There's simply no feasible way to withdraw the funds on lower end puzzles like #66. It will be snatched up by bots. Not maybe, but it's 100% guaranteed. There will be hundreds of withdrawal transactions with varying fees all battling each other. You will simply be left in the dust.
I don't understand how it is possible for that to happen as long as the solver does not disclose the private key. I mean, #64 and #65 both have unknown keys, right? I'm surprised why you are asking such weird questions these days, do you really not know anything about these things, or maybe you are someone else who is running the "@NotATether" account. Anyway, if you mean to ask who solved puzzles 64 and 65, let me tell you, puzzle 64 was solved by some unknown person and puzzle 65 was not solved but swept by the creator himself, I want to tell the creator to put it back in puzzle 65 or in my wallet  |
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use this code to "predict"
Hexadecimal: 0x11f774e94c1ec000 # puzzle 62 0x363d541eb611abee
Hexadecimal: 0x7d556bf6f89d2c00 # puzzle 63 0x7cce5efdaccf6808
Hexadecimal: 0xe7655f0b50acf800 # puzzle 64 0xf7051f27b09112d4
Hexadecimal: 0x1a78fd44662532000 # puzzle 65 0x1a838b13505b26867
Hexadecimal: 0x290860e0f31602000 # puzzle 66 ?
Yep. Some results get kinda close, others not much. Could accuracy increase as we append new keys to the sequence? Can we improve this script somehow? spline_rep = splrep(x_values_known, sequence_decimal, k=2) I've messed with other values for k but 2 seems to yield better results. Index 3: Predicted = 0xd, Actual = 0x8, Error = 5.0
Index 4: Predicted = 0x5, Actual = 0x15, Error = 15.83333333333334
Index 5: Predicted = 0x30, Actual = 0x31, Error = 0.2857142857142918
Index 6: Predicted = 0x5c, Actual = 0x4c, Error = 16.04901960784312
Index 7: Predicted = 0x63, Actual = 0xe0, Error = 124.75357443229606
Index 8: Predicted = 0x202, Actual = 0x1d3, Error = 47.404329004328815
Index 9: Predicted = 0x31c, Actual = 0x202, Error = 282.866702978386
Index 10: Predicted = 0x13c, Actual = 0x483, Error = 838.5322535426649
Index 11: Predicted = 0x9e5, Actual = 0xa7b, Error = 149.13061026670766
Index 12: Predicted = 0x1403, Actual = 0x1460, Error = 92.41323240818747
Index 13: Predicted = 0x2241, Actual = 0x2930, Error = 1774.144396004227
Index 14: Predicted = 0x4a1b, Actual = 0x68f3, Error = 7895.604944862127
Index 15: Predicted = 0xd8f3, Actual = 0xc936, Error = 4029.671642258196
Index 16: Predicted = 0x14745, Actual = 0x1764f, Error = 12041.382349990992
Index 17: Predicted = 0x2784f, Actual = 0x3080d, Error = 36797.025408181595
Index 18: Predicted = 0x59719, Actual = 0x5749f, Error = 8826.371450069128
Index 19: Predicted = 0x8b61a, Actual = 0xd2c55, Error = 292410.36592774664
Index 20: Predicted = 0x1af328, Actual = 0x1ba534, Error = 45579.312763758004
Index 21: Predicted = 0x30fdc8, Actual = 0x2de40f, Error = 203193.17374296952
Index 22: Predicted = 0x4360b7, Actual = 0x556e52, Error = 1183130.437051029
Index 23: Predicted = 0x955cee, Actual = 0xdc2a04, Error = 4640021.909114862
Index 24: Predicted = 0x1ce3cea, Actual = 0x1fa5ee5, Error = 2892282.0998125
