:)
Posted on 2002-05-08 18:14:35 by Brad
hi

please see no aggresivity in my messages =)
just that prime numbers are known since antiquity, so i doubt that you found some new way of seeing and thinking that apply to prime and fit so well the RSAP. both of us know you have no algorithm for efficient factorization. apparently, some people on this board, as brad, are quite convinced that you made some major discoveries. anyway, if you' re point was to make us think on a problem, and try to look at easy and simpliest or obvious solutions, that might be a way of thinking, anyway, life is hard, life is complex =) i don' t think this method would have a real impact on whatever. but well, you just come and say 'i made some major discovery, i won' t tell you how it works, i won' t show you evidence that it work, i will just mumble some illogical disconnected sentences on 11 and 17 and applause anyone that say anything on these numbers', so don' t take us for morons =)
Posted on 2002-05-08 20:20:33 by roy
In net destuctive tecniques it is called trolling.
What about some math training.
One can submit bullsh*t math theory (let it be easy for starter)
the others try to prove it wrong.
But the false theory must be submitted with some prove.
No more those "Great Ice" mistick statements :)

Bull**** math theory #1.
Lemma:
(True part)
IF
1
All composite numbers can be present as product of primes.
(Gauss - main arithm theorem)
2
All composite numbers can be present of product of to numbers
each one > 1.
(followed from 1.)
Then
(False part)
3
~One 3d of natural numbers are primes
Proof:
assume we have equaly odd and even numbers
(for integers at any end of monotonic seq numofeven = numofodd(-1) for natural numofeven = numofodd (+1))
(combinatoric bullsh*t part)
True part
Then if we randomly take 2 numbers to make product it has equal chance to be odd and even.
Now let's have a look when product of two numbers is odd and
when it's even


1st_mult 2d_mult product
odd odd odd
even even even
even odd even
odd even even

Convert it to true table let's odd = 1 even = 0
then we can convert the implication into comjunction (and)
(if even = 1 odd = 0 we get disjunction(or))


1st 2nd res
1 1 1
0 0 0
1 0 0
0 1 0


So only one multiplications of 4 possibilities gives us odd.
but we know that we have same number of odds as even
How the rest odds were produced.
Wow we get to bullsh*t discovery!!! I'm proud to put my
abrivated piece of nothing (c) here!

The rest are PRIMES!
Look again (we extend the tbl to make equal odds and primes)
:


1st 2d res
? ? 1
? ? 1
1 1 1
0 0 0
0 1 0
1 0 0


Now we can see that we have equal number of even and odds
but first two are not product of two numbers > 1.
The only explonation why he have them - they are primes
that can be represented in product only as X=X*1.
there are to cases of such in the table of 6
6\2 = 1\3
So 1 third of all existing integers are PRIMES.
The bullsh*t theory proved.

(Bullsh*t Discovery made By the Svin, under insperation of
this thread)

I'm sorry for my math English (nobody gave me primary math book in English yet so I could learn math expresions commands in it ;)
Now prove it wrong.
It's very easy for those who has knowlege at least of primary scool.



But I am not going into proving right or wrong of Giovanni statements.
Just 'cause he stated nothing.
Posted on 2002-05-08 22:36:57 by The Svin
That's Bullsh*t :grin:
.
.
.
.
.
.
.
.
.
Just kidding :)
Posted on 2002-05-08 22:43:22 by Brad
Of course,:)
But I supplied proof with it. In a way it supposed to talking of
math discoveries.
The question is if you can prove it wrong.
At least it is easier than prove wrong anything that wasn't said.
Posted on 2002-05-08 22:50:11 by The Svin
Alex, too true....

but think about how many similar studies were made about Fermat's.....some led to other things...and some didn't....but most
sharpened skills.....

:) B
Posted on 2002-05-08 22:56:14 by Brad
Brad,
I don't mind any theory including false ones.
Assume that I didn't state that my theory is false.
People could learn many things checking it math part.
'Cause it has it. And even though at the end they will prove that it is wrong they can make some usefull notions that can lead to something usefull.
But what notions you can make about what wasn't said.
All his theory ended to mistical constant which one can guess.
Nobody know how.
If one can guess the mistical constant - why the one can not guess just one of primes used to make product :)
Then other prime would be just devident of key and guessed prime :)

OK, I'm out :)
I not supposed to be involved in the first place.
Posted on 2002-05-08 23:10:10 by The Svin
Alex,

Yes.... I think I agree.....but, what's puzzling me is that....
we can remove question in the first digits...so we should be able to reduce the equation one digit....if so ... we can solve it... I think...It reminds me of Newtons, original discription of calculas....
the one where he explained how something is geometrically increasing while it is algebraically decreasing....I think ????

