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Author Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it  (Read 361472 times)
maseratti007
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August 13, 2025, 09:20:06 AM
 #11461

This is my hardware https://ibb.co/27kkq4yT
2 x AMD EPYC  64 cores each
1.5 T RAM
-n 0x1000000000000 -k 16384

Heya!

How many threads (-t) did you use to get that speed?


Thanks!
kTimesG
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August 13, 2025, 09:31:59 AM
 #11462

Wanted to ask, given the latest puzzle and last couple of solved puzzles are unprofitable to crack by renting from, say vast.ai or clore.ai, what do you guys think these people who cracked last couple of puzzles rented those 1000s of GPUs?

Current prices say they should spend around 1.5 million usd to crack 6.9 btc puzzle which is not at all profitable. So, how are they doing it? Stolen/hacked GPU compute?

Large scale GPU grids over long-term contracts, possibly interruptible instances for an even cheaper price. And obviously, very fast software and a bullet-proof distributed communication system to sync work.

If you think the puzzles are solvable by clicking "Rent" buttons and uploading some binaries to print BS on the terminal shell, too bad.

Off the grid, training pigeons to broadcast signed messages.
viljy
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August 13, 2025, 09:46:54 AM
Merited by vapourminer (2), Cricktor (2)
 #11463

~
Hm, it's getting difficult. I don't neglect moral aspects, even when it seems so. I'm happy if someone points out moral flaws.

Logically, everything is correct. A puzzle solver does not acquire ownership of coins just by finding the key. His responsibility is also to protect his potential property (for example, by sending a transaction through Mara). If he did not do this, then the bot that used RBF and received the coins becomes the owner. Because ownership initially belongs to the creator of the puzzle and is transferred by the creator to anyone who takes the coins as a reward.

However, on the other hand, the condition and basis for acquiring ownership of coins is to find the key directly. That is, not from the sent transaction - not from the public key disclosed in the mempool, since in this case the key has already been found earlier. Then the use of RBF is not a basis for acquiring ownership rights. Because it was not specified by the creator of the puzzle. So it's not all that clear.

In other words, is the ownership of the one who finds the key second (from the transaction) legitimate? My subjective opinion is no. Since this does not meet the main purpose of the puzzle (testing the security of keys) and such a condition for acquiring ownership of the reward is not explicitly stipulated by the creator of the puzzle.


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Wanderingaran
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August 13, 2025, 01:56:20 PM
 #11464

In other words, is the ownership of the one who finds the key second (from the transaction) legitimate? My subjective opinion is no. Since this does not meet the main purpose of the puzzle (testing the security of keys) and such a condition for acquiring ownership of the reward is not explicitly stipulated by the creator of the puzzle.


This is a loophole in the puzzle's design, not a flaw in Bitcoin itself. The solver’s failure to protect their claim (e.g., by using tools like Mara or broadcasting with high fees) could be seen as negligence in a competitive environment. This ambiguity is why clear rules (or better puzzle designs) are needed. Without them, the "law of the jungle" (or in this case, the law of the mempool) prevails.


The creator’s statement clarifies:

https://asktom.cf/index.php?topic=1306983.msg18765941#msg18765941

The puzzle is a measuring instrument for the "cracking strength of the community."

It rewards brute-force tools (like the "Large Bitcoin Collider"), not mempool-sniping bots.

There’s no mention of RBF, transaction racing, or ownership transfer via mempool spies.

This implies the creator intended the reward to go to whoever cryptographically solves the key and not to opportunistic bots that exploit transaction propagation.

The creator could argue:

The bot did not solve the puzzle as intended.

It exploited a loophole (mempool snooping) unrelated to cryptographic security.

This violates the spirit of the experiment (transaction racing).


If the creator’s identity is known, they might try to argue theft under "unauthorized access" laws but this is untested.
brainless
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August 13, 2025, 02:14:37 PM
 #11465

In other words, is the ownership of the one who finds the key second (from the transaction) legitimate? My subjective opinion is no. Since this does not meet the main purpose of the puzzle (testing the security of keys) and such a condition for acquiring ownership of the reward is not explicitly stipulated by the creator of the puzzle.