Index 25: Predicted = 0x3b79f62, Actual = 0x340326e, Error = 7826676.155909941
Index 26: Predicted = 0x4992721, Actual = 0x6ac3875, Error = 34804051.331749916
Index 27: Predicted = 0xc998ee1, Actual = 0xd916ce8, Error = 16244230.842530549
Index 28: Predicted = 0x181a56c4, Actual = 0x17e2551e, Error = 3670438.391939521
Index 29: Predicted = 0x25955523, Actual = 0x3d94cd64, Error = 402618432.6683471
Index 30: Predicted = 0x82c6e33c, Actual = 0x7d4fe747, Error = 91683829.12309027
Index 31: Predicted = 0xd6239bac, Actual = 0xb862a62e, Error = 499185022.823071
Index 32: Predicted = 0xe9b22d07, Actual = 0x1a96ca8d8, Error = 3216669648.649395
Index 33: Predicted = 0x371532851, Actual = 0x34a65911d, Error = 653104948.3604355
Index 34: Predicted = 0x5949f8bc7, Actual = 0x4aed21170, Error = 3855448663.1672974
Index 35: Predicted = 0x5af449ee4, Actual = 0x9de820a7c, Error = 17972423575.53308
Index 36: Predicted = 0x1391413025, Actual = 0x1757756a93, Error = 16210213485.866455
Index 37: Predicted = 0x2dbf7270dd, Actual = 0x22382facd0, Error = 49513939981.24811
Index 38: Predicted = 0x2886559a9b, Actual = 0x4b5f8303e9, Error = 149672520013.72742
Index 39: Predicted = 0x98c8113e3b, Actual = 0xe9ae4933d6, Error = 347459810714.7615
Index 40: Predicted = 0x20b05d20b7e, Actual = 0x153869acc5b, Error = 788113342243.4756
Index 41: Predicted = 0x1696ccb8e1d, Actual = 0x2a221c58d8f, Error = 1343066079089.604
Index 42: Predicted = 0x50b26b4d29f, Actual = 0x6bd3b27c591, Error = 1864358884081.7373
Index 43: Predicted = 0xdef4cb268c7, Actual = 0xe02b35a358f, Error = 83326651591.11523
Index 44: Predicted = 0x1875de8190b7, Actual = 0x122fca143c05, Error = 6898060186802.75
Index 45: Predicted = 0x1230f00b831f, Actual = 0x2ec18388d544, Error = 31407275332132.58
Index 46: Predicted = 0x689e837c30da, Actual = 0x6cd610b53cba, Error = 4636639038432.0
Index 47: Predicted = 0xcd26aa6013f0, Actual = 0xade6d7ce3b9b, Error = 34358976174165.78
Index 48: Predicted = 0xec974b5f9e34, Actual = 0x174176b015f4d, Error = 148984356258072.8
Index 49: Predicted = 0x2d6a754d3f7ab, Actual = 0x22bd43c2e9354, Error = 187823628313687.25
Index 50: Predicted = 0x2b7ce396f3181, Actual = 0x75070a1a009d4, Error = 1293723206998098.5
Index 51: Predicted = 0x11abcd83d36e18, Actual = 0xefae164cb9e3c, Error = 757478132797404.0
Index 52: Predicted = 0x18b4f3390decf8, Actual = 0x180788e47e326c, Error = 190672196844172.0
Index 53: Predicted = 0x2258a643d0ea28, Actual = 0x236fb6d5ad1f43, Error = 306834910754076.0
Index 54: Predicted = 0x31634c7beb9390, Actual = 0x6abe1f9b67e114, Error = 1.6143936485346692e+16
Index 55: Predicted = 0xf7c9efde77f4e0, Actual = 0x9d18b63ac4ffdf, Error = 2.55276090216256e+16
Index 56: Predicted = 0xaaf00881ca1cd0, Actual = 0x1eb25c90795d61c, Error = 9.013109354212386e+16
Index 57: Predicted = 0x48bd5d0ecad0f40, Actual = 0x2c675b852189a21, Error = 1.2761382323882934e+17
Index 58: Predicted = 0x2e13f10ba558040, Actual = 0x7496cbb87cab44f, Error = 3.175539853443205e+17
Index 59: Predicted = 0x10359b6a07b1e600, Actual = 0xfc07a1825367bbe, Error = 3.2969207850953344e+16
Index 60: Predicted = 0x1c178523462fa100, Actual = 0x13c96a3742f64906, Error = 5.984454014555484e+17
