Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 272. (Read 229433 times)

Simply you posted that values and I have seen them before you obfuscated.


If you mean: retrieve 'a' and 'b' you used from 'w1' and 'w2' is not possible, because given w1 and w2 there is no only a couple (a,b) but many many couples with a/b = w1 and b/a = w2  mod p

For example, for each value of a:

b = w2 * a mod p   because  b = (b/a) * a   mod p


For all these couples (a,b) you have that a/b = w1 mod p and b/a = w2 mod p.

Besides if you provide w1, you don't need to provide w2 too, because w2 is (w1)^-1 mod p    --> w2 = 1/w1

w1 = 41689968412277401815192940146121615515073176981576112778517871276338175969527
w2 = pow(w1, p-2, p)
w2
94599322545067982882578206403521588574602656593708975301033069273870448067861

You have to provide w1 and a or b , but from w1 and w2 you cannot get a specific 'a' and 'b'.

In other terms you are providing only a single information (w1) and you want to know 2 values, it is like to resolve a single equation with 2 variables, like y = 2 * x    (in your case b = w2 * a mod p)

there is no a single solution, but infinite couples (1,2), (2,4), (3,6) .....   there is no way to guess the couple you used!

It's not that easy!
The correct is:
a = 76470300715912249562689990107401687364194232406198996658976353330269918489458
b = 64658408237276871767689061520961436408509493287485285377611016482361694763299

I need a tool that finds the a and b value correctly

One question ... how did you discover the value of a and b?

Discover this!

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = ??
b = ??

w = GF(p)

w (((a / b)))
w (((b / a)))

w= 41689968412277401815192940146121615515073176981576112778517871276338175969527
w= 94599322545067982882578206403521588574602656593708975301033069273870448067861

I will send you 1 BTC if you find the value a and b

I've never used python ... I'm not good at programming or math.
But I really appreciate your suggestion ... If I achieved my goal ... I will send you 1 BTC

If you use python:

copy this file "test.py"


and then in the terminal:

python test.py

I'm not good at math ... But I don't think it's that easy! Can you create a tool that I can test? If you really did it ... I'll send you 3 BTC

Well, that's easy:
So, an obvious solution is a=12447032699845648078645791161909514142990644957498005805208944683777961822095, b=1

So, w has to values? or maybe is it w1 and w2? That is very confusing.

It got better?

w1 =12447032699845648078645791161909514142990644957498005805208944683777961822095

w2 = 66620152837833785920928131416087065201280002472666144035333386572317622196480

So, w has to values? or maybe is it w1 and w2? That is very confusing.

Sorry for the post off topic ... but can someone develop a tool to solve this?

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = ?
b = ?

w = GF(p)

w (((a / b)))
w (((b / a)))

w= 12447032699845648078645791161909514142990644957498005805208944683777961822095
w= 66620152837833785920928131416087065201280002472666144035333386572317622196480


If you can develop any tool that does this ... I will send you 3 BTC

The puzzle is very complex and not everyone can find the answer to it, to solve it you need to be really smart and experienced and maybe it is not enough.

Happy New Year 2020 to all,

anything new under the sun regarding the project, maybe some gpu pollard.
or

our most active and full of knowledge personality @MrFreeDragon @bounty0z, do you like comments and your suggestion for forum readers,about your test completion with me
waiting your comments , thankx

can try (who has nothing to do)) as I wrote before https://bitcointalksearch.org/topic/m.51089662 with 3 digits.

we need divisible by 3 numbers.

67 1BY8GQbnueYofwSuFAT3USAhGjPrkxDdW9 | 147573952589676412927
68 1MVDYgVaSN6iKKEsbzRUAYFrYJadLYZvvZ  | 295147905179352825855
69 19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG   | 590295810358705651711

147 573 952 589 676 412 927
295 147 905 179 352 825 855
590 295 810 358 705 651 711

the spread there is from 0 to 3000 mainly (can be cut to 2500)

i.e. take 2^4096 cut the first 3000 digits, generate randomly 7 by 3 (0,0,0,0,0,0,0-3000,3000,3000,3000,3000,3000,3000) and read on

2^4096
2^4097
2^4098
2^4099
etc...

from 2^4096 to 2^2000000 (cut off up to 3000 and saved to file) takes up disk space about 6gb.

64      gb       2^192000000‬
128    gb       2^384000000
256    gb       2^768000000
512    gb       2^1536000000‬
1024  gb       2^3072000000
2048  gb       2^6144000000‬
4096  gb       2^12288000000
8192  gb       2^24576000000
16384 gb      2^49152000000‬

at 4096gb 2^12288000000 you need a 4 terabyte hard drive for the file

147573952589676412927
295147905179352825855
590295810358705651711
                12288000000

that is, we take our random 7 to 3 eg 4,153,2001,1578,87,589,1111 and read their positions from the file from 2^4096 to 2^12288000000.

4,153,2001,1578,87,589,1111 and read their positions from the file from 2^4096 to 2^12288000000.
264,15,2111,2178,487,1589,1112 and read their positions from the file from 2^4096 to 2^12288000000.
46,633,5,2331,1375,1589,96 and read their positions from the file from 2^4096 to 2^12288000000.

can search 2 (or 3) puzzle immediately, not 1 at a time in between 147573952589676412927-295147905179352825855, 295147905179352825855-590295810358705651711.

but there is a nuance, 1) it generate this text file trimmed 2^... 2) the speed of the program.

@mrfreeDragon and @bounty0z
both have tested my research, dear both freinds, do u like to explain public, what you learn in my test

You are right. This "puzzle" is impossible to solve, this is the reason why people created it, to prove that is impossible and that Bitcoin is safe.

dear bro
do you like test and proof that my 110bit pubk to 100 bit pubk, mean 10 bit down PM for take test
thankx

I don't think it's possible to solve this puzzle. It's not possible for me. Making money easy with Bitcoin excites everyone. Nothing easy. Nobody gives anything to anyone for free. At least the puzzles can be used to advertise bitcoin.