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    Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 324374 times)

    • digaran
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    July 15, 2023, 03:25:03 AM
     #2921
    Quote from: homosapienstrader on July 15, 2023, 02:50:07 AM
    Puzzle 129, 134, 139, 144, 149, 154, 160 are already solved, this means that this puzzle is rigged, but anyway, is it harder to solve puzzle 66 than mining a block?
    I just wonder, why did you start with 129 and not 66 or 67? And what do you mean by this puzzle is rigged, how exactly is it rigged?
    To answer your question, we can't really compare mining with key brute forcing, but I'd say if you spend 3 bitcoins to rent GPUs, you might be able to solve #66 and grab 3.6 free  bitcoins, but I'd wait for experts to calculate the exact cost.😉

    Ps all puzzles from 161 up to 256 are already solved, dis rigged puzzle!😅
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  • homosapienstrader
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    July 15, 2023, 04:29:03 AM
     #2922
    Quote from: unpluggedcoin on July 15, 2023, 03:13:12 AM
    Quote from: homosapienstrader on July 15, 2023, 02:50:07 AM
    Puzzle 129, 134, 139, 144, 149, 154, 160 are already solved, this means that this puzzle is rigged, but anyway, is it harder to solve puzzle 66 than mining a block?

    Why on earth you said that above puzzles are already solved, however they aren't? We've been trying to solve a single puzzle 124 for past 8 months are so and now you suddenly claimed that all puzzles with pubkeys are solved. Are you checking their balances on Ethereum or some other cheap drug Blockchain?

    Just look the hashs on blockchain, all mentioned hashs already have outs, how could a hash with unknown privatekey has outs?

  • nc50lc
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    July 15, 2023, 06:09:00 AM
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     #2923
    Quote from: homosapienstrader on July 15, 2023, 04:29:03 AM
    Quote from: unpluggedcoin on July 15, 2023, 03:13:12 AM
    Quote from: homosapienstrader on July 15, 2023, 02:50:07 AM
    Puzzle 129, 134, 139, 144, 149, 154, 160 are already solved, this means that this puzzle is rigged, but anyway, is it harder to solve puzzle 66 than mining a block?
    -snip-
    Just look the hashs on blockchain, all mentioned hashs already have outs, how could a hash with unknown privatekey has outs?
    Puzzle numbering starts with '1' (1BgGZ)
    But in Bitcoin's blockchain, output_index starts with '0'.
    So you're looking at the wrong puzzles, you must be talking about puzzle number 130, 135, 145, 150, 155, 161 respectively.

    The reason why those have spent outputs already is because the owners decided to reveal the public keys of puzzles divisible by '5' as "bonus".
    They didn't spent the puzzle's UTXO but sent and spent another set.
    For 161, 161~256 are spent because the owner (not the OP) agreed to a comment that those are computationally impossible to bruteforce.

  • vhh
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    July 15, 2023, 12:57:34 PM
     #2924
    Quote from: homosapienstrader on July 15, 2023, 04:29:03 AM
    Quote from: unpluggedcoin on July 15, 2023, 03:13:12 AM
    Quote from: homosapienstrader on July 15, 2023, 02:50:07 AM
    Puzzle 129, 134, 139, 144, 149, 154, 160 are already solved, this means that this puzzle is rigged, but anyway, is it harder to solve puzzle 66 than mining a block?

    Why on earth you said that above puzzles are already solved, however they aren't? We've been trying to solve a single puzzle 124 for past 8 months are so and now you suddenly claimed that all puzzles with pubkeys are solved. Are you checking their balances on Ethereum or some other cheap drug Blockchain?

    Just look the hashs on blockchain, all mentioned hashs already have outs, how could a hash with unknown privatekey has outs?

    Did you at least bother to use the search function in this thread to read why there are outs in those transactions?

