Here I have repeated the stages as before to prove my overall formula for the 3×3 inset squares.
After multiplying the corners of the 3×3 squares I then took the lowest away from the highest. This number is always 40.
a×(a+22) - (a+2) × (a+20)
[a²+22a+40]-[a²+22a]
a²+22a+40 -
a²+22a
0+ 0+ 40
=40
4×4-
41×74=3034
44×71=3124
3124-3034=90
7×40=280
10×37=370
370-280=90
1×34=34
4×31=124
124-34=90
67×100=6700
70×97=6790
6790-6700=90
Here I have repeated the stages as before to prove my overall formula for the 4×4 inset squares.
After multiplying the corners of the 4×4 squares I then took the lowest away from the highest. This number is always 90.
a×(a+33) = (a+3) × (a+30)
[a²+33a+90] -[a²+33a]
a²+33a+90 -
a²+33a
0+0+90
=90
To create my overall formula for the difference I will need to use letters to represent certain numbers. In this case n represents the width of the square and a represents the number in the top left hand corner of the square.
Using previous formulas I can see that in the first section:
a×(a+11(n-1)) – ((a+(n-1)×(a+10(n-1)))
This cancels down to
(a²+11a(n-1))-(a²+11a(n-1)+10(n-1)²)
To work out my overall formula I will subtract the first part of the Formula from the other:
a²+11a(n-1) -
a²+11a(n-1) )+10(n-1) ²
0 + 0 + 0 +10(n-1)²
=10(n-1)2
To check this formula I will place numbers into it and work it out.
a=55
n=2
55²+(11×55)(2-1)+10(2-1)² -
55²+(11×55)(2-1)
= 0 + 0 + 0 + 10 ×1²
10×1²=10
The number in the 2×2 squares was 10 so this proves my formula correct.
To extend my work I will create a different sized number grid and calculate a formula that will work for the difference in any sized square in any sized grid.
As in my original grid I have marked out individual squares and will now calculate the formula for them
2×2
1×9=9
2×8=16
16-9=7
6×14=84
7×13=91
91-84=7
25×33=825
26×32=832
832-825=7
36×44=1584
37×43=1591
1591-1584=7
41×49=2009
42×48=2016
2016-2009=7
a×(a+8)-(a+1)×(a+7)
[a²+8a+7]-[a²+8a]
a²+8a+7 -
a²+8a
0 +0 +7
=7
3×3
1×17=17
3×15=45
45-17=28
10×26=260
12×24=288
288-260=28
29×45=1305
31×43=1333
1333-1305=28
33×49=1617
35×47=1645
1645-1617=28
5×21=105
7×19=133
133-105=28
a×(a+16)-(a+2) ×(a+14)
[a²+16a+28]-[a²+16a]
a²+16a+28 -
a²+16a
0 + 0+28
=28
4×4
1×25=25
4×22=88
88-25=63
22×46=1012
25×43=1075
1075-1012=63
25×49=1225
28×46=1288
1288-1225=63
a×(a+24)-(a+3)×(a+21)
[a²+24a+63]-[a²+24a]
a2+24a+63 -
a²+24a
0+0+63
=63
Before creating an overall formula I will create a formula for my 7×7 grid. This will allow me to create my overall formula:
a×(a+8(n-1)) – ((a+(n-1)×(a+7(n-1)))
This cancels down to
(a²+8a(n-1))-(a²+8a(n-1)+7(n-1)²)
To work out my overall formula I will subtract the first part of the Formula from the other:
a²+8a(n-1)+7(n-1)² -
a²+8a(n-1)
0 + 0 + 0 + 7(n-1)²
=7(n-1)2
To check my formula I will place numbers in it:
a=5
n=7
5²+(5×8)(n-1)+7(n-1)²
5²+(5×8)(n-1)
0 + 0 + 0 + 7(n-1)²
7(n-1)² = 7×1²=7
The number in the 2×2 squares is 7 so this proves my formula correct.