Index 61: Predicted = 0x11f774e94c1ec000, Actual = 0x363d541eb611abee, Error = 2.6137405792622295e+18
Index 62: Predicted = 0x7d556bf6f89d2c00, Actual = 0x7cce5efdaccf6808, Error = 3.801338671410483e+16
Index 63: Predicted = 0xe7655f0b50acf800, Actual = 0xf7051f27b09112d4, Error = 1.1258296599663145e+18
Index 64: Predicted = 0x1a78fd44662532000, Actual = 0x1a838b13505b26867, Error = 4.753071358864998e+16 import numpy as np
from scipy.interpolate import splrep, splev
from decimal import Decimal
sequence = [1, 3, 7, 8, 21, 49, 76, 224, 467, 514, 1155, 2683, 5216, 10544, 26867, 51510, 95823, 198669, 357535, 863317, 1811764, 3007503, 5598802, 14428676, 33185509, 54538862, 111949941, 227634408, 400708894, 1033162084, 2102388551, 3093472814, 7137437912, 14133072157, 20112871792, 42387769980, 100251560595, 146971536592, 323724968937, 1003651412950, 1458252205147, 2895374552463, 7409811047825, 15404761757071, 19996463086597, 51408670348612, 119666659114170, 191206974700443, 409118905032525, 611140496167764, 2058769515153876, 4216495639600700, 6763683971478124, 9974455244496707, 30045390491869460, 44218742292676575, 138245758910846492, 199976667976342049, 525070384258266191, 1135041350219496382, 1425787542618654982, 3908372542507822062, 8993229949524469768, 17799667357578236628, 30568377312064202855]
initial_points = 3
results = []
for i in range(initial_points, len(sequence)):
x_values_known = np.arange(i)
sequence_decimal = [Decimal(value) for value in sequence[:i]]
spline_rep = splrep(x_values_known, sequence_decimal, k=2)
predicted_next_number = splev(i, spline_rep)
actual_value = sequence[i]
predicted_hex = hex(int(predicted_next_number))
actual_hex = hex(actual_value)
results.append({
'index': i,
'predicted': predicted_hex,
'actual': actual_hex,
'error': abs(float(predicted_next_number) - actual_value)
})
for result in results:
print(f"Index {result['index']}: Predicted = {result['predicted']}, Actual = {result['actual']}, Error = {result['error']}")
Everything is useless, I have tried many predictions like this, none of them worked  |
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Happy 10x Anniversary. On this inspiring day, will Satoshi increase the value of my wallet tenfold and provide renewed motivation?  |
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e54acb08cf7e7d9be0102e2914d1a4eb643f5df386e67bb4be1bad5a05a53879
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 In Sequence Some Sample Founds : FOUND!!!
PrivateKey= 00000000000000000000000000000000000000000000000000000000002de40f
Address =1CfZWK1QTQE3eS9qn61dQjV89KDjZzfNcv
Minikey = S11111111111111144Xkpy
FOUND!!!
PrivateKey= 0000000000000000000000000000000000000000000000000000000001fa5ee5
Address =15JhYXn6Mx3oF4Y7PcTAv2wVVAuCFFQNiP
Minikey = S1111111111111122378PV
FOUND!!!
PrivateKey= 0000000000000000000000000000000000000000000000000000000006ac3875
Address =128z5d7nN7PkCuX5qoA4Ys6pmxUYnEy86k
Minikey = S1111111111111128JJjjq
FOUND!!!
PrivateKey= 00000000000000000000000000000000000000000000000000000000001ba534
Address =14oFNXucftsHiUMY8uctg6N487riuyXs4h
Minikey = S111111111111113345mux
FOUND!!!
