I am going to try and work out the formula by at first seeing what would happen if n was 1:
So the final outcome would be:
1x12=12
2x11= 22
22-12= 10
This can be proved with other 2 x 2 boxes. For example:
34 x 45= 1530
44 x 35= 1540
15- 1530 = 10
Equation.
To find n and prove that any number can be selected and its 2 x 2 box can be calculated I can use this method which I have developed in to a formula so any number can be found.
Therefore:
n(n+11) = x
(n+10)(n+1)= y
y– x = 10
PROVE AND TEST using the number 28 as n.
28(28+11)
28 x 39 = 1092
(28+10)(28+1)
38 x 29 = 1102
1102-1092= 10
Simplify.
(n +10)(n +1)= y
I have switched the two separate
Parts of the equation so that y will come
first in the final calculation which will
make the subtraction easier.
n (n+11)= x
y – X = 10
y= n + 10(n+1) – x= n (n+11)
test: n = 13
y = 13+10 (13+1) – x= 13(13+11)
y = 23 x 14 – x = 13 x 24
y = 322 – x = 312
322 – 312 = 10
The formula to work out the product of a 2 x 2 box is:
y= n + 10(n+1) – x= n (n+11)
3 x 2 boxes.
I will now use a similar method to prove that a 2 x 3 box will always produce 20.
The numbers highlighted in yellow are the numbers that I need to use.
14 x 35 = 490
34 x 15 = 510
510 – 490 =20
(This box is a different way round so I will expand my boxes across instead of down to make an equation although both = 20)
45 x 57 = 2565
55 x 47 = 2585
2585 – 2565 = 20
This proves that any randomly selected number is likely to produce 20. For my research I also did the numbers 92, 41 and 7. I will use the same method so the top left hand box will be n.
X= n(n+2)
Y= n+10(n+12)
Y – x
Simplify:
y= n + 10(n+2) – x= n (n+12) = 20
Test and prove.
Y= 32+10(32+2) – x= 32(32+12)
Y= 42 x 34 – x = 32 x 44
Y=1428 – x = 1408
1428-1408= 20
My formula is proved correct.
2 x 4 boxes.
For the past two formulas I have used a similar method and they have both been correct so I predict that for 2 x 4 boxes the equation will be:
y= n + 10(n+3) – x= n (n+13) = 30
Test and prove.
Y= 32 +10(32+3) – x = 32(32+13) = 30
Y= 42 x 35 - x =32 x 45
Y = 1470 – x = 1440
1470 – 1440 = 30
My predicted formula was correct.
Equation for all box sizes and numbers.
I would now like to calculate a formula that would mean I would not need to work out an equation for each different box size.
These are the equations I have so far:
2 x 2 boxes:
y= n + 10(n+1) – x= n (n+11) =10
2 x 3 boxes:
y= n + 10(n+2) – x= n (n+12) = 20
2 x 4 boxes:
y= n + 10(n+3) – x= n (n+13) = 30
These equations are all very similar so I want to work out if there is a formula for which to devise them.
I will call the size of the box B. for example 2 x B will be the number that one wishes to use to multiply with 2 to make the box size.
The part of the equation which needs to change is:
n+3/2/1 and n+13/12/11
I have noticed that this number will always be one less than the number that is ‘xing’ by 2 to create the box.
So, the part of the formula could be:
n+(B-1) and the other part could be n+ (B+9)
therefore the formula would be:
y= n+10(n+(B-1)) – x = n (n+(B+9))
So assuming n was 7 and B was 3 the equation would be:
y= 7+10(7+ (3-1)) – x = 7 (7 + (3+9))
y= 17 (7+2) – x = 7 (7+12)
y= 17 x 9 – x =7 x 19
y= 153 – x= 133
153- 133 = 20
This proves that my formula correctly shows that one can work out the total of a box of whichever size from just one equation.
7 x 19 = 133
17 x 9 = 153
153 – 133= 20.
