= 52-1 + 52+1 + 52-10 + 52+10
= 4x52
= 208
If X=16 then
[(16-1) + (16+1)] + [(16-10) + (16+10)]
= 16-1 + 16+1 + 16-10 + 16+10
= 4x16
= 64
This tells me that whenever I use the above formula and replace X with any number (apart from the outside edge numbers) I always get 4xX as an answer. Therefore this is a master formula for this shape and this grid.
c)
[(X+g) - (X-g)] – [(X+1) - (X-1)]
= [X+g – X +g] – [X+1 – X+1]
= x +g –x +g –x-1 +x – 1
=2g – 2
If I replace X, I get the following results.
If X = 33 then
[(33+10) - (33-10)] – [(33+1) - (33-1)]
= [33+10 – 33+10] – [33+1 – 33+1]
= 2x10–2
=18
If X=83 then
[(83+10) - (83-10)] – [(83+1) - (83-1)]
= [83+10 – 83+10] – [83+1 – 83+1]
= 2x10–2
= 18
If X=49
[(49+10) - (49-10)] – [(49+1) - (49-1)]
= [49+10 – 49+10] – [49+1 – 49+1]
= 2x10–2
= 18
This tells me that whenever I use the above formula, the solution is 2g – 2, hence I always get 18 as an answer. Therefore this is a master formula for this shape and this grid.
I will now use a different grid size to prove the
above formulas with grid size 6x10
a)
[(X-1) (X+1)] – [(X+g) (X-g)]
= (X²-1) – (X²-g²)
= X²-1-x²+g²
= g²-1
If I replace the centre number, I get the following results.
If X=13 then
[(13-1) (13+1)] – [(13+6) (13-6)]
= (13²-1) – (13²-6²)
= 13²-1-13²+6²
= 6²-1
= 35
If x=77 then
[(77-1) (77+1)] – [(77+6) (77-6)]
= (77²-1) – (77²-6²)
= 77²-1-77²+6²
= 6²-1
= 35
If x=36 then
[(36-1) (36+1)] – [(36+6) (36-6)]
= (36²-1) – (36²-6²)
= 36²-1-36²+10²
= 6²-1
= 35
This tells me that whenever I use the above formula, the solution is g²-1, hence I always get 35 as an answer. Therefore this is a master formula for this shape and this grid.
b)
[(X-1) + (X+1)] + [(X-g) + (X+g)]
= X-1 + X+1 + X-g + X+g
= 4x
If I replace the centre number, I get the following results.
If X=87 then
[(87-1) + (87+1)] + [(87-6) + (87+6)]
= 87-1 + 87+1 + 87-6 + 87+6
= 4x87
= 348
If X=52 then
[(52-1) + (52+1)] + [(52-6) + (52+6)]
= 52-1 + 52+1 + 52-6 + 52+6
= 4x52
= 208
If X=16 then
[(16-1) + (16+1)] + [(16-6) + (16+6)]
= 16-1 + 16+1 + 16-6 + 16+6
= 4x16
= 64
This tells me that whenever I use the above formula, the solution is always 4X. and if I replace X with any number (apart from the outside edge numbers) I always get 4xX as an answer.
Therefore this is a master formula for this shape both the grid sizes 10x10 and 6x10.
c)
[(X+g) - (X-g)] – [(X+1) - (X-1)]
= [X+g – X+g] – [X+1 – X+1]
= 2g – 2
If I replace X, I get the following results.
If X = 33 then
[(33+6) - (33-6)] – [(33+1) - (33-1)]
= [33+6 – 33+6] – [33+1 – 33+1]
= 2x6–2
=10
If X=83 then
[(83+6) - (83-6)] – [(83+1) - (83-1)]
= [83+6 – 83+6] – [83+1 – 83+1]
= 2x6–2
= 10
If X=49
[(49+6) - (49-6)] – [(49+1) - (49-1)]
= [49+6 – 49+6] – [49+1 – 49+1]
= 2x6–2
= 10
This tells me that whenever I use the above formula , the solution is always 2g – 2, hence I always get 10 as an answer.
This is a master formula for this shape and this grid.
Proving above formulas with grid size 4x10
[(X-1) (X+1)] – [(X+g) (X-g)]
= (X²-1) – (X²-g²)
= X²-1-x²+g²
= g²-1
If I replace the centre number, I get the following results.
If X=13 then
[(13-1) (13+1)] – [(13+4) (13-4)]
= (13²-1) – (13²-4²)
= 13²-1-13²+4²
= 4²-1
= 15
If x=77 then
[(77-1) (77+1)] – [(77+4) (77-4)]
= (77²-1) – (77²-4²)
= 77²-1-77²+4²
= 4²-1
= 15
If x=36 then
[(36-1) (36+1)] – [(36+4) (36-4)]
= (36²-1) – (36²-4²)
= 36²-1-36²+4²
= 4²-1
= 15
This tells me that whenever I use the above formula, the solution is g²-1, Hence I always get 15 as an answer. Therefore this is a master formula for this shape and this grid.