PrivateKey= 00000000000000000000000000000000000000000000000000000000002de40f
Address =1CfZWK1QTQE3eS9qn61dQjV89KDjZzfNcv
Minikey = S1111111111111137WWbfk If someone has already tried something like this, please let me know so we can avoid wasting time. |
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There is no pattern.
 Does that mean you just pressed CTRL+F and searched for '11', and a pattern like this appeared? Am I right? seems interesting  |
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13zb1hQbWV 50 pieces.
13zb1hQbWVs 2 pieces.
13zb1hQbWV 10 char match in ripemd160 hash will be little reference significance #0x20000000000000000 ~ 0x40000000000000000 have 10000+ private key has 10 char match compressed address 13zb1hQbWVs 11 char match #0x20000000000000000 ~ 0x40000000000000000 have 200~250 private key has 11 char match compressed address 13zb1hQbWVsc 12 char match #0x20000000000000000 ~ 0x40000000000000000 have 3~4 private key has 12 char match compressed address I found 13zb1hQbWVsSAaNvpKfmc43NsuAj33LyRM in range 0x20000000000000000 ~ 0x40000000000000000 last month But I still don't know which range I should jump and keep search Are you sure about this? Where did you get this information from? Will you share the source of this information? Hi all. In principle, I didn’t plan to write here, I’ll just write as a participant.
You are welcome bro  |
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SHA256(privateKey + "some secret value") Truly, this turned out to be an incredibly straightforward process;Lol:D I realize that I was needlessly complicating things. That's the true advantage of sharing whatever thoughts occupy your mind—people make it more easier.  Thanks I receive numerous messages in my inbox from individuals who are diligently attempting to solve Puzzle 66 and choose to share their ideas with me. However, I've noticed that not a single post from them exists within the community. I would advise them to come here and share their ideas; doing so will undoubtedly be beneficial for them.
Edit: In the midst of our conversation, another thought crosses my mind: why not explore the another crude measuring instrument, of the hashing strength of the community within a 66-bit range? Hence, this is the SHA256 hash: '7e4cb2ad57ba006e5991ad3289b51744ee2d1a02812905a0b64f17c22d978ce0', In bitrange '20000000000000000:40000000000000000', where the puzzle 66 private key exists in hexadecimal format with leading '0x', without any additional text or message or password. (Something valuable is hidden here)... In my opinion, hashing alone might be an exceptionally fast process, potentially solvable in matter of few minutes ??  Something like this... import hashlib
target = '7e4cb2ad57ba006e5991ad3289b51744ee2d1a02812905a0b64f17c22d978ce0'
for i in range (2**65,2**66):
key = hex(i)
sha = hashlib.sha256(key.encode()).hexdigest()
if sha == target:
print("\nfound",key, sha)
break
else:
print(key, sha, end='\r')
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Whoever manages to obtain the private key for Puzzle 66.
Is requested to encrypt it using this script and post the encrypted message here. Post only the encrypted message, and keep the salt and password to yourself.
Then, proceed to complete your transaction process after few days. If, for any reason, the transaction fails to validate and someone else hijacks it,
You are encouraged to post your salt and password here as evidence that you had solved Puzzle 66.
This will enable the entire community to potentially pursue legal action against the thief. Therefore, anyone attempting to undermine someone else's hard work is hereby warned.
@alberto or any other member, if you discover any vulnerabilities here or are capable of cracking this script, please inform us. Additionally, if there are any alternative implementations, kindly share those as well.