Brad
Posted on 2002-05-08 23:17:43 by Brad
Giovanni,

Lets try an easy one, like 36853, or 10147....

can you work it through?

B
Posted on 2002-05-08 23:29:40 by Brad
You see, false theory but with math part can lead to something.

As far as I know right conclusion for the above paradox wasn't
published yet, so if you write it you can make publication(I'm not supposed to help you - but good luck)
Hint - the result is very close to known statistics(~5th is p).

As to number of products of possible pares of monotonic part
natural row with even elements the q of all prdct\ prdct of odd
can be described as
F(n)=((2n(n+1)/(n/2(n/2+1))
I don't transformate it for clarity.
n is number of elements in part natural row
in part 2,3,4,5,6,7 n = 6

and the function behaves strange jumping as frog without
obvious correlation.
Posted on 2002-05-08 23:42:16 by The Svin



Your 17's design is impressive. Did you notice that you wrote down a lot of 17's, all the same, while eventually you could vary the 11 to 111 or 23 or something else. Did you notice you do actually multiply always the same number?

Now, when you pass to something more interesting, you'll see more relationships, since a multiple of a number is a sequence of the same number one below the other, and summed. 17 * 5:

17
17
17
17
17
---
??

Sorry I'm too tired to calc this :) Anyway, this is our *vague* constant, which becomes rapidly important as products grow. Let's say you get m which ends by 117809 (lifewire sample), the reverse engined last digit, despite of the 3*3 or 7*7 or 1*9, will be always 9. The second from right is 0, which teorically could come from 1+9, and if it does, the stage obove is the same as the one below, and their endings are 19. Do you see what I mean?

x19
xx19
------
xx09



Hmmm.... an early-out algorithm? To reject a possible pairing of primes by multiplying the first couple bits, then checking them immediately? So no other digits need to be multiplied? Probably not...

You know, I had this book once which sounded suspiciously like it was able to factor stuff easily... There were procedures there for adding and multiplying stuff by hand, faster than inputting them on calculator... and they all had something to do with the digits, the representation of numbers...

I have got an inkling of a suspicion of a possibility that I actually get what you're thinking....
Posted on 2002-05-09 00:17:03 by AmkG
The Svin is right, "Here's my theory, it's you that have to prove me wrong" is religion, not science.

Thanks God I'm atheist..
Posted on 2002-05-09 04:26:10 by Maverick
Hi, people,

today I have to work it all, I've no free time at all. I appreciate a lot your cowork, especially the doubting part. This evening I will print out all new messages, and check them out. I will have to code something, too (which I actually hate), to see what you meant (Svin). My brain is not so binary as it looks like :)

I want to point out some question which came into my mind yesterday. If we had a number ending by 5, usually we could say it will be divisible by 5, because our math experience of relatively small numbers tells us, it will do so. But could we be sure about it, if we have an extremely large number, such as going from my workstation to the sun, using an Arial 8 pt character and tight (no) spacing? No we can't, because something could happen to this number, which is actually out of our rational believes. It could be, for example, be bent into space and time. A sympatic guy told us this some time ago...

Roy pointed out why I would (could) work against my own interests. Science is not bound to personal interests. If someone found a way around his own believes, he will change his own needs and adapt. It might be all wrong, but at least he didn't fright off. But for me it's not even a question of courage. It's just and simply interesting to follow an input. For example, I didn't try many numbers, but some, enough to be sure it works, and I didn't spare myself. But you're right, too, because it is like lo play chess against yourself, you'll win always and you can choose which color will win :)

Giovanni
Posted on 2002-05-09 04:34:32 by sch.jnn
What about cracking a small key then? You can't say no to this.. otherwise a lot of people will really think you're trolling us.

No 2048 bit key.. a small one, just enough to prove yourself half right. Of course possibly not of 4 bit, anybody can do that with his mind.