  • neirobobi
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    July 17, 2023, 08:46:05 AM
    Last edit: July 17, 2023, 09:46:24 AM by neirobobi
     #2925
    Hey people!
    I'm using Altcrack to find 66 but it only searches consecutively through 1 or some other number. Tell me if there is a program that can search for 66 puzzles by sorting through all the range options by mask.

    for example this mask (decimal): XXXXXX123456XXXXXXXX

    Now I have to run Altcrack 368,935 times consecutively to check the range for this mask, but this is very long. Is there any other way?

  • rosengold
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    July 17, 2023, 03:43:56 PM
     #2926
    I built my own private brute forcer that make all calculations with pure cuda threads. actually it uses a single gpu, but if anyone with more than 10 good gpus on a windows enviroment would like to join forces we could try our luck together and split the prize.

    my brute forcer is different because it uses a different logic while searching sequentially.

    if you have 10+ gpus and would like to try DM me.

    Actually I'm not interested to share my code. or software. thanks.

  • digaran
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    July 17, 2023, 04:15:25 PM
     #2927
    Quote from: rosengold on July 17, 2023, 03:43:56 PM
    I built my own private brute forcer that make all calculations with pure cuda threads. actually it uses a single gpu, but if anyone with more than 10 good gpus on a windows enviroment would like to join forces we could try our luck together and split the prize.

    my brute forcer is different because it uses a different logic while searching sequentially.

    if you have 10+ gpus and would like to try DM me.

    Actually I'm not interested to share my code. or software. thanks.
    You are not sharing it, so there is no way for interested parties to test it's performance, who would gamble their GPUs blindly? Ask that from yourself and happy hunting!😉
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  • brainless
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    July 17, 2023, 04:23:35 PM
     #2928
    who can solve
    p = 115792089237316195423570985008687907852837564279074904382605163141518161494335
    a = 1099511627776
    b = 115792089237316195423570985008687907852837564279074904382605163141005436653346
    c = (a-b) %p
    result = 1612236468765

    in pubkey

    p =115792089237316195423570985008687907852837564279074904382605163141518161494335
    a = 02feea6cae46d55b530ac2839f143bd7ec5cf8b266a41d6af52d5e688d9094696d
    b = 02746bd76e07a0dbbcc610245439ee1db94f73b70df43bc543d4046ebe119ad6b3
    c = (a-b) %p
    result = 02b21dd66bfde832c2dae35688c0e15b91b274ec018e2c14e23f1ca7cb32fcca73

    substract formula
    p = int(2**256 - 2**32 - 977)
    x1 =  # fill pubkey1-x
    y1=  # fill pubkey1-y
    x2=  # fill pubkey2-x
    y2=  # fill pubkey2-y



    dx = (x1 - x2) % p
    dy = (y1 - (-y2)) % p
    c = dy * gmpy2.invert(dx, p) % p
    Rx = (c*c - x2 - x1) % p
    Ry = (c*(x2 - Rx) - y2) % p
    print (Rx , Ry)
    print (hex(Rx) , hex(Ry))


    if you have alternate formula for adjust with mod p, apply and check for get acurate result in pubkey
    13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF

  • rosengold
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    July 17, 2023, 06:37:41 PM
     #2929
    Quote from: digaran on July 17, 2023, 04:15:25 PM
    Quote from: rosengold on July 17, 2023, 03:43:56 PM
    I built my own private brute forcer that make all calculations with pure cuda threads. actually it uses a single gpu, but if anyone with more than 10 good gpus on a windows enviroment would like to join forces we could try our luck together and split the prize.

    my brute forcer is different because it uses a different logic while searching sequentially.

    if you have 10+ gpus and would like to try DM me.

