A Numerical Black Hole

D

DeMan06

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Jul 3, 2009
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Alright... so I was reading during a meeting the other day and came across this interesting phenomenon.

Pick any number (Ex 9,745,229)

Write down first, the number of even digits, then the number of odd digits, then the total number of digits
(for above example: 347 for the above number; 3 even digits, 4 odd, 7 total digits)

Do the same thing for this new number
(so now it would be 123 again, 1 even, 2 odd, 3 total)

Do it again... and you'll find you again get 123... this is a blackhole, you cannot escape 123 by doing the same process; and you can get there with ANY number if you repeat this process, usually only a couple of times unless the number is extremely large; then you have to repeat it a few extra times... but you'll always get sucked into the 123 black hole.
 
Well, the process you are using is somewhat rigged. It could be easily explained with modular arithmetic, and it may just be a property of base 3 number systems.

What is constant is that after the first iteration you get always a 3 digit number. So that gives you the 3 at the end of 123.

Consider numbers of 7 digits, therefore after the first iteration you get an odd number always since it ends in 7. And since 7 is odd itself all you really need to look is at one of the digits. So you get a variation of 347 or 437. Say i pick for example 2,222,222. Then you get 707. If you do another iteration you get 023 or 123 if you consider 0 to be even, although 0 does not have a parity since all numbers divide 0. So your conjecture ends there or you have to say that on numbers of all even digits or all odd digits you have 023 not 123. Take also 9,999,999. Then you get 077, which becomes 023 again.

To get even fancier if you have a 7 digit number then you can have:

  • 7 even 0 odd. These numbers iterate down to 023. 707 becomes 023 which becomes 113 on another iteration, and 033 on another and 023 on another and cycles between 033 and 023.
  • 6 even 1 odd. You get 617 which becomes 123.
  • 5 even 2 odd. You get 527 which becomes 123.
  • 4 even 3 odd. You get 437 which becomes 123.
  • 3 even 4 odd. You get 347 which becomes 123.
  • 2 even 5 odd. You get 257 which becomes 123.
  • 1 even 6 odd. You get 167 which becomes 123.
  • 0 even 7 odd. These numbers iterate down to 023. 077 becomes 023 and same as above.


Take an 8 digit number.

  • 8 even 0 odd. These numbers iterate down to 023. 808 becomes 203 which becomes 113 which becomes 033 and 023 and 113 and 033.
  • 7 even 1 odd. You get 718 which becomes way 123.
  • 6 even 2 odd. You get 628 which becomes 303 and 023 then 113 then 033 then 023 then 113.
  • 5 even 3 odd. You get 538 which becomes 123 as above.
  • 4 even 4 odd. You get 448 which becomes 303 as above.
  • 3 even 5 odd. You get 358 which becomes 123 as above.
  • 2 even 6 odd. You get 268 which becomes 303 as above.
  • 1 even 7 odd. You get 178 which becomes 123 as above.
  • 0 even 8 odd. These numbers iterate down to 088. 088 becomes 203 and 113 and 033 and 023 and 113.

So the conjecture holds for numbers of odd digit length and that are not all even digits or all odd digits. For numbers of even digit length it holds less than half the time, in the case of 8 digits 5/4, and in general gets more complicated and you have to start looking at powers of 2 which divide the number since for 4 digit numbers you get a ratio that is reversed than for 8 since 2^3 divides 8 and 2^2 divides 4.

It is a fun exercise. Could play with this for a few hours without bringing in advanced mathematics. More advanced math could kill this problem quite fast.

If you like this get Euler's Elements of Algebra. The footnotes are full of tricks like this, and most likely including this one.

[ame]http://www.amazon.com/Elements-Algebra-Leonhard-Euler/dp/1899618791/ref=sr_1_9?ie=UTF8&s=books&qid=1247980463&sr=8-9[/ame]
 
Last edited:
Frank Booth said:
Too much pwnage to repost
Click to expand...

I think I need some extra ultra industrial strength aspirin to get over the headache I got from reading this.

I think this still won't be enough
Asprin+Paperweight.jpg
 
Frank Booth said:
Well, the process you are using is somewhat rigged. It could be easily explained with modular arithmetic, and it may just be a property of base 3 number systems.

What is constant is that after the first iteration you get always a 3 digit number. So that gives you the 3 at the end of 123.

Consider numbers of 7 digits, therefore after the first iteration you get an odd number always since it ends in 7. And since 7 is odd itself all you really need to look is at one of the digits. So you get a variation of 347 or 437. Say i pick for example 2,222,222. Then you get 707. If you do another iteration you get 023 or 123 if you consider 0 to be even, although 0 does not have a parity since all numbers divide 0. So your conjecture ends there or you have to say that on numbers of all even digits or all odd digits you have 023 not 123. Take also 9,999,999. Then you get 077, which becomes 023 again.

To get even fancier if you have a 7 digit number then you can have:

  • 7 even 0 odd. These numbers iterate down to 023. 707 becomes 023 which becomes 113 on another iteration, and 033 on another and 023 on another and cycles between 033 and 023.
  • 6 even 1 odd. You get 617 which becomes 123.
  • 5 even 2 odd. You get 527 which becomes 123.
  • 4 even 3 odd. You get 437 which becomes 123.
  • 3 even 4 odd. You get 347 which becomes 123.
  • 2 even 5 odd. You get 257 which becomes 123.
  • 1 even 6 odd. You get 167 which becomes 123.
  • 0 even 7 odd. These numbers iterate down to 023. 077 becomes 023 and same as above.


Take an 8 digit number.

  • 8 even 0 odd. These numbers iterate down to 023. 808 becomes 203 which becomes 113 which becomes 033 and 023 and 113 and 033.
  • 7 even 1 odd. You get 718 which becomes way 123.
  • 6 even 2 odd. You get 628 which becomes 303 and 023 then 113 then 033 then 023 then 113.
  • 5 even 3 odd. You get 538 which becomes 123 as above.
  • 4 even 4 odd. You get 448 which becomes 303 as above.
  • 3 even 5 odd. You get 358 which becomes 123 as above.
  • 2 even 6 odd. You get 268 which becomes 303 as above.
  • 1 even 7 odd. You get 178 which becomes 123 as above.
  • 0 even 8 odd. These numbers iterate down to 088. 088 becomes 203 and 113 and 033 and 023 and 113.

So the conjecture holds for numbers of odd digit length and that are not all even digits or all odd digits. For numbers of even digit length it holds less than half the time, in the case of 8 digits 5/4, and in general gets more complicated and you have to start looking at powers of 2 which divide the number since for 4 digit numbers you get a ratio that is reversed than for 8 since 2^3 divides 8 and 2^2 divides 4.

It is a fun exercise. Could play with this for a few hours without bringing in advanced mathematics. More advanced math could kill this problem quite fast.

If you like this get Euler's Elements of Algebra. The footnotes are full of tricks like this, and most likely including this one.

http://www.amazon.com/Elements-Alge...=sr_1_9?ie=UTF8&s=books&qid=1247980463&sr=8-9
Click to expand...
o____________0
 
....wow sgt buzzkiller. thanks.... lol joking, thats a pretty cool little thing your found.... makes my head hurt from reading it...or was it the smoothie im drinking...idk.
 
Well... 0 is generally taken to be an even number which is why what I said holds...

The normal definition of an even number (n) is n=2k; where K is an integer. Since 0 is an integer, and 2*0 = 0; 0 must be even.
 
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