As before I will create an overall formula for the difference to enable me to find an algebraic rule for any sized square in any sized number grid.
X=Top left hand square
n= Size of square
g= Size of grid
To find my final formula I will multiply the opposite corners like I have in all my previous calculations. I will then subtract the two formulas to give me my overall formula for the difference.
x×(x+(g+1)(n-1) - x2 + x(g+1)(n-1)
x2+xg(n-1) + x (n-1)
(x+(n-1)) × (x+g(n-1)
x2 + xg(n-1) + x(n-1)+g(n-1)2
x2 + xg(n-1)+x(n-1)+g(n-1)2
x2 + xg(n-1)+x(n-1) -
0 + 0 + 0 +g(n-1)2
Final formula is g(n-1)2
To check this I will now add some numbers into this formula:
g=7
n=3
x=4
7(3-1)2
3-1=2
22 = 4
7×4=28
4×20=80
6×18=108
108-80=28
I now know that my formula is correct.
To further my work I will now create a rectangle and find a general rule for the difference in any sized rectangle in any sized grid, following the same process as before.
2×3
1×32=32
2×31=62
62-32=30
29×60=1740
30×59=1770
1770-1740=30
65×96=6240
66×95=6270
6270-6240=30
39×70=2730
40×69=2760
2760-2730=30
I will now create an algebraic rule for the 2×3 rectangles.
a×(a+31) – (a+1) × (a+30)
a2+31a – a2+31a+30
a2+31a+30 –
a2+31a
0 + 0 + 30
=30
4×2
7×25=175
10×22=220
220-175=45
34×52=1768
37×49=1813
1813=1768=45
87×105=9135
90×102=9180
9180-9135=45
76×94=7144
79×91=7189
7189-7144=45
I will now create an algebraic rule for the 4×2 rectangles.
a×(a+18)-(a+3) ×(a+15)
a2+18a – a2+18a+45
a2+18a+45 –
a2+18a
0 + 0 + 45
I will now create a rule for the difference in any sized rectangle in any sized grid.
For this section
a represents the number in the top left hand corner of the rectangle
A represents the width of the grid
B represents the length of the grid
Y represents the width of the rectangle
Z represents the length of the rectangle.
a×(a+B(A-1)+(Z-1)) – a+(Z-1)) × (a+B(A-1))
a2+aB(A-1)+a(Z-1)+(Z-1)B(A-1)
a2+aB(A-1)+a(Z-1) –
0 + 0 + 0 +(Z-1)B(A-1)
=B(Z-1)(A-1)
=15(Z-1)(A-1)
Looking at this formula and my previous work with squared I can see that the general formula for the difference could be:
B(Z-1)(A-1)
I will now prove this by creating a 7×5 grid and using the same process before, make up a general rule for any sized rectangle in any sized grid.
2×3
1×16=16
2×15=30
30-16=14
18×33=594
19×32=608
608-594=14
a×(a+9) – (a+2)×(a+7)
a2+9a – a2+9a+14
a2+9a+14
a2+9a –
0+0+14
=14
6×2
15×27=405
22×20=440
440-405=35
22×34=748
29×27=783
783-748=35
a×(a+12) – (a+5)×(a+7)
a2+12a – a2+12a+35
a2+12a+35
a2+12a –
0+0+35
=14
I will now create my overall formula for any sized rectangle in any sized grid.
The letters represent the same as previously.
a×(a+B(A-1)+(Z-1)) – a+(Z-1)) × (a+B(A-1))
a2+aB(A-1)×a(Z-1)+(Z-1)B(A-1)
a2+aB(A-1)×+a(Z-1) –
0 + 0 + 0 +(Z-1)B(A-1)
=B(Z-1)(A-1)
Therefore I conclude that my overall formula for the difference is B(Z-1)(A-1). I will check this by inserting numbers into my formula.
B=7
Z=6
A=2
B(Z-1)(A-1)
7×(6-1)(2-1)
=7×(5×1)
=7×5=35
This proves my formula for the difference correct as in the 6×2 grids the difference was always 35.