But about your:
I want to point out some question which came into my mind yesterday. If we had a number ending by 5, usually we could say it will be divisible by 5, because our math experience of relatively small numbers tells us, it will do so. But could we be sure about it, if we have an extremely large number, such as going from my workstation to the sun, using an Arial 8 pt character and tight (no) spacing? No we can't, because something could happen to this number, which is actually out of our rational believes. It could be, for example, be bent into space and time. A sympatic guy told us this some time ago...

Stai davvero esagerando.. ti vuoi fare prendere per il culo da mezzo mondo?? ;(
A Giova'.. e daje 'n tajo... ;)
Posted on 2002-05-09 04:42:41 by Maverick
Actually, it can be proven algebraically, that a decimal integer that ends in 5 is divisible by 5. It helps to use sigma notation to handle arbitrarily large (but finite) numbers.

The rule doesn't work for hexadecimal integers. (Example: 15h = decimal 21.)
Posted on 2002-05-09 04:59:45 by tenkey
hi

you say maybe a big number ending by 5 (in base 10) could not be divisible by 5.... well, it' s easily provable that it' ll be divisible :
let A this number, (x1)(x2)...(xn)5 be the representation of the number in base 10, then A=5 + (xn)*10 + ... + (x1)*10^n, so we have A= 5*(1 + (xn)*2 + ... + (x1)*2*10^(n-1)), so A is divisible by 5. now, if you don' t even believe that number finishing by 5 can be divisible by 5, because you only tested this properties on some number, consider what you said just after : you said you tested your (strange) factorization algorithm on some numbers, just enough for you to be convinced of the efficiency of it. now, i' m sure you doubt if your algo works for *all* numbers (since your algo seems to rely on strange theory, and division by 5 does not rely on strange theory), so why don' t you show your algorithm (in fact i know why, you just can' t show it, because there is no algorithm =)). anyway, on the question of whether you should release it or not, like svin, i think you should, since we would proove you that it doesn' t works as you want =)
Posted on 2002-05-09 05:02:23 by roy
a number ending by 5, usually we could say it will be divisible by 5, because our math experience of relatively small numbers tells us,

In decimal numeric system?!
I'm shoked.
Go to book store and by good arithmetic book
for primary school.
We know about it not because of our "limited math experience
of little numbers"
This conclusion is not result of "observation" numbers,
but of nature of positioned numeric system.
It doesn't matter how long decimal number is,
the positioned system is sum elements representing by digits
power of radix * q (digit)
were power = zero based position
q = digit.
for example
1234 =
10^3*1 + 10^2*2 + 10^1*3+10^0*4 = 1000 + 200 +30 +4
if x mulptiple y then x*q mulptiple y.
Thus if 10 multiple 5 then any intereger power of 10 multiple 5.
And so with any interger power of 10 * any natural number.
So only question is if 10^0*x is multiple by 5. Where x <=9 (biggest q)
among all 10 possibilities there are only 2: 0 and 5 that multiple by 5
That's why any decimal number with 5 at the end is mulptuple - cause
the rest part is multiple for sure - cause the rest part itself by nature is
created as integer power of 10 and for that only is multiple by 5.

Imaging you have boxes and each box contents 10 tins. And you have
only full boxes.
Do you need to know how many boxes you have to say that all camulating
number of tins is multiple 10?
And 5? And 2?

The same with numbers.
Posted on 2002-05-09 05:14:31 by The Svin
pretending that he coded revolutionary algo but still not very sure about his elementary maths...loool
Posted on 2002-05-09 05:33:58 by DZA
Sheeeeesh!!!! :grin:

Giovanni is merely referencing the Theory of Realitivity......

Look, this thread has proven a couple things,

1.

If you spend all your time trying to factor large numbers with paper and pencil.....you will need to release this energy by creating large colorful discriptions of very simple things...:)

2.

If you don't present large colorful discriptions of very simple things, then it's likely you won't get much response to a partially complete concept....

3. Alex, you got to admit that: some of Giovanni's posts sound a little like some of yours....quote, something like "I'm just trying to get you all to think"....

and All and All, I have much enjoyed The Svin's posts as well as these....


Brad :)
Posted on 2002-05-09 06:34:10 by Brad
Hi,

I must admit I've been a bit angry about some messages, but for my fortune I do have the algo. Here it is: who understands how this works, will understand why 10147 could *not* be solved by assuming endings 1*7. I left over something, which has to be solved by your own, first because I am really off-shore with my work timings, and second because you'll have to do something.