Just the first person who solves Puzzle 66 issues the first 32 bytes + 32 bytes of the public key and replaces it with *********** example: 023b1ac48deb1302bbc9bd302104e6747************************** The first person who solves Puzzle 66 can post it in this thread or a YouTube video... record the time It can prove that he is the first person to discover the private key of Puzzle 66... and transfer 6.6 BTC away the next day... But of course thousands of programs will leak the public key within 1 minute after the transaction, and snatch this bonus No one Really legal possession.... Even the creator saatoshi_rising cannot sue because he has already released the BTC ...as game bonuses This is also a good idea, but when you publicly post either the beginning or the end part of the 32-bit public key of puzzle 66, I am completely sure that Satoshi himself will sweep puzzle 66 .. just like he has done with puzzle 120 and puzzle 125.. and he has not even spent those coins yet, which proves that he has done it,, but if we post a lot of encrypted messages in which fake and real are mixed,, then Satoshi will not even know, and after the public key is exposed, it will be known who has posted the real key. Whether someone does it or not, I am definitely going to do it. I have scanned 12% so far, but if luck doesn't support us, what can we do? I am telling everyone to do this because I am very curious to see this process, and conducting this process silently would only perpetuate confusion surrounding the issue forever. |
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This will enable the entire community to potentially pursue legal action against the thief. Therefore, anyone attempting to undermine someone else's hard work is hereby warned.
Wake up Hey, you well-dwelling toad, wake up a bit. A Bitcoin address isn't entirely anonymous. At the very least, we can demand some percentage from that person. Oh sorry, now I understand. That was an April Fool's joke. No offense, it was a good one  Upon reviewing his many posts, It's evident that he has nothing to do besides poking his nose into others' business. He hasn't shared any ideas or help for the community so far, except for poking others.This person demonstrates a highly negative and narrow-minded attitude. I'm fully confident that he hasn't spent a single penny to solve this puzzle, but if he gets something for free, he's ready to grab it like a dog. If he had worked hard and invested money, he would have been worried about not wasting his effort . He would be concerned about ensuring that their efforts and funds are not wasted and would seek ways to avoid such wastage. Whether it works or not, This post is intended for those who have invested significantly here. For you, this may indeed be considered a joke but not for all. Kindly refrain from nonsense talk and let the community members decide whether this method is appropriate or not. After all, I am not imposing it on them. |
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from cryptography.fernet import Fernet
from cryptography.hazmat.backends import default_backend
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.kdf.pbkdf2 import PBKDF2HMAC
import base64
import os
import getpass
# Function to generate a key from the password
def key_from_password(password, salt):
kdf = PBKDF2HMAC(
algorithm=hashes.SHA256(),
length=32,
salt=salt,
iterations=100000,
backend=default_backend()
)
key = base64.urlsafe_b64encode(kdf.derive(password.encode()))
return key
# Encrypt the message
def encrypt_message(message, password):
salt = os.urandom(16)
key = key_from_password(password, salt)
fernet = Fernet(key)
encrypted = fernet.encrypt(message.encode())
return encrypted, salt
# Decrypt the message
def decrypt_message(encrypted_message, password, salt):
key = key_from_password(password, salt)
fernet = Fernet(key)
try:
decrypted = fernet.decrypt(encrypted_message).decode()
return decrypted
except Exception as e:
return "Error: " + str(e)
# Main interface
if __name__ == "__main__":
choice = input("Do you want to encrypt or decrypt a message? (encrypt/decrypt): ")
if choice == "encrypt":
secret_message = input("Enter your secret message: ")
password = getpass.getpass("Enter a password for encryption: ")
encrypted_message, salt = encrypt_message(secret_message, password)
print("Your encrypted message is:")
print(encrypted_message.decode())
print("Your salt (needed for decryption) is:")
print(base64.b64encode(salt).decode())
elif choice == "decrypt":
encrypted_message_input = input("Enter the encrypted message: ")
salt_input = base64.b64decode(input("Enter the salt: "))
password = getpass.getpass("Enter the password for decryption: ")
decrypted_message = decrypt_message(encrypted_message_input.encode(), password, salt_input)
print("Your decrypted message is:")
print(decrypted_message)
else:
print("Invalid choice. Please type 'encrypt' or 'decrypt'.")