    Actually I'm not interested to share my code. or software. thanks.
    You are not sharing it, so there is no way for interested parties to test it's performance, who would gamble their GPUs blindly? Ask that from yourself and happy hunting!😉

    of course we will run some tests on a private enviroment to check performance but I'm seriuously thinking about rent some on vast.ai and hunt by myself

  • digaran
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    July 17, 2023, 07:59:21 PM
     #2930
    Quote from: rosengold on July 17, 2023, 06:37:41 PM
    Quote from: digaran on July 17, 2023, 04:15:25 PM
    Quote from: rosengold on July 17, 2023, 03:43:56 PM
    I built my own private brute forcer that make all calculations with pure cuda threads. actually it uses a single gpu, but if anyone with more than 10 good gpus on a windows enviroment would like to join forces we could try our luck together and split the prize.

    my brute forcer is different because it uses a different logic while searching sequentially.

    if you have 10+ gpus and would like to try DM me.

    Actually I'm not interested to share my code. or software. thanks.
    You are not sharing it, so there is no way for interested parties to test it's performance, who would gamble their GPUs blindly? Ask that from yourself and happy hunting!😉

    of course we will run some tests on a private enviroment to check performance but I'm seriuously thinking about rent some on vast.ai and hunt by myself
    You should at least post your results, such as key per second, your hardware etc, this way everyone can see how your tool is performing, besides there is no use in hiding a tool from the public, it's not like a program which you could brute force bitcoin keys and steal coins, all the programs are useless, they are only good for 65 bit, above that range, you'd need to add more hardware.

    A breakthrough would be when someone develops a program that could brute force 10 times faster with a single GPU, till then feel free to do as you want, remember sharing is caring.😉
    🖤😏

  • asbani
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    July 18, 2023, 12:20:15 AM
     #2931
    Quote from: neirobobi on July 17, 2023, 08:46:05 AM
    Hey people!
    I'm using Altcrack to find 66 but it only searches consecutively through 1 or some other number. Tell me if there is a program that can search for 66 puzzles by sorting through all the range options by mask.

    for example this mask (decimal): XXXXXX123456XXXXXXXX

    Now I have to run Altcrack 368,935 times consecutively to check the range for this mask, but this is very long. Is there any other way?

    There's an open source software which scans ranges that you specify if that's what you're asking. I don't know if I can post links or not but search bitcrackrandomiser on GitHub, it uses BitCrack and it also blacklists defeated ranges.

  • neirobobi
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    July 18, 2023, 10:09:56 AM
     #2932
    Quote
    I don't know if I can post links or not but search bitcrackrandomiser on GitHub, it uses BitCrack and it also blacklists defeated ranges.
    - here I need to run BitCrack 368,935 times in a row to check the range for this mask, but it's very long. I need another way.

  • james5000
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    July 18, 2023, 02:53:00 PM
     #2933
    Well, facing the solution of the number 125 right after the number 120 I would like to join forces with a cuda programmer so that we can together implement my technique and together discover the keys of the numbers 66, 67 and 68

  • pvynsbztutfjyxpc
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    July 18, 2023, 07:28:46 PM
     #2934
    Hi,

    I have an RTX-4070TI and I want to attempt to crack puzzle #66. BitCrack or any other (CUDA) program uses old CUDA toolkit versions. I want to use Docker images available here: https://hub.docker.com/r/nvidia/cuda/tags so that I don't have to install CUDA toolkit in my system. Is there any solution? Any help is greatly appreciated, thanks.

  • Miracle*
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    July 19, 2023, 01:26:42 AM
     #2935
    Does anyone ever use privatekeys.pw/scanner ? is it real ?
    Too many scam attempt in this forum. Always use Escrow !!!
    My shop ( US bank withdrawal service minimum balance $1500  ) : https://bt.irlbtc.com/view/1515926.0
    Tip me ?    1BMakzEVtiJNAxPH99i46Tm8QjQy9WKmmR

  • unclevito
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    July 19, 2023, 01:46:57 AM
     #2936
    Quote from: VinIVaderr on July 18, 2023, 06:14:38 PM
    I haven't found any previous posts about brute forcing the base58 compressed key. So I thought I would share a simple code example.
    The prefix in the code below is hexadecimal "1"  with address 1BgGZ9tcN4rm9KBzDn7KprQz87SZ26SAMH
    If you delete the last character 'n' and run the script you will see the correct output. It may take a few tries. From puzzle 65 it can be determined that the base58 compressed key changes at Y. If you delete everything up to q you can search a range(19) and with much brute forcing we could create the key for puzzle 66.

