10(b-1) ² this is my new formula.
I will test this formula on two box sizes I already have the results for:
E.g. 3x3 and 4x4 box size (see page 2).
10(3-1) ² = 40
10(4-1) ² =90 my new formula works.
I predict that the 6x6 square in a 10x10 grid will be 250 by using this formula:
-
10(b-1) ²
-
=10(6-1) ²
- =10x25
- =250
My prediction is right.
I predict that the result for an 8x8 square in a 10x10 grid will be 490 by using this formula:
-
10(b-1) ²
-
=10(8-1) ²
- =10x49
- =490
My prediction is right.
I predict that the result for a 10x10 square in a 10x10 grid will be 810 by using this formula:
-
10(b-1) ²
-
=10(10-1) ²
- =10x81
- =810
My prediction is right.
Conclusion
I found a new formula which will find the difference of the two opposing corners on a 10x10 grid for any square shape. Now that I have worked out a quadratic formula for the squares on a 10x10 grid, I can investigate further to see if I can work out a formula for a different sized number grid. I will have to use the same process as before.
Firstly I am going to start with an 8x8 grid and pick up 4 different squares and I will start with the 2x2 square. Then I move on and use the 3x3, 4x4 and the 5x5.
I have noticed that the products difference of 2x2 squares in an 8x8 grid equal to 8. I predict if I move the 2x2 square down I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+1)(n+8)=n²+8+9n
n(n+9)=n²+9n
Products difference is equal to (n²+8+9n) – (n²+9n) =8
In the same grid I will now work out a 3x3 square.
I have noticed that the products difference of 3x3 squares in an 8x8 grid equal to 32. I predict if I move the 3x3 square down I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+2)(n+16)=n²+32+18n
n(n+18)=n²+18n
Products difference is equal to (n²+32+18n) – (n²+18n) =32
In the same grid I will now work out a 4x4 square.
I have noticed that the products difference of 4x4 squares in an 8x8 grid equal to 72. I predict if I move the 4x4 square down I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+3)(n+24)=n²+72+27n
n(n+27)=n²+27n
Products difference is equal to (n²+72+27n) – (n²+27n) =72
I have put my results in a table and I am now going to try to predict the 5x5 square in an 8x8 grid.
8, 32, 72, 128,
+24 +40 +56
+16 +16
nth term= 8n²
The n is not the box size because for example if I put the 3x3 square in an 8x8 grid I will get 72. Unfortunately, this formula does not work but if I minus the box size by one I will then get 32 which is the right answer.
8(b-1) ² this is my new formula.
I will test this formula on two box sizes I already have the results for:
8(2-1)2 = 8
8(3-1)2 =32 my new formula works.
I predict that the 5x5 square in an 8x8 grid will be 128 by using this formula:
-
8(b-1) ²
-
=8(5-1) ²
- =8x16
- =128
My prediction is right.
I predict that the result for a 7x7 square in an 8x8 grid will be 288 by using this formula:
-
8(b-1) ²
-
=8(7-1) ²
- =8x36
- =288
My prediction is right.
I predict that the result for an 8x8 square in an 8x8 grid will be 392 by using this formula:
-
8(b-1) ²
-
=8(8-1) ²
- =8x49
- =392
My prediction is right.
I found a new formula which will find the difference of the two opposing corners on an 8x8 grid for any box size. I also found a new formula which will find any box size in any number grid which is:
g (b-1) ²
g = the grid size so if I want to find the formula of:
-
10x10 grid = 10(b-1) ²
-
8x8 grid = 8(b-1) ²
I will choose at random from previous work and apply this formula.
E.g. 5x5 box size in a 10x10 grid (see page 3).
E.g. 3x3 box size in an 8x8 grid (see page 7).
My new formula works. Now I am going to try this formula on a 7x7 grid and see if it works.
E.g. 4x4 box size in a 7x7 grid:
This proves that the new formula works.
Conclusion
I have found a new formula which will find the difference of the two opposing corners in any box size and in any number grid.
Now that I have worked out these new formulas, I can extend my investigating further to see if I can work out a formula for a rectangle shape in a 10x10 grid. I will have to use the same process as before, although it is slightly different.
I am going to start with a 10x10 grid and pick up 4 different rectangles and I will start with the 2x3 rectangle. Then I move on and use the 3x4, 4x5 and the 5x6.
I have noticed that the products difference of 2x3 rectangles in a 10x10 grid equal to 20. I predict if I move the 2x3 square to the left and down I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+1)(n+20)=n²+20+21n
n(n+21)=n²+21n
Products difference is equal to (n²+20+21n) – (n²+21n) =20
In the same grid I will now work out a 3x4 rectangle.
