Dissected Number Law: Math Poetry & Formula

[Antuan]

HELLO ! THIS IS MY FIRST POST HERE.


I spend countless hours playing with numbers and found a general formula to describe the following examples:

777666 = 999 (777-666) + 666777

2318 = 99 (23-18) + 1823

439 = 99 (4-9) + 934

57392673 = 9999 (5739-2673) + 26735739

As you can see, the general idea is to split the number apart right through the middle and redefine it with its dissected parts. I called it Dissected Number Law, since all positive integers can be re-expressed like this.

I also call it Mathematical Poetry, since these expressions have a " rhyming " feel to them.

443223 = 999 (443-223) + 223443

3962972522222222 = 99999999 ( 39629725 - 22222222 ) + 2222222239629725

Id like your comments on this and if you want to check the formula, view my youtube video at:

http://www.youtube.com/v/yMraoZZhzZ0

**In the video, my mathematical language could have been the usual, but it would probably take me longer to explain it.

 

[Kittel Knight]

I spend countless hours playing with numbers and found a general formula to describe the following examples:
...
Id like your comments on this...

Sincerely? ...:yuck:
Well, I guess you wouldn't like my comments...

 

[Antuan]

Well, I guess you don't like mathematics at all.

 

[ramsey2879]

HELLO ! THIS IS MY FIRST POST HERE.


I spend countless hours playing with numbers and found a general formula to describe the following examples:

777666 = 999 (777-666) + 666777

2318 = 99 (23-18) + 1823

439 = 99 (4-9) + 934

57392673 = 9999 (5739-2673) + 26735739

As you can see, the general idea is to split the number apart right through the middle and redefine it with its dissected parts. I called it Dissected Number Law, since all positive integers can be re-expressed like this.

I also call it Mathematical Poetry, since these expressions have a " rhyming " feel to them.

443223 = 999 (443-223) + 223443

3962972522222222 = 99999999 ( 39629725 - 22222222 ) + 2222222239629725

Id like your comments on this and if you want to check the formula, view my youtube video at:

http://www.youtube.com/v/yMraoZZhzZ0

**In the video, my mathematical language could have been the usual, but it would probably take me longer to explain it.

I think it is a cute number play, at least on even par with many other little tibits of multiplication and addition that I seen.

 

[Focus]

I think if you spend the time you spent on this well you could have solved the RH. It looks pretty neat though but I am not really a fan of number playing like this.

 

[Antuan]

I think it is a cute number play, at least on even par with many other little tibits of multiplication and addition that I seen.

Thank you ! Other than cute, maybe, just maybe there is a use for this simple concept. One little tibit of multiplication and addition turns out to be the "method to complete the square", so could you imagine that sometimes this little ideas can solve big problems. Who knows ? Its simply another insight...some sort of "factorization" that applies to ALL NUMBERS >0 using ROWS OF 9's.

1234567890987654321 = 9999999999 (123456789-987654321) + 9876543210123456789

Has anyone said it before ?

 

[alphachapmtl]

Interesting, I had never seen this.
Upon transformation it is obvious, but is still a nice curiosity.
--------------------------------
example:
11172325 = 9999 * (1117-2325) + 23251117

--------------------------------
Let Rn = n-digit Repunit (so R4=1111)
If a and b are n-digit positive integers,
9..9 = 9 * (1..1) = 9 * Rn
--------------------------------
general formula:
ab = (9*Rn) * (a - b) + ba
--------------------------------

proof: (ab and ba are juxtapositions)
ab = a *10^n +b
ab = a * (1+ 9*Rn) + b
ba = b *10^n +a
ba = b * (1+ 9*Rn) + a
(ab - ba) = a * (1+ 9*Rn) + b - b * (1+ 9*Rn) - a
(ab - ba) = (a - b) * (1+ 9*Rn) + (b - a)
(ab - ba) = (a - b) * (1+ 9*Rn) - (a - b)
(ab - ba) = (a - b) * (1+ 9*Rn - 1)
(ab - ba) = (a - b) * (9*Rn)
(ab - ba) = (9*Rn) * (a - b)
ab = (9*Rn) * (a - b) + ba
--------------------------------

 

[Antuan]

example:
11172325 = 9999 * (1117-2325) + 23251117

--------------------------------
Let Rn = n-digit Repunit (so R4=1111)
If a and b are n-digit positive integers,
9..9 = 9 * (1..1) = 9 * Rn
--------------------------------
general formula:
ab = (9*Rn) * (a - b) + ba

Your proof is nice, but could a more conventional mathematical proof be provided ?