Example:Encrypted message is: gAAAAABmCsIhWZe1vm94Ma4p6v3hv8LmP5JuvhsA1qI65TK_X2LolyNqEFr2y2yZyORoi5KuVubjyay nScTIfAKrqBka17jslzoHfVTB7-c1fgx-qjbjTbDEDUG5vlCE0qy5uOgqBYG4jt5b9gxnhBdVLL3qc2e7nw==-c1fgx-qjbjTbDEDUG5vlCE0qy5uOgqBYG4jt5b9gxnhBdVLL3qc2e7nw== Salt (needed for decryption) is: IJZVtFg2B1u5r/A1bsk/TA== The password for decryption is: password
Whoever manages to obtain the private key for Puzzle 66. Is requested to encrypt it using this script and post the encrypted message here. Post only the encrypted message, and keep the salt and password to yourself. Then, proceed to complete your transaction process after few days. If, for any reason, the transaction fails to validate and someone else hijacks it, You are encouraged to post your salt and password here as evidence that you had solved Puzzle 66. This will enable the entire community to potentially pursue legal action against the thief. Therefore, anyone attempting to undermine someone else's hard work is hereby warned. @alberto or any other member, if you discover any vulnerabilities here or are capable of cracking this script, please inform us. Additionally, if there are any alternative implementations, kindly share those as well. |
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What if I can open all of them at once? Would that not overwhelm the miners? They can't be watching all of them.
What if I just sweep all of them at the same time? What if I know how to generate all of them, the same way he did? Where I have all of them, all at once? What then? It's like being frozen and unable to walk from the casino to the parking lot just to get back to your car after everyone around you heard you won... How do you get back to your car safely?
I need help and I don't know where to find it...
It's been 3 days now and I still can't come up with a safe way to get back to the mother fucking car...
~Ką
(Does anyone live in Southeast Missouri? I will meet you, pay you, to get this over with. We all know how much is here, you can have 40%.)
Hi..Digaran..Meet you after a long time..How are you..Lol |
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I did about 10 attempts today, trying to change already broadcasted transaction, sent from bitcoin-core with RBF flag off.
All attempts fail. First transaction always win.
Then we must do kind of kung-fu. Using older Electrum versions than 4.4.0 with RBF flag off.  I already installed and ready to finalize.. needs only 1 more thing  |
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if someone found 66 PK, can share with me, for withdraw, i know secure way to transfer, other you have seen above test, and definatly, slowly slowly lot of people would have 66 PK, but no one take step for attempt withdraw For a drowning person, even a straw is a great support. I will definitely contact you. When everything is dependent on fate, it's best that we only utilize this code. There's no need for undue force with the computer, no burden on the machine, no excessive worry about electricity costs, and most importantly, without any concern for double spending.  2^129 - 2^130 | PUB : 0383af736b4eefaa2f697e63b4530129ff34ff945be353ee498b0473238ea91ef4 | PVK : 0x2264f44bd23a25b4a5b8390e650cf9cae
2^134 - 2^135 | PUB : 03a987392d57dd82a9ee630f99a75b3925ca5417869bbdfc76b6f59997e558ee73 | PVK : 0x7a5330f4174f496ca6c4364888e9aef5db
2^139 - 2^140 | PUB : 0389d6d9b49a09988a70238b4c79639ef2881c686134075a485843e4814e36320d | PVK : 0xdffa0bf08d3584c46a48904da5df25016af
2^144 - 2^145 | PUB : 0226d47bfd7e231e9c123683137bf0f34522c03a8604fee9a9f626f0334daf9c0b | PVK : 0x167b8d6bec4db640ca2560f120f0a072bf658
2^149 - 2^150 | PUB : 0363abdd6ae3bc62c6ccf3d34d22b44d90f710fa99931d566a6a8c0da68acca933 | PVK : 0x3e1954a5af437841b085ba454c8b9e1149268a