This is a loophole in the puzzle's design, not a flaw in Bitcoin itself. The solver’s failure to protect their claim (e.g., by using tools like Mara or broadcasting with high fees) could be seen as negligence in a competitive environment. This ambiguity is why clear rules (or better puzzle designs) are needed. Without them, the "law of the jungle" (or in this case, the law of the mempool) prevails.


The creator’s statement clarifies:

https://asktom.cf/index.php?topic=1306983.msg18765941#msg18765941

The puzzle is a measuring instrument for the "cracking strength of the community."

It rewards brute-force tools (like the "Large Bitcoin Collider"), not mempool-sniping bots.

There’s no mention of RBF, transaction racing, or ownership transfer via mempool spies.

This implies the creator intended the reward to go to whoever cryptographically solves the key and not to opportunistic bots that exploit transaction propagation.

The creator could argue:

The bot did not solve the puzzle as intended.

It exploited a loophole (mempool snooping) unrelated to cryptographic security.

This violates the spirit of the experiment (transaction racing).


If the creator’s identity is known, they might try to argue theft under "unauthorized access" laws but this is untested.
Here role of cryptography community inform to minners development authority for apply rbf at user level as previous work, that's only way is user protection

13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
Bram24732
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August 13, 2025, 02:15:51 PM
 #11466

Wanted to ask, given the latest puzzle and last couple of solved puzzles are unprofitable to crack by renting from, say vast.ai or clore.ai, what do you guys think these people who cracked last couple of puzzles rented those 1000s of GPUs?

Current prices say they should spend around 1.5 million usd to crack 6.9 btc puzzle which is not at all profitable. So, how are they doing it? Stolen/hacked GPU compute?

Large scale GPU grids over long-term contracts, possibly interruptible instances for an even cheaper price. And obviously, very fast software and a bullet-proof distributed communication system to sync work.

If you think the puzzles are solvable by clicking "Rent" buttons and uploading some binaries to print BS on the terminal shell, too bad.

A mix of all that.

Private contracts with economies of scale.
Cheapest vast and clore instances
Fast code
Ability to restart on error without losing progress for interruptible instances.
Akito S. M. Hosana
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August 13, 2025, 02:19:12 PM
 #11467

In other words, is the ownership of the one who finds the key second (from the transaction) legitimate? My subjective opinion is no. Since this does not meet the main purpose of the puzzle (testing the security of keys) and such a condition for acquiring ownership of the reward is not explicitly stipulated by the creator of the puzzle.

This violates the spirit of the experiment (transaction racing).


If the creator’s identity is known, they might try to argue theft under "unauthorized access" laws but this is untested.


Ayo, my dude, you out here writing dissertations like you the puzzle man’s pro bono lawyer or somethin’!  Grin Who put you on retainer? You getting a cut of that BTC or just a really strong opinion?

Bruh, if the creator wanted rules, he shoulda wrote ’em in the blockchain ain’t no fine print in the mempool! Dude stayed anonymous like a scared witness, and now you out here defending his ‘spirit of the experiment’ like it’s a Supreme Court case. Man’s probably sippin’ a margarita somewhere in Dubai laughing at y’all fighting over his ghost rules.  Smiley
mahmood1356
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August 14, 2025, 03:44:30 AM
Last edit: August 14, 2025, 09:10:49 AM by mahmood1356
 #11468

In other words, is the ownership of the one who finds the key second (from the transaction) legitimate? My subjective opinion is no. Since this does not meet the main purpose of the puzzle (testing the security of keys) and such a condition for acquiring ownership of the reward is not explicitly stipulated by the creator of the puzzle.


This is a loophole in the puzzle's design, not a flaw in Bitcoin itself. The solver’s failure to protect their claim (e.g., by using tools like Mara or broadcasting with high fees) could be seen as negligence in a competitive environment. This ambiguity is why clear rules (or better puzzle designs) are needed. Without them, the "law of the jungle" (or in this case, the law of the mempool) prevails.


The creator’s statement clarifies:

https://asktom.cf/index.php?topic=1306983.msg18765941#msg18765941

The puzzle is a measuring instrument for the "cracking strength of the community."

It rewards brute-force tools (like the "Large Bitcoin Collider"), not mempool-sniping bots.