 

[dodo]

It's just simple algebra. If one number is of the form

a . 10^n + b​

and the other is

b . 10^n + a​

then the difference between the two is

a . 10^n + b - b . 10^n - a
= 10^n . (a - b) - (a - b)
= (10^n - 1) . (a - b)​

 

[Antuan]

Dodo Re: Dissected Number Law

--------------------------------------------------------------------------------
It's just simple algebra. If one number is of the form
a . 10^n + b
and the other is
b . 10^n + a
then the difference between the two is
a . 10^n + b - b . 10^n - a
= 10^n . (a - b) - (a - b)
= (10^n - 1) . (a - b)

I don't see the connection. So what would be your general formula for any positive integer X defined with this concept ? i.e., considering X a number composed of two parts...X = ?

REMEMBER ALL POSITIVE INTEGERS CAN BE REDEFINED THIS WAY:
93 = 9 (9-3) + 39

PLUS WE COULD ALSO CHANGE SIGNS TO SAY:
93 = 11(9+3) - 39

***MAKE SURE IT APPLIES TO LARGE NUMBERS***
EXAMPLE: 483501 = 999 (483-501) + 501483

COULD THERE BE A CONVENTIONAL WAY TO EXPRESS THIS GENERAL FORMULA ?

 

[dodo]

I called the two parts "a" and "b". In your example with 483501, the two parts are a=483 and b=501, and the whole number is

a . 1000 + b​

or

a . 10^3 + b​

using 10^3 to represent "10 raised to the 3rd power", which is 1000.

The reversed number would be

b . 1000 + a = 501483.​

And the difference between the two, 483501 - 501483, would be (copying from the last line in post #9),

(10^3 - 1) . (a - b)
= (1000 - 1) (483 - 501)
= 999 (483 - 501).​

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

So, in this "vocabulary", a formula for the entire expression would read

a . 10^n + b = (10^n - 1) (a - b) + b . 10^n + a​

of which a particular case is

483501 = 999 (483 - 581) + 501483​

 

[Antuan]

Does anybody know how to describe with algebra a number X such that it represents the first half of the digits that compose any integer Y ? For examp. let's say Y=3245 then X should be 32. Could you make such a formula that applies to all numbers ?

*Please, without extrange "vocabulary"...simply algebra

 

[alphachapmtl]

if the number Y has 2*n digits, X=int(Y/(10^n))
where 2*n=1+int(log(Y))
so X=int(Y/(10^((1+int(log(Y)))/2)))
here int() is the integer part, and log is in base 10

 

[Antuan]

if the number Y has 2*n digits, X=int(Y/(10^n))
where 2*n=1+int(log(Y))
so X=int(Y/(10^((1+int(log(Y)))/2)))
here int() is the integer part, and log is in base 10

Thanks Alpha, but could you add an example, say Y=4356 ...
...please pluck the numbers in your equation and demonstrate how it works...

 

[alphachapmtl]

if the number Y has 2*n digits, X=int(Y/(10^n))
where 2*n=1+int(log(Y))
so X=int(Y/(10^((1+int(log(Y)))/2)))
here int() is the integer part, and log is in base 10

let Y=4356
then #digits=2*n=1+int(log(4356))=1+int(3.63908787)=1+3=4
so 2*n=4, so n=2
then X=int(Y/(10^n))=int(4356/100)=int(43.56)=43

 

[Antuan]

You did it ! That's fantastic Alpha ! Now THATS what I'm talking about...real algebra !

Just one last thing...What modification should that equation have for it to define another variable, say Z, such that it represents the last half of the digits that compose any integer Y ?

like in your demonstration if Y=4356 then this Z should be 56...

 

[alphachapmtl]

if we know Y=4356 and X=43, then Z=4356-(43*100)=56, so Z=Y-(X*10^n)

Y has 2n digits, X has n digits, Z has n digits
Y=XZ (juxtaposition, not product)
int() is the integer part
log is in base 10

if the number Y=XZ has 2*n digits, X=int(Y/(10^n))
where 2*n=1+int(log(Y))
so n = (1+int(log(Y)))/2

so X = int(Y/(10^n))
so X = int(Y/(10^((1+int(log(Y)))/2)))
and Z = Y - ( X * (10^n) )
so Z = Y - ( int(Y/(10^n)) * (10^n) )
so Z = Y - ( int(Y/(10^((1+int(log(Y)))/2))) * (10^((1+int(log(Y)))/2)) )

written in full, this looks terribly complicated but it really isn't.
Basically, we use log(N) to count the digits of N
log(10)=1, log(100)=2, log(1000)=3, etc
so 1<= log(10 to 99) <2
and 2<=log(100 to 999) <3
and ...
so #digit(N)=int(1+log(N))=1+int(log(N))
in our problem, N==Y and #digit(N)==2*n

 

[Antuan]

Beautiful !
Alphachampmtl, it has been a pleasure to experience your talent.