2^154 - 2^155 | PUB : 035bebccb841e7b812fcfbc5fe86d9f9549e68d3d3da1bffd613e8a3960ca26ddc | PVK : 0x7a7179d2027c342b336fb8fd92347d50d937d1e
2^159 - 2^160 | PUB : 022cc56adb8c5385aefbb3cd38dfc2294560c4f73ad4562f68916535d6ab8ebf66 | PVK : 0xc09735b95a670ac005324cb8ddcf3576df7b3f1e
from bit import *
import random
L = [
'03633cbe3ec02b9401c5effa144c5b4d22f87940259634858fc7e59b1c09937852',
'02145d2611c823a396ef6712ce0f712f09b9b4f3135e3e0aa3230fb9b6d08d1e16',
'031f6a332d3c5c4f2de2378c012f429cd109ba07d69690c6c701b6bb87860d6640',
'03afdda497369e219a2c1c369954a930e4d3740968e5e4352475bcffce3140dae5',
'03137807790ea7dc6e97901c2bc87411f45ed74a5629315c4e4b03a0a102250c49',
'035cd1854cae45391ca4ec428cc7e6c7d9984424b954209a8eea197b9e364c05f6',
'02e0a8b039282faf6fe0fd769cfbc4b6b4cf8758ba68220eac420e32b91ddfa673'
]
while True:
seed = random.getrandbits(161)
random.seed(seed)
for exp in range(129, 164, 5):
a = random.randrange(2**exp, 2**(exp+1))
pk = Key.from_int(a)
key = Key.from_int(a)
pub = pk._pk.public_key.format(compressed=True)
addrpub = pub.hex()
if addrpub in L:
print('================found=====================\n')
f=open(u"Key_Found.txt","a")
f.write("Seed : " + str(seed) + "\n")
f.write("Prvk : " + hex(a) + "\n")
f.close()
input('================Exit=====================\n')
exit()
#if addrpub.startswith('03633cbe3') or addrpub.startswith('02145d26') or addrpub.startswith('031f6a33') or addrpub.startswith('03afdda4') or addrpub.startswith('03137807') or addrpub.startswith('035cd185') or addrpub.startswith('02e0a8b0'):
print(f"2^{exp} - 2^{exp+1} | PUB : {addrpub} | PVK : {hex(a)}", end = '\n')
If fortune is on your side, a single core of a computer is sufficient for you.. Good...Luck...Luck...And....Only....Luck |
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Can you crack and post the PVK of that signed address??
-----BEGIN BITCOIN SIGNED MESSAGE-----
I also own this address
-----BEGIN SIGNATURE-----
13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD
H8SlTYJ7a/Mp5cra9VzqgDFMGRQUYfA5NLrSCb0GkwbeEDqx8vKWGYWX3YmiqIU8nl6THdprDK/k34y1GQrFFDk=
-----END BITCOIN SIGNED MESSAGE-----
WTF, I'm utterly astounded to realize that the smaller keys are not secure by any means, and searching for the larger ones is proving to be an insurmountable challenge for us. Given the magnitude of this predicament, it seems wise for us to disengage from this futile endeavor. The journey thus far has been both eye-opening and taxing.   |
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It is enough to mark the tx as without RBF and the winner gets it all!
Stop spreading bs.
This is not bullshit, as some others users already mentioning, Doesn't matter if you mark it with or without RBF, the transaction always can be repleaced and that only depends of Node configuracions to accept it or reject it. The only safest way to move the founds is ask to a miner if they mined your transaction without broadcast it (previously to the block begin mined), also it is even better if you are a miner and you include the transaction without broadcast it before it is mined. Thanks Alberto, then I have to sell the key to the miner who bids the highest. Whats the best way for me to prove that I have the key, without disclosing the private key and public key to a miner ?  Address = 13zb1hQbWVuYdZoAkztVrNrm65aReL2pYD Message = I own this address Signature = HxSDzJj3BhS8lh7qoU1L9zJsymorFkJv8/roi8o6+269PEDPr8psAE1QEOerWAWKv/ypXss7hhXEe0FutUR4b0s= https://bitaps.com/signatureI am still curious to know if someone is able to crack these signatures as well Thanks Alberto, then I have to sell the key to the miner who bids the highest.