[(X-1) + (X+1)] + [(X-g) + (X+g)]
= X-1 + X+1 + X-g + X+g
= 4x
If I replace the centre number, I get the following results.
If X=87 then
[(87-1) + (87+1)] + [(87-4) + (87+4)]
= 87-1 + 87+1 + 87-4 + 87+4
= 4x87
= 348
If X=52 then
[(52-1) + (52+1)] + [(52-4) + (52+4)]
= 52-1 + 52+1 + 52-4 + 52+4
= 4x52
= 208
If X=16 then
[(16-1) + (16+1)] + [(16-4) + (16+4)]
= 16-1 + 16+1 + 16-4 + 16+4
= 4x16
= 64
This tells me that whenever I use the above formula, the solution is always 4X ,and if I replace X with any number (apart from the outside edge numbers) I always get 4xX as an answer.
Therefore this is a master formula for this shape, and grid sizes 10x10, 6x10 and 4x10.
c)
[(X+g) - (X-g)] – [(X+1) - (X-1)]
= [X+g – X+g] – [X+1 – X+1]
= 2g – 2
If I replace X, I get the following results.
If X = 33 then
[(33+4) - (33-4)] – [(33+1) - (33-1)]
= [33+4 – 33+4] – [33+1 – 33+1]
= 2x4–2
=6
If X=83 then
[(83+4) - (83-4)] – [(83+1) - (83-1)]
= [83+4 – 83+4] – [83+1 – 83+1]
= 2x4–2
= 6
If X=49
[(49+4) - (49-4)] – [(49+1) - (49-1)]
= [49+4 – 49+4] – [49+1 – 49+1]
= 2x4–2
= 6
This tells me that whenever I use the above formula, the solution is always 2g – 2, hence I always get 4 as an answer. Therefore this is a master formula for this shape and this grid.
Overall so far, the only formula that works for all three grid sizes (4x10, 6x10 and 10x10) to give the same answer is:
[(X-1) + (X+1)] + [(X-g) + (X+g)]
= X-1 + X+1 + X-g + X+g = 4x (addition formula)
All other formulas work but give a different figure but the same algebraic solution.
For the subtraction formula the solution is g²-1 and for multiplication it is 2g-2.
Dsfsd
Fsd
Fsd
F
Sdf
Sd
F
Sdf
Dsf
Sdf
Proving that all Algebraic formulas work with all grid sizes
Grid size 4x10:
If X=10 and g=4
The number above x is X-g because 10-4=6
The number below x is X+g because 10+4=14 .
The number to the left is (X-1) because
14-1=13.
The number to the right is (X+1) because 14+1=15.
Grid size 8x10
If X=10 and g=4
The number above x is X-g because 20-8=12 and g=8.
The number below x is X+g: 20+8=28 and g=8
The number to the left is (X-1) because 20-1=19.
Grid size 6x10
If X=39 and g=6
The number to the right is (X+1) because 20+1=21
The number above x is X-g because 39-6=33 and g=6.
The number below x is X+g because 39+6=45 and g=6.
The number to the left is (X-1) because 39-1=38.
The number to the right is (X+1) because 39+1=40.
This shows that these
are the master formulas for all values of (g) with this shape.
To show that this works, If X=8 for g=4 then:
The number above x is X-g because 10-4=6
The number below x is X+g because 10+4=14.
The number to the left is (X-1) because 14-1=13.
The number to the right is (X+1) because 14+1=15.
This was found to be correct.
Furthering Investigation
To further my investigation, I will now use a cross shape instead of plus shape as shown below.
Prediction
X=center number.
g=numbers across ahorizontal axis.
e.g.
g=10
I predict that whichever grid size I use the:
number above left always = X- (g+1),
number above right always = X-(g-1),
number on the bottom left always = X+(g-1)
and number on the bottom right always = X+(g+1).
Therefore I predict this is a master formula for every number I pick.
Each formula in the saquares will give you the number depending on what X and g equal.
Algebraic Investigation
I am going to investigate 3 different formulas for a 10x10 grid size using the assigned cross shape.
