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.
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
Aug3-08, 07:33 PM
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
Aug3-08, 10:10 PM
Well, I guess you don't like mathematics at all.
ramsey2879
Aug7-08, 05:51 PM
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.
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
Aug7-08, 10:57 PM
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
Aug8-08, 06:38 PM
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.
Interesting, I had never seen this.
Upon transformation it is obvious, but is still a nice curiosity.
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example:
11172325 = 9999 * (1117-2325) + 23251117
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Let Rn = n-digit Repunit (so R4=1111)
If a and b are n-digit positive integers,
9..9 = 9 * (1..1) = 9 * Rn
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general formula:
ab = (9*Rn) * (a - b) + ba
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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
Sep21-08, 10:54 PM
example:
11172325 = 9999 * (1117-2325) + 23251117
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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
Sep22-08, 05:01 AM
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
Sep22-08, 06:18 PM
Dodo Re: Dissected Number Law
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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
Sep22-08, 08:49 PM
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