Whats the best way for me to prove that I have the key, without disclosing the private key and public key to a miner ?
There is no safe way to do that, even signature text messages include the public key. Can you crack and post the PVK of that signed address?? |
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Please advise me on which wallet to use where the facility to enable and disable RBF (Replace-By-Fee) is available. My fully synchronized Bitcoin Core wallet has become corrupted, and IDK, Why Electrum has removed the RBF option.
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happy birthday,
Thanks Mostly, I work on puzzle 66, but today, being a special day, I was trying some of my old scripts, among which was a batch (.bat) file that I occasionally run. Today, When I executed it, I saw something like Start:2C15823997A13A9000000000000000000
Stop :2C15823997A13A9FFFFFFFFFFFFFFFFFF
Keys :1
Number of CPU thread: 0
Range width: 2^72
Jump Avg distance: 2^36.04
Number of kangaroos: 2^20.25
Suggested DP: 13
Expected operations: 2^37.12
Expected RAM: 707.9MB
DP size: 13 [0xFFF8000000000000]
GPU: GPU #0 NVIDIA GeForce RTX 3060 Ti (38x0 cores) Grid(76x128) (102.0 MB used)
SolveKeyGPU Thread GPU#0: creating kangaroos...
SolveKeyGPU Thread GPU#0: 2^20.25 kangaroos [4.2s]
[999.21 MK/s][GPU 999.21 MK/s][Count 2^36.91][Dead 1][02:27 (Avg 02:29)][483.1/610.4MB]
Key# 0 [1S]Pub: 0x03633CBE38F52C67DED3104637FAF4B05ABC85C6D9815DB628DF18719051FB8B3B
Priv: 0x2C15823997A13A99FE40A3DC9797A1F2E
Done: Total time 02:37 Which made me quite happy. However, this joy was short-lived because the batch file was originally created to check some sample public keys during an experiment, Following this realization I felt quite disheartened in that sorrow, I typed the above message. |
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I am the creator.
You are quite right, 161-256 are silly. I honestly just did not think of this. What is especially embarrassing, is this did not occur to me once, in two years. By way of excuse, I was not really thinking much about the puzzle at all.
I will make up for two years of stupidity. I will spend from 161-256 to the unsolved parts, as you suggest. In addition, I intend to add further funds. My aim is to boost the density by a factor of 10, from 0.001*length(key) to 0.01*length(key). Probably in the next few weeks. At any rate, when I next have an extended period of quiet and calm, to construct the new transaction carefully.
A few words about the puzzle. There is no pattern. It is just consecutive keys from a deterministic wallet (masked with leading 000...0001 to set difficulty). It is simply a crude measuring instrument, of the cracking strength of the community.
Finally, I wish to express appreciation of the efforts of all developers of new cracking tools and technology. The "large bitcoin collider" is especially innovative and interesting!
I need some fund to continue for my discoveries and projects, or else I'll have to abandon these endeavors entirely. It's quite distressing for me, but I've been left with no choice. The creator's support would mean the world to me as I strive to keep my work alive. As you mentioned, "it is simply a crude measuring instrument, of the cracking strength of the community." However, if all members of the community continue to leave their tasks incomplete due to constraints, your measuring instrument will repeatedly break. Please understand that in this community, 99% of members may have limited strength, but they put in a tremendous amount of effort. Anyways... Today is my birthday |
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hi there,
could we get a copy? hmm if not ok, but want to try your nicely made software piece,
you have a github.. or so. anyway thanks a lot.
This involves many third-party commands, here the output of a local server, .exe, and several self-made modules are being taken on a one display, so it is difficult to share it at the moment. I will make it easier and upload on GitHub in the future |
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