I have noticed that the products difference of 3x4 rectangles in a 10x10 grid equal to 60. I predict if I move the 3x4 rectangle up I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+2)(n+30)=n²+60+32n
n(n+32)=n²+32n
Products difference is equal to (n²+60+32n) – (n²+32n) =60
In the same grid I will now work out a 4x5 rectangle.
I have noticed that the products difference of 4x5 rectangles in a 10x10 grid equal to 120. I predict if I move the 4x5 square up, I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+3)(n+40)=n²+120+43n
n(n+43)=n²+43n
Products difference is equal to (n²+120+43n) – (n²+43n) =120
In the same grid I will now work out a 5x6 rectangle.
I have noticed that the products difference of 5x6 rectangles in a 10x10 grid equal to 200. I predict if I move the 5x6 rectangle to the left I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+4)(n+50)= n²+200+54n
n(n+54)=n²+54n
Products difference is equal to (n²+200+54n) – (n²+54n) =200
I have put my results in a table and I am now going to try to predict the 6x7 rectangle in a 10x10 grid.
I am going to look again to my box sizes and how I got the results.
I found a new formula which will find the difference of the two opposing corners in a 10x10 grid for any rectangle shape which is:
(c-1) x (d-1) x10
I am going to try this formula on the 6x7 rectangle in a 10x10 grid and see if it works.
-
By using this formula:(c-1)x(d-1)x10
- 5x6x10
- =300
My new formula works.
Now I will test this formula on two box sizes I already have the results for:
E.g. 3x4 box sizes (see page 12).
- 2x3x10
- =60 this formula works.
E.g. 4x5 box sizes (see page 13).
- 3x4x10
- =120 this formula works.
Conclusion
I found a new formula which will find the difference of the two opposing corners on a 10x10 grid for any rectangle shape. Now that I have worked out a quadratic formula for a rectangle on a 10x10 grid, I can investigate further to see if I can work out a formula for an 8x8 number grid. I will have to use the same process as before.
I am going to start with an 8x8 grid and pick up 4 different rectangles and I will start with the 2x3 rectangle. Then I move on and use the 3x4, 4x5 and the 5x6.
I have noticed that the products difference of 2x3 rectangles in an 8x8 grid equal to 16. I predict if I move the 2x3 rectangle down I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+2)(n+24)=n²+48+26n
n(n+26)=n²+26n
Products difference is equal to (n²+48+26n) – (n²+26n) =48
In the same grid I will now work out a 3x4 rectangle.
I have noticed that the products difference of 3x4 rectangles in an 8x8 grid equal to 48. I predict if I move the 3x4 rectangle up and to the right I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+2)(n+30)=n²+60+32n
n(n+32)=n²+32n
Products difference is equal to (n²+60+32n) – (n²+32n) =60
In the same grid I will now work out a 4x5 rectangle.
I have noticed that the products difference of 4x5 rectangles in an 8x8 grid equal to 96. I predict if I move the 4x5 rectangle down I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+3)(n+32)=n²+96+35n
n(n+35)=
Products difference is equal to (n²+96+35n) – (n²+35n) =96
In the same grid I will now work out a 5x6 rectangle.
I have noticed that the products difference of 5x6 rectangles in an 8x8 grid equal to 160. I predict if I move the 5x6 rectangle up I will get the same answer.
My prediction is right. I am going to use algebra to test my results.
(n+4)(n+40)= n²+160+44n
n(n+44)=n²+44n
Products difference is equal to (n²+160+44n) – (n²+44n) =160
I found a new formula which will find the difference of the two opposing corners in an 8x8 grid for any rectangle shape which is:
(c-1) x (d-1) x8
I also found a new formula which will find any rectangle shape in any number grid which is:
(c-1) x (d-1) x g
g = the grid size so if I want to find the formula of:
- 10x10 grid = (c-1) x (d-1) x10
- 8x8 grid = (c-1) x (d-1) x8
I will choose at random from previous work and apply it to this formula:
E.g. 5x6 box size in a 10x10 grid (see page 14).
This new formula works.
E.g. 3x4 box sizes in an 8x8 grid (see page 17).
The new formula works.
Now I am going to try this formula on a 7x7 grid and see if it works.
E.g. 5x6 box size in a 7x7 grid:
This proves that the new formula works.
Conclusion
I found a new formula which will find any rectangle shape in any number grid.
In this project I hoped to extend the investigation further in order to find more interesting patterns, which can be relevant to the task.