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| 11 MAY 2004 at 3:16pm |
| Deleted User | http://www.justadventure.com/cgi-bin/yabb/YaBB.cgi?board=GeneralTrivia;action=display;num=1063321015
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| 11 MAY 2004 at 4:57pm |
| Deleted User | What? There's no proof of the above statement in that thread.
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| 11 MAY 2004 at 5:16pm |
| Deleted User | I just meant that they were related. I thought you might not have seen the other thread and you would find it interesting..
*sigh*
That's what I get for posting. Sorry. :-[
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| 11 MAY 2004 at 6:01pm |
| Deleted User | More like that's what you get for being cryptic. What you get being a question. It's not like I chewed your head off.
I did find it interesting, though, so thanks.
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| 11 MAY 2004 at 6:15pm |
MichalNGrand Inquisitor
Posts : 7058
Joined: 14 SEP 2003
Status : Online | Originally Posted By Unity (11 MAY 2004 12:43pm)
Why?
First let's suppose that 0 is "evenly divisible" by any integer to get rid of edge cases (like 1 or 222)...
A 2-digit number can be represented as 10x + y where x and y are the digits. If we swap the digits and subtract the numbers, we'll have
10x + y - (10y + x) = 9x - 9y
lo and behold, the result will be divisible by 9.
For numbers with more digits, we are only interested in the swappable ones (the others will just cancel out on subtraction), and those can be represented as (10^a)x + (10^b)y, where a and b are non-negative integers, and a > b. Now the generic equation is
(10^a)x + (10^b)y - ((10^a)y - (10^b)x) = x(10^a - 10^b) - y(10^a - 10^b)
so the question is, will 10^a - 10^b always be divisible by 9?
Since a > b, the answer is yes. If a is, say, 4, then b can be 3, 2, 1 or 0; 10^a is 10^4 is 10000, and 10^b can be either of 1000, 100, 10 or 1. The numbers 9000, 9900, 9990 and 9999 are all divisible by 9.
I forgot my sig.
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| 11 MAY 2004 at 6:24pm |
| Deleted User | That's it!
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| 11 MAY 2004 at 6:36pm |
ElfstoneGuild Master
Posts : 5892
Joined: 4 NOV 2002
Status : Online | Ah BJ, it's THAT thread!
I'm still wondering about that. I think I've seen some sort of explanation recently for why those numbers will loop or narrow down to one "magic" number eventually.
Petter told me I need to prove that. Sorry, I can't. But it looks very likely. Of course, that's no proof at all.
[b]playing[/b]: Destination Treasure Island (done in two sittings, but it's nice), Syberia (ho-hum), Dracula: Last Sanctuary (on hold)&&[b]reading[/b]: even more study papers&&[b]listening to[/b]: [url=http://www.last.fm/user/Brax82/]this and that[/url], plus [url=http://www.musicovery.com/]Musicovery[/url]&&[b]TV favorites[/b]: (currently) Pushing Daisies, Chuck, Journeyman (cancelled! grrr...), Heroes&& all-time) 24, Stargate SG1, X-Files, Lost, House
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| 13 MAY 2004 at 5:04pm |
| Deleted User | Hehe, I have tried myself without much luck. These sort of things usually recovers some deeper knowledge of algebra. You know, rings and fields, prime number theory, combinatorics and such things... I mean, I've studied these things, but I understand the definitions and not much more. It takes an intuitive understanding and quite a bit of talent to solve these sort of problems by yourself. Many similar problems, easy to understand, are extremely tough to solve and challenges the best mathematicians in the world. Stuff that is of this nature is still largely not fully understood. Maybe there's an easy solution in that case, but I haven't found it.
An interesting note: Some Indian mathematicians recently proved (it may have been last year, I'm not sure) that prime factorization of numbers can be achieved in polynomial time in the size of the number. The degree of the polynomial is very high (something like the order n^12) and the process is very advanced, but it is still somewhat worrying considering that the difficulty of primality testing this is a foundation for modern chryptology. A computer can catch up with this computational bound. Of course, the size of the number is easy to increase to reduce the risk, but it would be desirable to find something with a higher asymptotical bound, like 2^n or n!
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