(52*96)- (56*92) = -160
I predict that the result will be -160
Example 5
(25*69)- (29*65) = -160
I will now use algebra to try an prove my result
Algebraic equation
(x)*(x+44) – (x+4)*(x+40)
X² + 44x – x² + 44x + 160
λ – (λ + 160) = -160
Predicting a 6*6 square
I can now see a pattern emerge from the results:
So I predict that a 6*6 square will give the result -250. I noticed that all the results were negative square numbers multiplied by ten. I made this prediction using the tabulate above using the previous results.
Considering the results are all consecutive square number it gave me reason to believe that the 6*6 square would give an answer on -250. I can also prove this:
The tabulate shows me that the results are all multiples of (-10) that are multiplied by consecutive sq numbers. So there is further evidence that the 6*6 square will give -250
Example 1
(2*57)- (7*52) = -250
Example 2
(45*100)- (50*95) = -250
Those examples proves me theory was correct. I will try to use algebra to prove me theory.
Algebraic equation
(x)*(x + 55) – (x + 5)*(x + 50)
X² + 55x - x² + 55x +250
λ – (λ + 250) = -250
My Prediction for an ‘N*N’ square:
My Prediction for an ‘N*N’ square: -10(n-1)². This equation can only suit the given rule if 2>n<10.
This equation is only used for predicting the result of the square.
Example 1
With a 7*7 grid I can predict using the N*N equation that the result will be -360.
(34*100)- (40*94) = -360
I worked the formula out by placing the results of the other grid into a diagram
With my previous outcomes I can see that the equation should produce a multiple of -10. The previous products all are square number. This gives reason to say that it is (n-1²).
Algebraic working:
‘X’ being the given number and ‘n’ being the square of the length of one of the squares.
Example1
Using the given rule, this gives us
X[X + 11(n-1)] - [(X + n-1) (X+10(n-1)]
Taking the out ‘[ ]’s leave…
X(X+11N-11) – (X+N-1) (X+10N-10)
Equals…
X²+11NX-10x – (X²+11NX-11X- +10N²-10N-10N+10)
Simplifies to…
-10n²+20n-10
And again simplifies to…
-10(n²+2n-1)
This finishes as…
-10(n-1)²
This proves my hypothesis.
This works because the product of top right multiplied by the bottom left will always have a common difference to the top left multiplied by the bottom right in squares. This means that you can calculate a ‘n’ equation by applying the above grid. As we can simplify the equation this tells that with all 2x2 – 10x10 on a 10*10 Master grid, the equation to apply is -10(n-1)² when we it is (Top left x Bottom right) – (Top right x Bottom left)
Rectangles
I am about to relate my formula used with squares to fit with a rectangle. I will only increase a variable slowly so that I can read the difference easily and state a hypothesis. I will then prove the hypothesis using algebra and the use the same method for the other variable. The first variable will be the width of the rectangle, which I will increase by 1 for each example until I can see a common difference.
2 down*3 wide grids
Example 1
(1*13)- (3*11) = -20
Example 2
(24*36)- (26*34) = -20
2 down*4 wide grids
Example 1
(1*14)- (4*11) = -30
Example 2
(55*68)- (58*65) = -30
2 down*5 wide grids
Example 1
(1*15)- (5*11) = -40
Example 2
(66*80)- (70*76) = -40
2 down* 6 wide grids
Example1
(1*16)- (6*11) = -50
Example 2
(73*88)- (78*83) = -50
Observing these worked examples, the variable- width has a common difference and can be seen in the diagram
As the width increases by 1, the difference goes up by -10
So I predict that in the next 2x7 grid the difference will go up by -10
2 down * 7 wide
(4*20)- (10*14) = -60
Example 2
(33*49)- (39*43) = -60
So my prediction was right, I will now try and prove my theory using algebra.
Algebraic equation
‘X’ here represents a number starting the rectangular grid.
(X * [x+16])- ([x+6] * [x+10])
(X²+16x)- (x²+16x+60)
X²+16x- x²+16x+60
16x – 16x+60
=-60
As we have now proved it with an algebraic equation for the width we can now move on and work out the formula for the length down.
3 down *2 wide
(1*22)- (2*21) = -20
Example 2
(26*47)- (27*46) = -20
4 down *2 wide
(5*36)- (6*35) = -30
Example 2
(8*39)- (9*38) = -30
5 down * 2 wide
(24*65)- (25*64) = -40
Example 2
(8*49)- (9*48) = -40
6 down * 2 wide
(49*100)- (50*99) = -50
Example 2
(5*56)- (6*55) = -50
Observing these worked examples, the variable- width has a common difference and can be seen in the diagram
As the width increases by 1, the difference goes up by -10
So I predict that in the next 7*2 grid the difference will go up by -10.
7 down *2 wide
(7*68)- (8*67) = -60
Example 2
(33*94)- (34*93) = -60
Algebraic equation
(X*[x+61])- ([x+1]*[x+60])
(X²+61x)- (x²+61x+60)
X²+ 61x – x² + 61x + 60
X² - x²+ 60
= -60
Here M represent the width an N represents the length and x represents
= {x[x+ (m-1) +10(n-1)]} – {[x+ (m-1)] [x+10(n-1)]}
= [x(x+m-1+10n-10)] – [(x+m-1) (x+10n-10)]
= (x²-x+mx+10nx-10x) – (x²+10nx-10x+mx+10nm-10m-x-10n+10)
= - (10nm-10m-10n+10)
= -10nm+10m+10n-10
= -10(nm-m-n+1)
= -10(n-1) (m-1)
The square formula is different to the rectangle formula because the square formula is: -10(n-1)² which can be seen as -10[(n-1) (n-1)] .In that formula you see that it is ‘-10’ multiplied by 2 lengths but in the case of a square both length and width are the same resulting, only the need for one letter to represent the both of them. Whereas in the rectangle, it has a variable length compared to its width. Resulting in the need for another letter: to represent the change in the width/length. So basically because the rectangle hasn’t got equal sides we have to add an ‘m’ to represent the width and keep the ‘n’ representing the length. They are similar because to predict an outcome you use the -10 multiplied by both its width minus 1 and its length minus 1 but again in the squares case both length and width are the same.
