I shall try to use the formula:
(N+3)x(N+30)-N(N+33) = 90
(N2+30N+3N+90)-(N2+33N) = 90
N2+33N+90-N2-33N = 90
Therefore 90 = 90
5x5 boxes
(55x91)-(51x95) = 160
(26x62)-(22x66) = 160
Lorenzo Brusini
(10x46)-(6x50) = 160
(40x76)-(36x80) = 160
Proving algebraically the 5x5 boxes:
The answers for 5x5 boxes all answer to 160. I shall try to use the algebraic box once again:
I shall try to use the formula:
(N+4)x(N+40)-N(N+44) = 160
(N2+40N+4N+160)-(N2+44N) = 160
N2+44N+160-N2-44N = 160
Therefore 160 = 160
NxN boxes
If we say that N=1 then 2= N+1, 11 would equal N+10 and 12 would equal N+11. Because we can say that the difference between 2 and 1 multiplied, with 11 and 1 multiplied, is equal to the difference in the square. So the difference between 1 and 2 and 10 and 2 can be expressed in the formula, in a 10x10 grid:
Where: N= the smallest number
X = Length of the square
Lorenzo Brusini
I shall put this into the formula I used before for all of the other square grids:
(N+10(X-1)(N+(X-1))-(N(N+10(X-1)+X-1))
(N+10X-10)(N+X-1)-(N(N+10X-10+X-1)
N2+XN-N+10XN+10X2-10X-10N-10X+10-N2-10XN+10N-XN+N
10X2-20X+10
This can be factorised to make:
10(X-1)2
I shall try this formula to make sure it is correct. I shall use it for each type of box:
Looking at the results of the formula it shows that it works and can be used for any square in a 10x10 grid.
Rectangles
Introduction: for my second variable I shall investigate boxes that are not the same in size for each length.
2x3 boxes:
(40x48)-(38x50) = 20
(55x63)-(53x65) = 20
(4x12)-(2x14) = 20
(85x93)-(83x95) = 20
Lorenzo Brusini
Proving algebraically the 2x3 boxes:
The answers for 2x3 boxes all answer to 20. I shall try to use the algebraic box once again where n is the smallest number in the box:
I shall try to use the formula:
(N+2)x(N+10)-N(N+12) = 20
(N2+10N+2N+20)-(N2+12N) = 20
N2+12N+20-N2-12N = 20
Therefore 20 = 20
2x4 boxes:
(15x22)-(12x25) = 30
(20x27)-(17x30) = 30
(68x75)-(65x78) = 30
(45x52)-(42x55) = 30
Proving algebraically the 2x4 boxes:
The answers for 2x4 boxes all answer to . I shall try to use the algebraic box once again where n is the smallest number in the box:
I shall try to use the formula:
(N+3)x(N+10)-N(N+13) = 30
(N2+10N+3N+30)-(N2+13N) = 30
N2+13N+30-N2-13N = 30
Therefore 30 = 30
Lorenzo Brusini
2x5 boxes:
(18x24)-(14x28) = 40
(60x66)-(56x70) = 40
(49x55)-(45x59) = 40
(90x96)-(86x100) = 40
Proving algebraically the 2x5 boxes:
The answers for 2x5 boxes all answer to 40. I shall try to use the algebraic box once again where n is the smallest number in the box:
I shall try to use the formula:
(N+4)x(N+10)-N(N+14) = 40
(N2+10N+4N+40)-(N2+14N) = 40
N2+14N+40-N2-14N = 40
Therefore 40 = 40
3x4 boxes:
(46x63)-(43x66) = 60
(75x92)-(72x95) = 60
(7x24)-(4x27) = 60
(36x53)-(33x56) = 60
Lorenzo Brusini
Proving algebraically the 3x4 boxes:
The answers for 3x4 boxes all answer to 60. I shall try to use the algebraic box once again where n is the smallest number in the box:
I shall try to use the formula:
(N+3)x(N+20)-N(N+23) = 60
(N2+20N+3N+60)-(N2+23N) = 60
N2+23N+60-N2-23N = 60
Therefore 60 = 60
3x5 boxes:
(58x74)-(54x78) = 80
(17x33)-(13x37) = 80
(6x22)-(2x26) = 80
(40x56)-(36x60) = 80
Proving algebraically the 3x5 boxes:
The answers for 3x5 boxes all answer to 80. I shall try to use the algebraic box once again where n is the smallest number in the box:
Lorenzo Brusini
I shall try to use the formula:
(N+4)x(N+20)-N(N+24) = 80
(N2+20N+4N+80)-(N2+24N) = 80
N2+24N+80-N2-24N = 80
Therefore 80 = 80
4x5 boxes:
(47x73)-(43x77) = 120
(26x53)-(22x57) = 120
(70x96)-(66x100) = 120
(39x65)-(35x69) = 120
Proving algebraically the 4x5 boxes:
The answers for 4x5 boxes all answer to 120. I shall try to use the algebraic box once again where n is the smallest number in the box:
I shall try to use the formula:
(N+4)x(N+30)-N(N+34) = 120
(N2+30N+4N+120)-(N2+34N) = 120
N2+34N+120-N2-34N = 120
Therefore 120 = 120
Lorenzo Brusini
AxB Boxes:
I have figured out the formula for the rectangles by looking at the difference between the top left number, the top right number and the bottom left are directly proportional to the bottom right number. This means that if I take a random box it will follow like this:
(19-14) + (44-14) = d
d = the difference between the top left number and bottom right number.