    Python3 Code:
    import secrets
    from bit import Key, PrivateKey

    address = "1"
    addr = ""

    while addr[:len(address)] != address:
        prefix = "KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qYjgd9M7rFU73sVHnoWn"
        base58_char = "123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz"
        text = ''.join(secrets.choice(base58_char) for i in range(1))
        data = prefix + text
        key = Key(data)
        addr = key.address


    f = open("key.txt", "a")
    f.write(data)
    f.close()

    print(data)
    print(key)
    print(addr)
    There is already a few programs like that for cpu and gpu. But finding 19 characters will take a lot of time or a lot of gpu's.   https://github.com/unclevito2017/Fialka

  • asbani
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    July 19, 2023, 06:36:53 AM
     #2937
    Quote from: Miracle* on July 19, 2023, 01:26:42 AM
    Does anyone ever use privatekeys.pw/scanner ? is it real ?

    Ofcourse it's real in a sense that it pulls the priv key, shows the pub key and address. But you can do with any program there's nothing special there. The number of pages is so huge that you hardly will find anything manually speaking. Now I don't know if there's a backdoor where the author will see what you search/look for, I cannot speak for this but for extra-safety do not insert your own private key on the site.

    There are many sites like this one too, check keys.lol - One guy made an experiment on keys.lol to see if the owner of the site logs/checks the searches whenever someone finds a private key right? This is a fun story, he literally created a new wallet and put in 0.01 bitcoin in it 4 years ago. Then what he did for laughs is "Searched his own private key" inside of the website keys.lol to see which page his wallet is on in that website - this experiment was to see if the owner will empty his wallet or not, then he posted his blockchain public key and told us all to keep an eye open for this one, if the wallet get swept then keys.lol is monitored. To my surprise I still keep checking the wallet, these 0.01 btc are still there.

    Which means Keys.lol do not log any searches so if you click "Random" and get lucky with a key nobody else will see it. But I cannot speak of privatekeys.pw myself, I can speak of the other one because I saw the experiment myself.

  • zahid888
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    July 19, 2023, 10:21:51 AM
    Last edit: July 19, 2023, 10:32:04 AM by zahid888
     #2938
    Quote from: Bulista on July 14, 2023, 09:16:55 AM
    Anyone thinks that with the help of some funds it can help to crack the case?
    I Can.. and will return 2x of your fund Smiley
    1BGvwggxfCaHGykKrVXX7fk8GYaLQpeixA

  • GR Sasa
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    July 19, 2023, 12:12:21 PM
     #2939
    Quote from: zahid888 on July 19, 2023, 10:21:51 AM

    I Can.. and will return 2x of your fund Smiley


    But how will you do it Zahid? What's your plan/goal

    Will you target 66 or 130?

    +rep for zahid

    He's trusted.

  • zahid888
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    July 19, 2023, 01:23:46 PM
     #2940
    Quote from: GR Sasa on July 19, 2023, 12:12:21 PM
    Quote from: zahid888 on July 19, 2023, 10:21:51 AM

    I Can.. and will return 2x of your fund Smiley


    But how will you do it Zahid? What's your plan/goal

    Will you target 66 or 130?

    +rep for zahid

    He's trusted.

    I will approach it using brute force, but with the aid of my own program that I have already developed. Let me offer a slight hint regarding my counting formula, which involves (LFSR counting), but definitely without counting power; I am still incomplete and I am specifically discussing pure counting, so obvious it's puzzle 66 first.
    1BGvwggxfCaHGykKrVXX7fk8GYaLQpeixA
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