Symmetry of Digit "2" In Squaring

If we square 11, it is very simple put 1(2*1) (12) get 121 same as square 12 put 1(2*2)(22) get 144 again for 13 we get 169 and for 14 we get 1 8 16=196 and so on.

When we go deep, we find that there is symmetry of two types

(2, 4, 6, 8, 10, 12, 14, 16, 18, 20 …. Diff is always 2) &

(1, 4 , 9 , 16, 25, 36, 49, 64, 81, 100 ) diff. is 3 5 7 9 11 13 15 17 19 and diff. of 3 5 7 9 11 always 2, so there is true symmetry .

Up to 19 it is right but at 20 how we can put 1 20 100 just because of symmetry.

 

• 112 = 1 2 1

• 122 = 1 4 4

• 132 = 1 6 9

• 142 = 1 8 16 = 100 + 80 + 16 = 196

• 152 = 1 10 25 = 100 + 100 + 25 = 225

• 162 = 1 12 36 = 100 + 120 + 36 = 256

• 172 = 1 14 49 = 100 + 140 + 49 = 289

• 182 = 1 16 64 = 100 + 160 + 64 = 324

• 192 = 1 18 81 = 100 + 180 + 81 = 361

• 202 = 1 102 = 1 20 100 = 100 + 200 + 100 = 400

• 212 = 1 112 = 1 22 121 = 100 + 220 + 121 = 441

• 222 = 1 122 = 1 24 144 = 100 + 240 + 144 = 484

• 232 = 1 132 = 1 26 169 = 100 + 260 + 169 = 529

• 242 = 1 142 = 1 28 196 = 100 + 280 + 196 = 576

• 252 = 1 152 = 1 30 225 = 100 + 300 + 225 = 625

• 262 = 1 162 = 1 32 256 = 100 + 320 + 256 = 676

• 272 = 1 172 = 1 34 289 = 100 + 34 + 289 = 729

• 282 = 1 182 = 1 36 324 = 100 + 360 + 324 = 784

• 292 = 1 192 = 1 38 361 = 100 + 380 + 361 = 841

• 302 = 1 202 = 1 40 400 = 100 + 400 + 400 = 900

 

There is a symmetry, a method, which is shown as below for 31, 41, 51 and so on.

• 312 = 1 212 = 1 42 (1 11)2 = 1 42 (1 22 121)

= 961 = 121 + 220 + 100

= 441 + 420 + 100 = 961

• 412 = 1 312 = 1 62 (1 21)2 = 162 (1 42) (1 11)2

= 1 62 (1 42) (1 22 121)

= 1 62 (961) = 961 + 620 + 100 = 1681

• 512 = 1 412 = 1 82 (1 31)2 = 1 82 (1 62) (1 21)2

= 1 82 (1 62) (1 42) (1 11)2

= 1 82 (1 62) (1 42) (1 22 121)

= 1 82(1681) = 1681 + 820 + 100 = 2601

• 612 = 1 512 = 1 102 (41)2 = 1 102 (1 82) (1 31)2

= 1 102 (1 82) (1 62) (1 21)2

= 1 102 (1 82) (1 62) (1 42) (1 11)2

= 1 102 (1 82) (1 62) (1 42) (1 22 121)

= 1 102 (2601) = 2601 + 1020 + 100 = 3721

• 712 = 1 612 = 1 122 (1 51)2 = 1 122 (1 102) (41)2

= 1 122 (1 102) (1 82) (1 31)2 = 1 122 (1 102) (1 82) (1 62) (1 21)2

= 1 122 (1 102) (1 82) (1 62) (1 42)(1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 22 121)

= 3721 + 1220 + 100 = 5041

• 812 = 1 712 = 1 142 (1 612) = 1 122 (1 51)2 = 1 122 (1 102) (41)2

= 1 122 (1 102) (1 82) (1 31)2 = 1 122 (1 102) (1 82) (1 62) (1 21)2

= 1 122 (1 102) (1 82) (1 62) (1 42)(1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 22 121)

= 5041 + 1420 + 100 = 6561

• 912 = 1 812 = 1 162(1 712 ) = 1 142 (1 612) = 1 122 (1 51)2

= 1 122 (1 102) (41)2 = 1 122 (1 102) (1 82) (1 31)2

= 1 122 (1 102) (1 82) (1 62) (1 21)2

= 1 122 (1 102) (1 82) (1 62) (1 42)(1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 22 121)

= 6561 + 1620 + 100 = 8281

• 1012 = 1 912 = 1 182(1 812) = 1 162(1 712 ) = 1 142 (1 612) = 1 122 (1 51)2

= 1 122 (1 102) (41)2 = 1 122 (1 102) (1 82) (1 31)2

= 1 122 (1 102) (1 82) (1 62) (1 21)2

= 1 122 (1 102) (1 82) (1 62) (1 42)(1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 11)2

= 1 122 (1 102) (1 82) (1 62) (1 42) (1 22 121)

= 8281 + 1820 + 100 = 10201

Edited December 21, 20169 yr by mathspassion

Sorry but I haven't read to the end of your longish post but immediately I would say that this is merely an artifact of multiplication and nothing deeper:

 

011 x

011

----

001 (units multiplied)

010 (top unit times lower tens)

010 (lower unit time top tens)

100 (tens multiplied)

===

121

[Leading zeros added to make it align]

 

You will always get the pattern of the unit squared + the tens time unit doubled + the tens squared

 

ie

019 x

019

----

081

090

090

100

===

361

Without having put much thought into it, is the symmetry you have discovered not independent of the base you are working in? So it's the same for octal or hexadecimal, or any base greater than 2.

It is also due to the "binomial theorem":
[math](x+ y)^n= \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} x^ky^{n-k}[/math]
Where [math]\begin{pmatrix}n \\ k\end{pmatrix}[/math] is the "binomial coefficient", [math]\frac{n!}{k!(n-k)!}[/math].
With x= y= 1 that becomes [math]2^n= \sum_{k=0}^\infty \begin{pmatrix}n \\ k\end{pmatrix}[/math]
and clearly the binomial coefficient is "symmetric":
[math]\begin{pmatrix}n \\ k \end{pmatrix}= \frac{n!}{k!(n-k)!}= \frac{n!}{(n- (n-k))!(n-k)!}= \begin{pmatrix}n \\ n-k\end{pmatrix}[/math]

Edited December 21, 20169 yr by Country Boy

yes know but if we go deep we get something new..........

 

now your turn....... find more.........

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