Despite of a huge load of work, I made it as complete as possible, and all the ones who understand *how* this works, pass immediatedly to Level 3 of this game. The others have to be patient until saturday, or even sunday, because the work I didn't do today is a leftover for the following. We have to respect times and that's it. The more patient will, sooner or later, understand how this algo works, and why.




10147

10147 1 * 7; 3 * 9

1 * 7 >> res tmp off
1 7 7
1 17 17
1 27 27
1 37 37
1 47 47 *
1 57 57
1 67 67
1 77 77
1 87 87
1 97 97
1 107 107

01 47 47 *
11 47 517
21 47 987
31 47 1457
41 47 1927
51 47 2397
61 47 2867
71 47 3337
81 47 3807
91 47 4277
101 47 4747 * //

01 047 47
01 147 147
01 247 247
01 347 347
01 447 447
01 547 547
01 647 647
01 747 747
01 847 847
01 947 947
01 1047 1047 * //

101 1047 105747 //

101 147 4747
101 247 24947 //

91 047 4277
91 147 13377 //

81 047 3807
81 147 11907 //

71 047 3337
71 147 10437 //

61 047 2867
61 147 8967 *
61 247 15067 //

61 147 8967
161 147 23667 //

51 047 2397 *
51 147 7497 *
51 247 12597 //

51 047 2397 *
151 047 7097 *
251 047 11797 //

51 1047 2397
51 2047 104397 //

151 047 7097
151 1047 158097 //

41 047 1927 *
41 147 6027 *
41 247 10127 *
41 347 14227 //

041 047 1927 *
141 047 6627 *
241 047 11327 //

041 047 1927
041 1047 42927 //

141 047 6627
141 1047 147627 //

041 147 6027
141 147 20727 //

041 247 10127
141 247 34827 //

31 047 1457 *
31 147 4557 *
31 247 7657 *
31 347 10757 //

31 047 1457 *
131 047 6157 *
231 047 10857 //

031 047 1457
131 1047 32457 //

131 047 6157
131 1047 137157 //

031 147 4557
131 147 19257 //

031 247 7657
131 247 32357 //

21 047 987 *
21 147 3087 *
21 247 5187 *
21 347 7287 *
21 447 9387 *
21 547 11487 //

021 047 987
121 047 5687 *
221 047 10387 //

121 047 5687
121 1047 126687 //

021 147 3087
121 147 17787 //

021 247 5187
121 247 29887 //

021 347 7287
121 347 41987 //

021 447 9387
121 447 54087 //

11 047 517 *
11 147 1617 *
11 247 2717 *
11 347 3817 *
11 447 4917 *
11 547 6017 *
11 647 7117 *
11 747 8217 *
11 847 9317 *
11 947 10417 //

011 047 517 *
111 047 5217 *
211 047 9917 *
311 047 14617 //

011 047 517
011 1047 11517 //

111 047 5217
111 1047 116217 //

211 047 9917
211 1047 220917 //

---

01 37 37 *
11 37 407 //
21 37 777 //
31 37 1147 *
41 37 1517 *
51 37 1887 //
61 37 2257 //
71 37 2627 //
81 37 2997 //
91 37 3367 //
101 37 3737 //

01 137 137 *
01 237 237 *
01 337 337 *
01 437 437 *
01 537 537 *
01 637 637 *
01 737 737 *
01 837 837 *
01 937 937 *
01 1037 1037 *

101 137 13837 //

---


01 27 27 *
11 27 297 //
21 27 567 //
31 27 837 //
41 27 1107 //
51 27 1377 //
61 27 1647 *
71 27 1917 //
81 27 2187 //
91 27 2457 //
101 27 2727 //

01 127 127 *
01 227 227 *
01 327 327 *
01 427 427 *
01 527 527 *
01 627 627 *
01 727 727 *
01 827 827 *
01 927 927 *
01 1027 1027 *



Giovanni

PS: Please be a little more patient! I do my best but sometimes I am off. Tomorrow I am surely off at all (at least you may tell what you want :) )

PATIENT TASKS: Find out where it is best to cut off. In some places the listings are extremely extended, even if not necessary. Where? Why? Did you see what happens? What relationship is there to the stupid 17 * 11 excercise?
Posted on 2002-05-09 11:20:08 by sch.jnn