Varying the size of the Master grid
2*2 square on a 8*8 grid
Example 1
(1*10)- (9*2) = -8
Example 2
(37*46)- (38*45) = -8
Example 3
(42*51)- (43*50) = -8
Example 4
(55*64)- (56*63) = -8
I predict that with all 2*2 squares the equation will always produce an answer of -10
Example 5
(26*35)- (27-34) = -8
I will now use algebra to prove my prediction
Algebraic equation: (x)*(x+9) – (x+1)*(x+8)
X²+9x – x²+x+8x+8
+9x – 9x+8
Λ – (Λ + 8) = -8
Therefore my theory was right.
3*3 square on a 8*8 grid
Example 1
1*19 – 3*17 = -32
Example 2
21*39 – 23*37 = -32
Example 3
46*64 – 48*62 = -32
Example 4
41*59 – 43*57 = -32
I predict that the next square will give me a result of -40
Example 5
12*30 – 14*28 = -32
I will now use algebra to try an prove my result
Algebraic equation: x*(x+18) – (x+2)*(x+16)
X²+ 18x – x²+2x+16x+32
+18x – 18x+32
Λ – (Λ + 32) = -32
4*4 squares on a 8*8 grid
Example 1
1*28 – 4*25 = -72
Example 2
19*46 – 22*43 = -72
Example 3
33*60 – 36*57 = -72
Example 4
20*47 – 44*23 = -72
I predict that the next square will give me a result of -72
Example 5
36*63 – 39*60 = -72
Algebraic equation: x*(x+27) – (x+3)*(x+24)
X²+ 27x – x²+3x+24x+72
+27x – 27x+72
Λ – (Λ + 72) = -72
Therefore my theory of the end results from the 4*4 squares proved right as shown above.
I can now see a pattern emerge from the results. So I predict that a 5*5 square will give the result -128. I noticed that all the results were the width and length minus 1 multiplied by negative ten. I made this prediction using the tabulate above using the previous results.
Considering the results are all consecutive square numbers, it gave me reason to believe that the 5*5 square would give an answer on -128.
Example 1
3*39 – 35*7 = -128
This result differs from the 10*10 master-grid because as above in the diagram on page 17 you can see that the results are negative 8 multiplied by (the width minus 1 multiplied by the length minus 1). You can see this illustrated on the tabulate above. Whereas, on the 10 by 10 Master-grid the result: is negative 10 multiplied by (the length minus 1 and the width minus 1). This means that the 2nd set of results will be smaller than the 1st set of results.
Finding a formula for the N*N chosen on an A*A Master grid.
Using this diagram displaying the previous results on an 8*8 grid and using the first n*n formula I am going to attempt change it to suit an A*A grid. The first formula was -10(n-1²) but as the first formula was on a 10*10 Master-grid that where the -10 at the front comes from. This in theory means that to change it to an 8*8 grid we would change the first number.
Example 1
So to predict the result of this square, we would use the formula -8(n-1²) which equals: -8(3*3).
That equals -72
4*4 square
19*46 – 43*22 = -72
We also have previous results to back that hypothesis. As a result of the correct answer produced from the following formula can be drawn up: -a (n-1²).
Algebraic working
‘X’ being the given number, ‘n’ being the square of the length of one of the squares and ‘a’ being the varying size of the Master grid.
Using the given rule, this gives us:
= {x[x+ (n-1) +a (n-1)]} – {[x+ (n-1)] [x+a (n-1)]}
Taking the out ‘{ }’s leave…
= [x(x+n-1+an-a)] – [(x+n-1) (x+an-a)]
Removing the ‘[ ]’leaves…
= (x²+nx-x+anx-ax) – (x²+anx-ax+nx+anm-an-x-an+a)
Extracting the last brackets leaves…
= - (anm-an-an+a)
And again simplifies to…
=-anm+an+an-a
=-a (nm-2n+1)
This finishes as…
-a (n-1²)
This proves my hypothesis.
Finding a formula for an M*N grid on an A*A Master grid.
As explained in the end of the rectangles section we can include an ‘m’ by removing an ‘n’. In this formula ‘m’ equals width, ’n’ equals the amount of rows down, ‘a’ equals the size of the Master grid and x equals the starting value on the grid.
= {x[x+ (m-1) +a (n-1)]} – {[x+ (m-1)] [x + a (n-1)]}
= [x(x+m-1+an-a)] – [(x+m-1) (x + an-a)]
= (x²-x+mx+anx-ax) – (x²+anx-ax+mx+anm-am-x-an+a)
= - (anm-am-an+a)
= - anm+am+an-a
= -a (nm-m-n+1)
= -a (n-1) (m-1)
This was achieve by replacing the 10, which corresponds to the size of the Master grid, with an ‘a’ using the working form rectangle. If we replace the 10 with the ‘a’s it doesn’t affect the formula it only means that everywhere a 10 appears we see an ‘a’ instead.
This formula is identical to the rectangular one on a 10*10 Master grid but with an ‘a’ instead of an -10. This is because it is the same but we have an ‘a’ representing the size of the master grid which in the 10*10 master grid was 10. But because on squares they have an equal length/width we don’t need an ‘m’ to represent the width. Leaving the formula as –a (n-1²) which means – (size of the Master grid) * (length-1²) which is similar to -10(n-1²).