 

[Antuan]

Ok. Now that Alphachapmtl has contributed with his ideas, I can now show a simple example in plain algebra of how we can have fun with the "dissected number phenomena" demonstrating just one of its conclusions. Here it goes:

(Y + Y/10^n) (X/(10^n) - X) / (Z/(10^2n) - Z) = (Y/10^n) (X/10^n) / (Z/10^2n) = XY/Z

where both Y & X have 2n digits
and Z has 4n digits
2n = 1 + int (log(x)) or 2n = 1 + int (log(y))
here int() is the integer part, and log is in base 10


THE FOLLOWING EXAMPLE IS SIMPLY ONE OF ITS INTERPRETATIONS...

(4356+5643)(1312-1213)/(65234779-47796523) = (43+56)(13-12)/(6523-4779)

Using juxtaposition's language we could say that A=43 B=56 C=13 D=12 E=6523 F=4779
so this shows in this example that: (AB+BA)(CD-DC)/(EF-FE) = (A+B) (C-D) / (E-F)

express your opinions...

 

[alphachapmtl]

(AB+BA)(CD-DC)/(EF-FE) = (A+B) (C-D) / (E-F)
Wow, that look neat!

(1234+3412) * (5678-7856) / (12345678-56781234) = (12+34) * (56-78) / (1234-5678)

(1234+3412) * (5678-7856) / (12345678-56781234) = 0.227722772..
(12+34) * (56-78) / (1234-5678) = 0.227722772..

 

[Antuan]

Alpha, I would certainly like to personally thank you and keep your contact in some way, my email is gabstudio@gmail.com and for anybody that would like to personally make any comments, you are all welcomed.

 

[HallsofIvy]

I think that was the problem everyone (like Kittel Knight) had from the start: you seem to equate "mathematics" with "arithmetic" and know very little about algebra. alphachapmtl gave you a very simple formula for your question and you immediately challenged him to apply to a specific number. Why did you not do it yourself? What alphachapmtl said at first was "real algebra". What he did in response to your challenge was arithmetic.

 

[LorenzoMath]

cool ,man! you know numbers, especially prime numbers, do sing! at least, according to a certain "celebrated" mathematician I know.

To see the general pattern, write, for example, 9999 as 10000-1 and use distribution law and stuff and calculate. you can see how the first four digits and the last four digits are canceled and/or add up.

 

[alphachapmtl]

A reference:
The USSR Olympiad Problem Book: Selected Problems and Theorems of Elementary Mathematics (Paperback)
by D. O. Shklarsky (Author), N. N. Chentzov (Author), I. M. Yaglom (Author)

https://www.amazon.com/dp/0486277097/?tag=pfamazon01-20

This book has some problems (with solutions) concerning arithmetic and digits of integers.
See problems 15-26 on page 11 and page 102.

 

[AmberCam]

HELLO ! THIS IS MY FIRST POST HERE.I spend countless hours playing with numbers and found a general formula to describe the following examples:

Great video demo!

This is great fun and you find mathematical secrets too!
See what poetry you can make of this.

From 4356

(4 + 3) + (5 + 6) = 18 = 1 + 8 = 9
(7) + (11) = 18
18 = 1 + 8 = 9

4 + 3 = 7

5 + 6 = 11
11 = 1+1 = 2

Ans: 7 + 2 = 92 + 4 + 6 + 8 = 20 = 2 + 0 = 2

2 + 4 = 6

6 + 8 = 14 = 1 + 4 = 5

Ans: 6 + 5 = 11
11 = 1 + 1 = 2

Posted in a hurry - hope no errors!

 

[alphachapmtl]

Interesting! Related to "digital roots" and "Casting out nines"

http://en.wikipedia.org/wiki/Digital_root
http://en.wikipedia.org/wiki/Casting_out_nines

 

[Antuan]

I think the best formula devised to explain this number phenomena was given to me by a friend in the following manner:

let M be the number of digits of A, and N be the number of digits of B.

AB = A*(10^N) + B = A*(1 + 10^N - 1) + B*(1 + 10^M - 10^M) = (B*10^M + A) + A*(10^N - 1) - B*(10^M - 1) = BA + A*(10^N - 1) +B*(1-10^M)

if N = M, that is A and B have same number of digits then
AB = BA + (10^N - 1)*(A-B)