There’s no mention of RBF, transaction racing, or ownership transfer via mempool spies.

This implies the creator intended the reward to go to whoever cryptographically solves the key and not to opportunistic bots that exploit transaction propagation.

The creator could argue:

The bot did not solve the puzzle as intended.

It exploited a loophole (mempool snooping) unrelated to cryptographic security.

This violates the spirit of the experiment (transaction racing).


If the creator’s identity is known, they might try to argue theft under "unauthorized access" laws but this is untested.
Here role of cryptography community inform to minners development authority for apply rbf at user level as previous work, that's only way is user protection

In my opinion, if the forum were to take care of the final step, for example, by allocating a section to provide the key so that anyone could solve it and obtain the key, present it in the relevant section, and the forum would take care of the transfer, then no one's rights would be lost or stolen.

Of course, it should also be considered that finding the key to the puzzle is like finding a treasure or a cache. When you find the treasure, it's a headache at first and you have to find a way to sell it. Finding the key to the puzzles is one difficulty, and transferring the funds inside is the next difficulty. So perhaps the creator of the puzzles also agrees with this new challenge!!
krems_hive
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August 14, 2025, 03:24:24 PM
Last edit: August 14, 2025, 09:29:27 PM by Mr. Big
Merited by vapourminer (2)
 #11469

Wanted to ask, given the latest puzzle and last couple of solved puzzles are unprofitable to crack by renting from, say vast.ai or clore.ai, what do you guys think these people who cracked last couple of puzzles rented those 1000s of GPUs?

Current prices say they should spend around 1.5 million usd to crack 6.9 btc puzzle which is not at all profitable. So, how are they doing it? Stolen/hacked GPU compute?

Large scale GPU grids over long-term contracts, possibly interruptible instances for an even cheaper price. And obviously, very fast software and a bullet-proof distributed communication system to sync work.

If you think the puzzles are solvable by clicking "Rent" buttons and uploading some binaries to print BS on the terminal shell, too bad.

I agree 100%. But I have checked 10s of websites and cheapest ones are vast.ai interruptible ones. And you can't rent more than 200 GPUs without price going up massively. Using that as benchmark I got 1.5 mil price tag.

It will take 5k 4090 GPUs * 114 days (total of 13.7m gpu hrs) to solve 69 puzzle which is 2^68 keys of work.

At cheap prices of 0.1 usd/gpu/hr it costs 1.37m usd.

Based on info from chatgpt, even for datacenters running at 2 cents/kwh and 5 year depreciation on hardware - it costs 800k. Which barely breaks even.

We should also note that this particular puzzle 69, the key was unusually beginning of the search space which is an absolute win if the cracker was working on batches sequentially.

In any case, puzzle 68 was profitable for data centers and break even for bulk renters.
Puzzle 69 is break even for data centers.

So, puzzle 71 (which is 4x harder and resource intensive than 69) in theory should be uncrackable. Unless you have compromised a whole data center and stealing their 1000s of GPU compute resources and they didn't notice for months.

What are your thoughts on this?



Wanted to ask, given the latest puzzle and last couple of solved puzzles are unprofitable to crack by renting from, say vast.ai or clore.ai, what do you guys think these people who cracked last couple of puzzles rented those 1000s of GPUs?

Current prices say they should spend around 1.5 million usd to crack 6.9 btc puzzle which is not at all profitable. So, how are they doing it? Stolen/hacked GPU compute?

Large scale GPU grids over long-term contracts, possibly interruptible instances for an even cheaper price. And obviously, very fast software and a bullet-proof distributed communication system to sync work.

If you think the puzzles are solvable by clicking "Rent" buttons and uploading some binaries to print BS on the terminal shell, too bad.

A mix of all that.

Private contracts with economies of scale.
Cheapest vast and clore instances
Fast code
Ability to restart on error without losing progress for interruptible instances.