19-14 = 5
44-14 = 30
30+5 = 35
49-14 = 35
Now this is only for this specific box so we can make this into an algebraic formula:
Where: A = The horizontal side
B = The vertical side
N = The smallest number in the box
I shall put this box into the formula I used for all the other boxes:
(N+A-1)(N+10(B-1))-N(N+10(B-1)+A-1)
=(N+A-1)(N+10B-10)-N(N+10B-10+A-1)
=N2+10BN-10N+AN+10AB-10A-N-10B+10-N2-10BN+10N-AN+N
= 10AB-10A-10B+10
This can be factorised to make:
10(A-1)(B-1)
Lorenzo Brusini
I shall show that this formula works by using it on every rectangular box:
Looking at the results of the formula it shows that it works and can be used for any rectangle in a 10x10 grid.
Different size grids
Introduction: For my third variable I will change the sizes of the grid from which I take the boxes. I shall take 5 different types of grids that will be: 6x6, 7x7, 8x8, 9x9. I shall take 2 types of square boxes that will be 2x2 and 4x4, and two types of rectangle boxes that will be 3x4 and 4x5 for every type of grid.
6x6 grids:
(16x21)-(15x22) = 6
(12x27)-(9x30) = 54
(24x33)-(21x36) = 36
(6x20)-(2x24) = 72
Lorenzo Brusini
Proving algebraically the 6x6 grids:
(N+1)x(N+6)-N(N+7) = 6
(N2+1N+6N+6)-(N2+7N) = 6
N2+7N+6-N2-7N = 6
Therefore 6 = 6
(N+3)x(N+18)-N(N+21) = 54
(N2+3N+18N+54)-(N2+21N) = 54
N2+21N+54-N2-21N = 54
Therefore 54 = 54
(N+3)x(N+12)-N(N+15) = 36
(N2+3N+12N+36)-(N2+15N) = 36
N2+15N+36-N2-15N = 36
Therefore 36 = 36
(N+4)x(N+18)-N(N+22) = 72
(N2+4N+18N+72)-(N2+22N) = 72
N2+22N+72-N2-22N = 72
Therefore 72 = 72
7x7 grids:
(33x39)-(32x40) = 7
(14x32)-(11x35) = 63
(33x44)-(30x47) = 42
(27x44)-(23x48) = 84
Lorenzo Brusini
Proving algebraically the 7x7 grids:
(N+1)x(N+7)-N(N+8) = 7
(N2+1N+7N+7)-(N2+8N) = 7
N2+8N+7-N2-8N = 7
Therefore 7 = 7
(N+3)x(N+21)-N(N+24) = 63
(N2+3N+21N+63)-(N2+24N) = 63
N2+24N+63-N2-24N = 63
Therefore 63 = 63
(N+3)x(N+14)-N(N+17) = 42
(N2+3N+14N+42)-(N2+17N) = 42
N2+17N+42-N2-17N = 42
Therefore 42 = 42
(N+4)x(N+21)-N(N+25) = 84
(N2+4N+21N+84)-(N2+25N) = 84
N2+25N+84-N2-25N = 84
Therefore 84 = 84
8x8 grids:
(24x31)-(23x32) = 8
(24x45)-(21x48) = 72
(22x35)-(19x38) = 48
(16x36)-(12x40) = 96
Lorenzo Brusini
Proving algebraically the 8x8 grids:
(N+1)x(N+8)-N(N+9) = 8
(N2+1N+8N+8)-(N2+9N) = 8
N2+9N+8-N2-9N = 8
Therefore 8 = 8
(N+3)x(N+24)-N(N+27) = 72
(N2+3N+24N+72)-(N2+24N) = 72
N2+24N+72-N2-24N = 72
Therefore 72 = 72
(N+3)x(N+16)-N(N+19) = 48
(N2+3N+16N+48)-(N2+19N) = 48
N2+19N+36-N2-19N = 48
Therefore 48 = 48
(N+4)x(N+24)-N(N+28) = 96
(N2+4N+24N+96)-(N2+28N) = 96
N2+28N+96-N2-28N = 96
Therefore 96 = 96
9x9 grids:
(42x50)-(41x51) = 9
(41x65)-(38x68) = 81
(9x24)-(6x27) = 54
(14x37)-(10x41) = 108
Lorenzo Brusini
Proving algebraically the 9x9 grids:
(N+1)x(N+9)-N(N+10) = 9
(N2+1N+9N+9)-(N2+10N) = 9
N2+10N+9-N2-10N = 9
Therefore 9 = 9
(N+3)x(N+27)-N(N+30) = 81
(N2+3N+27N+81)-(N2+30N) = 81
N2+30N+81-N2-30N = 81
Therefore 81 = 81
(N+3)x(N+18)-N(N+21) = 54
(N2+3N+18N+54)-(N2+21N) = 54
N2+21N+54-N2-21N = 54
Therefore 54 = 54
(N+4)x(N+27)-N(N+31) = 108
(N2+4N+27N+108)-(N2+31N) = 108
N2+31N+108-N2-31N = 108
Therefore 108 = 108
AxB grid:
Where: A = The horizontal side of the box,
B = The vertical side of the box,
N = The smallest number in the box,
G = The size of the grid.
I shall put this box into the formula I used for all the other boxes:
(N+A-1)(N+G(B-1))-N(N+1G(B-1)+A-1)
=(N+A-1)(N+GB-G)-N(N+GB-G+A-1)
=N2+GBN-GN+AN+GAB-GA-N-GB+G-N2-GBN+GN-AN+N
= GAB-GA-GB+G
This can be factorised to make:
G(A-1)(B-1)
Lorenzo Brusini
I shall show that this formula works by using it on every rectangular box:
Looking at the results of the formula it shows that it works and can be used for any square or rectangle in any size grid.
Lorenzo Brusini