Subtraction
(c-d) -(b-a)
[{X+(g+1)}- {X-(g+1)}] – [{X+(g-1)}-{X-(g-1)}]
= [X+g+1 – (x-g-1)] – [x+g-1 -(x-g+1)]
= X+g+1-X+g+1-X-g+1+X-g+1
=4
Addition:
(a+c+b+d)
[X-(g-1)] + [(X+(g+1)] +[ (X+(g-1)] +[ (X-(g+1)]
= (X-g+1) + (X+g+1) + (X+g-1) + (X-g-1)
= X-g+1+X+g+1+X+g-1+X-g-1
=4X
Multiplication
ab-cd
{X-(g-1)} {X+(g-1)} – {X+(g+1)} {X-(g+1)}
= [X²-(g-1)²] – [X²-(g+1)²]
= X²-(g-1)²-X²+ (g+1)²
= (g+1)² - (g-1)²
= g²+2g+1-g²+2g-1
=4g
Proving the above formulas with grid size 10x10
Subtraction
if X=68
[68+(10+1)- 68-(10+1)] – [68+(10-1)-(68-(10-1)
= [68+10+1 – (68-10-1)] – [68+10-1(68-10+1)]
= 68+10+1-68+10+1-68-10+1+68-10+1
=4
if X=72
[72+(10+1)- 72-(10+1)] – [72+(10-1)-(72-(10-1)
= [72+10+1 – (72-10-1)] – [72+10-1(72-10+1)]
= 72+10+1-72+10+1-72-10+1+72-10+1
=4
If X=28
[28+(10+1)- 28-(10+1)] – [28+(10-1)-(28-(10-1)
= [28+10+1 – (28-10-1)] – [28+10-1(28-10+1)]
= 28+10+1-28+10+1-28-10+1+28-10+1
=4
This tells me that whenever I use the above formula for different values of X the solution is always 4 no matter what the grid size. Therefore this is a master formula.
Sds
Sd
Sd
Addition
- If X=85
(85-(10-1) + (85+(10+1) + (85+(10-1) + (85-(10+1)
= (85-10+1) + (85+10+1) + (85+10-1) + (85-10-1)
= 85-10+1+85+10+1+85+10-1+85-10-1
=4x85
=340
- If X=22
(22-(10-1) + (22+(10+1) + (22+(10-1) + (22-(10+1)
= (22-10+1) + (22+10+1) + (22+10-1) + (22-10-1)
= 22-10+1+22+10+1+22+10-1+22-10-1
=4x22
= 88
- If X=35
(35-(10-1) + (35+(10+1) + (35+(10-1) + (35-(10+1)
= (35-10+1) + (35+10+1) + (35+10-1) + (35-10-1)
= 35-10+1+35+10+1+35+10-1+35-10-1
=4x35
=140
This tells me that whenever I use the above formula for different values of X the solution is 4X.
Multiplication
- If X=78
{78-(10-1)} {78+(10-1)} – {78+(10+1)} {78-(10+1)}
= [78²-(10-1)²] – [78²-(10+1)²]
= 78²-(10-1)²-78²+ (10+1)²
= (10+1)² - (10-1)²
= 10²+2x10+1-10²+2x10-1
=4x10
=40
c) If X=32
{32-(10-1)} {32+(10-1)} – {32+(10+1)} {32-(10+1)}
= [32²-(10-1)²] – [32²-(10+1)²]
= 32²-(10-1)²-32²+ (10+1)²
= (10+1)² - (10-1)²
= 10²+2x10+1-10²+2x10-1
=4x10
=40
- If X=54
{54-(10-1)} {54+(10-1)} – {54+(10+1)} {54-(10+1)}
= [54²-(10-1)²] – [54²-(10+1)²]
= 54²-(10-1)²-54²+ (10+1)²
= (10+1)² - (10-1)²
= 10²+2x10+1-10²+2x10-1
=4x10
=40
This tells me that whenever I use the above formula for different values of X the solution is 4g.
Gfh
Ghfg
Fghgf
Hfgh
Fgh
Gfh
G
G
G
G
G
G
G
g
Proving Algebraic formula works with all grid sizes
Grid size 6x10
If X=14
The top left will always be X-(g+1) because 14-(6+1) = 7
The top right will always be X-(g-1) because 14-(6-1)= 9
The bottom left will always equal X+(g-1) because 14+(6-1)=19
The bottom right will equal X+(g+1) because 14+(6+1)=21
Grid size 8x10
If X=14
The top left will always be X-(g+1) because 14-(8+1) = 5
The top right will always be X-(g-1) because 14-(8-1)= 7
The bottom left will always equal X+(g-1) because 14+(8-1)=21
The bottom right will equal X+(g+1) because 14+(8+1)=23
Grid size 4x10
If X=14
The top left will always be X-(g+1) because 14-(4+1) = 9
The top right will always be X-(g-1) because 14-(4-1)= 11
The bottom left will always equal X+(g-1) because 14+(4-1)=17
The bottom right will equal X+(g+1) because 14+(4+1)=19
This shows that these
are the master formulas for all grid size variations with this shape.
Conclusions:
In this investigation I found several different master formulas and one Universal formula. These are summarised in the table below.
g= grid size , X= center number of a cross
A master formula is a formula that works for a specific shape on all the three grid sizes ( 10 x 10, 6 x10 and 4 x 10) that I investigated.
A universal formula is a formula that can be used for all shapes and all grid sizes that I investigated.
I would have liked to further my investigation by using a three dimensional grid.
