44)
= 16 x 81 – 256
= 1040
The formula does work.
Therefore I can predict the formula to any power.
The rule is, the part inside the bracket always stays the same, (N² - N + 1)
The N term before the bracket has the power, the same as what you trying to find a formula for. This is the case for the power outside the bracket aswell. The – N term at the end of the equation has a power that is twice that of the power you are investigating.
So a formula for a grid with numbers to the power of 6 would be:
A = N6 (N² - N + 1) 6 – N12
Proof: I will check this for a 3 x 3 grid with numbers to the power of 6.
= 36 (3² - 3 + 1) 6 - 312
= 729 x 76 – 531441
= 85766121 – 531441
= 85234680
I will now do the calculation to check this:
(36 x 76) – (16 x 96)
= (729 x 117649) – (1 x 531441)
= 85766121 – 531441
=85234680
The answer is the same therefore the formula does work.
I can now use this to predict any power in a number grid.
Conclusion
Proving Formulae
I now have a set of formulae, which I can use to work out the answers to any size number grid. I have listed my formulae here and done two examples for each to prove once again that they work.
Square number grids
A = N³ – 2N² + N
3 x 3 grid – the answer should be 12:
= 3³ - 2(3²) + 3
= 27 – 18 + 3
= 12
5 x 5 grid – the answer should be 80:
= 5³ - 2(5²) + 5
= 125 – 50 + 5
= 125 – 45
= 80
Rectangular number grids
A = D (W²) + W – W² - WD
2 x 3 grid – the answer should be 4:
= 3(2²) + 2 – 2² - (2 x 3)
= 12 + 2 – 4 – 6
= 14 – 10
= 4
3 x 4 grid – the answer should be 18:
= 4(3²) + 3 – 3² - (3 x 4)
= 36 + 3 – 9 – 12
= 39 – 21
= 18
Sequences in square number grids
A = Increase²(N³ – 2N² + N)
2 x 2 grid with number increase of 4 – the answer should be 32:
= 4² x 2³ - 2(2²) + 2
= 16(8 – 8 + 2)
= 16 x 2
= 32
5 x 5 grid with number increase of 5 – the answer should be 2000:
= 5²(5³ - 2 x 5² + 5)
= 25(125 – 50 + 5)
= 25 x 80
= 2000
Sequences in rectangular grids
A = Increase²(D (W²) + W – W² - WD)
2 x 3 grid with number increase of 3 – the answer should be 36:
= 3²(3(2²) + 2 – 2² - 2 x 3)
= 9(12 + 2 – 4 – 6)
= 9 x 4
= 36
3 x 4 grid with number increase of 2 – the answer should be 72:
= 2²(4(3²) + 3 – 3² - 3 x 4)
= 4(36 + 3 – 9 – 12)
= 4 x 18
= 72
Decimals in square number grids
A= N³ – 2N² + N
100
3 x 3 grid – the answer should be 0.12:
= 3³ - 2(3²) + 3
100
= 27 – 18 + 3
100
= 12/100
= 0.12
7 x 7 grid – the answer should be 2.52:
= 7³ - 2(7²) + 7
100
= 343 – 98 + 7
100
= 252/100
= 2.52
Decimals in rectangular number grids
A = D (W²) + W – W² - WD
100
2 x 3 grid – the answer should be 0.04:
= 3(2²) + 2 – 4 – 6
100
= 4/100
= 0.04
3 x 4 grid – the answer should be 0.18
= 4 (3²) + 3 – 9 – 12
100
= 18/100
= 0.18
Decimal sequences in square number grids
A = Increase² (N³ – 2N² + N)
100
This formula is the same as the sequence and decimal formulae and therefore I do not need to check them again because I know they work. This is the case for all the other formulae below. All the parts in them have already been checked. Checking them again would just be a waste of time.
Decimal sequences in rectangular number grids
A = Increase²(D (W²) + W – W² - WD)
100
Squared numbers in square number grids
A = N6 + 2N4 – 2N5 – 2N³ + N²
Cubed numbers in square number grids
A = N³(N² – N + 1)³ - N6
Numbers to the power of 4 in square number grids
A = N4 (N² – N + 1) 4 – N8
When I made the difference table on pages 20 and 21 for the squared number grids it took me a long time but it didn’t give me much help to a formula. The best method I found for working out a formula was substituting the numbers in the grid for algebraic expressions. From this I was able to work out the expression for each corner and check it. I then put the four expressions into a formula and simplified it.
All the formulae I have, contain either of two main expressions: N³ – 2N² + N or
D (W²) + W – W² - WD.
This is the case for all the formulae except the ones for squared and cubed numbers.
In this investigation I have learnt how to use difference tables and how they help to work out formulae. I have learnt a lot about working with formulae and simplifying them.
There are many more extensions that can be investigated. Squared and Cubed decimals and squared sequences, for example. But there is simply not enough time to investigate all types. This is an endless investigation.
The main question was to investigate square grids with consecutive numbers. I believe I have gone a step further and investigated a wide range of extensions and given an equation for each, that works.
I drew all the diagrams that I did so that it was easier to work out the answer.
I have stuck to my plan all the way through except, in my plan I did not mention anything about extensions.
The part of the project I found most difficult was simplifying the equations for squared and cubed number grids. I could not simplify the formula for cubed numbers so I left it as it was.
All the calculations I have done took a long time but were worth doing in order to discover and prove a formula. The graph that I did on page 6 did not help whatsoever in my investigation, but until I did it I did not know that.
All the formulae I have, work and therefore I can use them to predict larger sized grids that are impossible to draw.