Bitcoin puzzle transaction ~32 BTC prize to who solves it

[bigvito19]

How do y'all know how many bits a public key is?

[MrFreeDragon]

How do y'all know how many bits a public key is?

It is not possible to know the private key bit length from the public key

[bounty0z]

How do y'all know how many bits a public key is?

It is not possible to know the private key bit length from the public key

MrFreeDragon, I have question for you if you can explain to me please !! how brainless can find that those address are 100bits ?
let think that he take the 110 address and try to add negative hex numbres to reduce the bit ,right !! but even that how he can be sure that the numbre that he add make address from 110bit to 100bit !!? that is really weird
the only thing that we know is 2^n = 2^(n-1) + 2^(n-2) .....+ 2^(n-n+1) + 2 !!
but if we have "n" =in range 110 and 111 that make impossible to know of even predict the bit of wallet.
if you have any idea please.

thanks

[mohdk52]

I do not like bitcoin puzzles, besides it's just a waste of time, no one will win anything.

[jameshugo17]

I have no way of solving this puzzle. Maybe this kind of competition should be a little different. For example, a program that organizes rewards for those who express bitcoin in social life can be made. I've been trying a lot like this last year. Conclusion frustration

[MrFreeDragon]

How do y'all know how many bits a public key is?

It is not possible to know the private key bit length from the public key

MrFreeDragon, I have question for you if you can explain to me please !! how brainless can find that those address are 100bits ?
let think that he take the 110 address and try to add negative hex numbres to reduce the bit ,right !! but even that how he can be sure that the numbre that he add make address from 110bit to 100bit !!? that is really weird
the only thing that we know is 2^n = 2^(n-1) + 2^(n-2) .....+ 2^(n-n+1) + 2 !!
but if we have "n" =in range 110 and 111 that make impossible to know of even predict the bit of wallet.
if you have any idea please.

thanks

It is not possible to know the bit length just from the public key. It is not possible. Probably the author of that argument has some additional details. But it is not possible to know it just from the public key.

Publlic key is a Point with x-coordinate up to almost 2^256 (actually the maximum value is module which is close to 2^256). Take the easy example with 1bit key (which is actually number 1). For private key 1, the public key is 0279be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798, and this number is something like 47% on the whole range of the keys, and it is also the number approximately 2^254.92, so it is 254bit number. Hence, for 1bit private key we have 254bit x-coordinate of the public key. It is not possible to know the bit length of the private key based just on the public key.

ECDSA is one way: it is possible to receive the public key from the private, bit not vice versa.

[bdivrik]

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.

[brainless]

How do y'all know how many bits a public key is?

It is not possible to know the private key bit length from the public key

MrFreeDragon, I have question for you if you can explain to me please !! how brainless can find that those address are 100bits ?
let think that he take the 110 address and try to add negative hex numbres to reduce the bit ,right !! but even that how he can be sure that the numbre that he add make address from 110bit to 100bit !!? that is really weird
the only thing that we know is 2^n = 2^(n-1) + 2^(n-2) .....+ 2^(n-n+1) + 2 !!
but if we have "n" =in range 110 and 111 that make impossible to know of even predict the bit of wallet.
if you have any idea please.

thanks

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

[itod]

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.

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.

[brainless]

How do y'all know how many bits a public key is?

It is not possible to know the private key bit length from the public key

MrFreeDragon, I have question for you if you can explain to me please !! how brainless can find that those address are 100bits ?
let think that he take the 110 address and try to add negative hex numbres to reduce the bit ,right !! but even that how he can be sure that the numbre that he add make address from 110bit to 100bit !!? that is really weird
the only thing that we know is 2^n = 2^(n-1) + 2^(n-2) .....+ 2^(n-n+1) + 2 !!
but if we have "n" =in range 110 and 111 that make impossible to know of even predict the bit of wallet.
if you have any idea please.

thanks

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

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

[Andzhig]

can try (who has nothing to do)) as I wrote before https://bitcointalk.org/index.php?topic=1306983.msg51089662#msg51089662 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.

[brainless]

How do y'all know how many bits a public key is?

It is not possible to know the private key bit length from the public key

MrFreeDragon, I have question for you if you can explain to me please !! how brainless can find that those address are 100bits ?
let think that he take the 110 address and try to add negative hex numbres to reduce the bit ,right !! but even that how he can be sure that the numbre that he add make address from 110bit to 100bit !!? that is really weird
the only thing that we know is 2^n = 2^(n-1) + 2^(n-2) .....+ 2^(n-n+1) + 2 !!
but if we have "n" =in range 110 and 111 that make impossible to know of even predict the bit of wallet.
if you have any idea please.

thanks

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

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

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

[dextronomous]

Happy New Year 2020 to all,

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

[sniveling]

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.

[bytcoin]

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

[GoldTiger69]

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= 66620152837833785920928131416087065201280002472666144035333386572317622196480
w= 12447032699845648078645791161909514142990644957498005805208944683777961822095


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

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

[bytcoin]

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= 66620152837833785920928131416087065201280002472666144035333386572317622196480
w= 12447032699845648078645791161909514142990644957498005805208944683777961822095


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

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

w1 =12447032699845648078645791161909514142990644957498005805208944683777961822095

w2 = 66620152837833785920928131416087065201280002472666144035333386572317622196480

[bytcoin]

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

It got better?

[j2002ba2]

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

Well, that's easy:

Code:

0 < b < p
a = b * 12447032699845648078645791161909514142990644957498005805208944683777961822095 (mod p)

So, an obvious solution is a=12447032699845648078645791161909514142990644957498005805208944683777961822095, b=1

[arulbero]

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

It got better?

Code:

p = 115792089237316195423570985008687907852837564279074904382605163141518161494337
a = 76470300715912249562689990107401687364194232406198996658976353330269918489458
b = 64658408237276871767689061520961436408509493287485285377611016482361694763299


b_inv =  pow(b, p-2, p)
w1 = a*b_inv % p
w1
12447032699845648078645791161909514142990644957498005805208944683777961822095


a_inv = pow(a, p-2, p)
w2 = b*a_inv % p
w2
66620152837833785920928131416087065201280002472666144035333386572317622196480