- On Tue, 2004-07-13 at 09:20, cino hilliard wrote:

> The pari program below produces primes. I produced primes for

In that case, it is almost certainly too hard to factor with present-day

> digitsqrpr(2000,m,150) and digitsqrpr(2000,m,200) and randomly selected a

> prime from each a,b

> output and multiplied them. I forgot what m I used or if it was the same for

> both or different as

> I did p = a*b and a prime() and then factor(p) and had to abort the

> session. I had copied p earlier

> to text to test for primality with another program which worked faster tham

> Pari. all I know is

> a is <= 150 digits and b is <=200 digits.

hardware and software. The values of a and b ma just be such that a

special-purpose algorithm such as P-1 may find them but the odds are

very much against it.

Almost certainly the easiest way to find the factors would be to

reverse-engineer the pseudorandom procedure used to generate the primes

in the first place.

Paul >From: Paul Leyland <pcl@...>

I did this and found the org number was copied and pasted wrong and was easy

>Reply-To: pcl@...

>To: cino hilliard <hillcino368@...>

>CC: primenumbers@yahoogroups.com

>Subject: Re: [PrimeNumbers] RE:[Factoring large numbers

>Date: 13 Jul 2004 10:38:32 +0100

>

>On Tue, 2004-07-13 at 09:20, cino hilliard wrote:

>

> > Pari. all I know is

> > a is <= 150 digits and b is <=200 digits.

>

>In that case, it is almost certainly too hard to factor with present-day

>hardware and software. The values of a and b ma just be such that a

>special-purpose algorithm such as P-1 may find them but the odds are

>very much against it.

>

>Almost certainly the easiest way to find the factors would be to

>reverse-engineer the pseudorandom procedure used to generate the primes

>in the first place.

>

to factor as was

pointed out by other posters. I did not set the limit in factor(n,{lim}) in

pari. Doing this

(14:56) gp > factor(p,1000)

time = 0 ms.

%25 =

[13 1]

[199 1]

[263 1]

[379 1]

[678026493414538148553665596445813830734630833053024846216579340872093413327647

4762803097949211213281819670661292330131124527435326156088028931580262368994401

6643351267973947046046313935094275418438526691021308861240797797863806746646690

8508141216855178178499487049221885287729920843955819202763969676507087051775192

055290647644770812521271 1]

So this mistake finds yet another large factor! I will play the lotto from

this list. :-)

The correct number I should have posted is

1220516103632118734239552093546852845556567859549464947942804246975191952318939

5382087627014892247240874902346477975756822244476470920475366468143120795553304

9000712508365396500993156592586695125254256067267320061792206044621345381741861

7266932089747110235311534348244823636951507811274424010127339017948672442184669

1782447097625239591063539221131129

One poster claims to factor 150,000 digit numbers. Here is a 350 digit

number. I do not see it any

different than the RSA challenges in this category. We will see.

So far, I and only I know the two factors. Gee, I don't know if I can handle

that much power :-)

Have fun,

CLH- I have been curious about trying to factor large numbers and on public

computers (my home computer is FAR too old!) I have not been able to

factor any number with more than 50 digits, whilst modern software can

usually find factors up to 40 or 45 digits (I found this when I saw how

dismal the prospects for fully factoring F12 [2^4096+1] in the near

future really are).

What sort of number would this new factoring method work upon? If it

could only work on a 150 digit number, it would be of little use for

factoring such numbers as Cunningham numbers. - --- In primenumbers@yahoogroups.com, "cino hilliard"

<hillcino368@h...> wrote:>

digit

> One poster claims to factor 150,000 digit numbers. Here is a 350

> number. I do not see it any

Try running a few ECM curves on F20 (20th Fermat Number) and you'll

> different than the RSA challenges in this category. We will see.

be attempting to factor a 315,653 digit number. Factoring of 150k

digit numbers is nothing new in the 21st Century. :)

As to the RSA comparison... When you're willing to put up $20k in

prize money, I suspect you'll get people trying to factor your

number. But if you do, make certain to post a reasonable challenge

and/or a reasonable reward. The next RSA challenge (RSA-640, $20k

prize) is only 193 digits with factors estimated at 85-88 digits.

While the RSA-1024 challenge is 309-digits with a $100k prize. So I

figure you'll need about $250k as a prize to get anyone to take your

challenge very seriously.