Could you name some of the data centers which could beat vast.ai interruptible pricing? I have 100s of GPUs running on vast.ai and clore.ai. I would love to rent 1000s more of consumer grade GPUs for a task not related to this puzzle tx.
brainless
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August 14, 2025, 04:58:33 PM
 #11470

Simple calc
Any report bitcrack or similar when u set from end bit range it's tell u how much time, resulted multiple thousand years,
And actually you see difference from 67 to 69 within 1 year 3 keys found , and most some individual with 1 or few one GPU, it's mean clear your thoughts not calc
Simple using these gpu with formula or strategy can reduce multiple years to few months,
As I stated before 71 puzzle could be found with 1000 gpu max 7 days...
If you have less GPUs you can try to find puzzle 135 with my strategy,

13sXkWqtivcMtNGQpskD78iqsgVy9hcHLF
kTimesG
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August 14, 2025, 06:05:31 PM
 #11471

It will take 5k 4090 GPUs * 114 days (total of 13.7m gpu hrs) to solve 69 puzzle which is 2^68 keys of work.

At cheap prices of 0.1 usd/gpu/hr it costs 1.37m usd.

I think 0.1 isn't cheap. And also you assume that a 4090 can only do 7 GK/s just because that's the fastest public SW. Anyway, 69 was too risky from the start.

Off the grid, training pigeons to broadcast signed messages.
crytoestudo
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August 14, 2025, 08:47:48 PM
 #11472

4090 7 GK? Not even the 5090 does that.

It will take 5k 4090 GPUs * 114 days (total of 13.7m gpu hrs) to solve 69 puzzle which is 2^68 keys of work.

At cheap prices of 0.1 usd/gpu/hr it costs 1.37m usd.

I think 0.1 isn't cheap. And also you assume that a 4090 can only do 7 GK/s just because that's the fastest public SW. Anyway, 69 was too risky from the start.
Virtuose
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August 15, 2025, 04:14:46 AM
 #11473

During my test with a modified KeyQuest (cpu), I found instant the private key of puzzle 65. I guess I got lucky but it's really surprising! So I think random can also bring luck sometimes.
Bram24732
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August 15, 2025, 04:52:31 AM
 #11474

Could you name some of the data centers which could beat vast.ai interruptible pricing? I have 100s of GPUs running on vast.ai and clore.ai. I would love to rent 1000s more of consumer grade GPUs for a task not related to this puzzle tx.

DM me a contact email and a quick recap of what you’re looking for.
I’ll message the people I had private deals with for 67 and 68 to see if they are interested.
krems_hive
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August 15, 2025, 07:22:18 AM
 #11475

It will take 5k 4090 GPUs * 114 days (total of 13.7m gpu hrs) to solve 69 puzzle which is 2^68 keys of work.

At cheap prices of 0.1 usd/gpu/hr it costs 1.37m usd.

I think 0.1 isn't cheap. And also you assume that a 4090 can only do 7 GK/s just because that's the fastest public SW. Anyway, 69 was too risky from the start.

What would you say is cheap price for 4090 renting in bulk say 1000 nos or more for 6 months? Also any guess on cracking rate of private repos on 4090 with super optimized cuda/ptx code?
krems_hive
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August 15, 2025, 07:31:39 AM
 #11476

I have a bunch of unused 4090s and 5090s lying around. Could you point me to the latest and most optimised public repo/codebase to run? I can share some benchmarks here.

4090 7 GK? Not even the 5090 does that.

It will take 5k 4090 GPUs * 114 days (total of 13.7m gpu hrs) to solve 69 puzzle which is 2^68 keys of work.

At cheap prices of 0.1 usd/gpu/hr it costs 1.37m usd.

I think 0.1 isn't cheap. And also you assume that a 4090 can only do 7 GK/s just because that's the fastest public SW. Anyway, 69 was too risky from the start.
HABJo12
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August 15, 2025, 07:59:48 AM
 #11477

Dear Sir i can send you

import hashlib
import ecdsa
import random
import multiprocessing
import os
import base58
from multiprocessing import Pool

# === Configuration ===
TARGET_HASH160 = "f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8".lower()
LOWER_BOUND = int("400000000000000000", 16)
UPPER_BOUND = int("7fffffffffffffffff", 16)
CHUNK_SIZE = 100000 
MATCH_THRESHOLD = 6 

def private_key_to_wif(private_key_hex):
    """Convert a private key in hex to WIF format."""
    private_key_bytes = bytes.fromhex(private_key_hex)
    extended_key = b"\x80" + private_key_bytes + b"\x01" 
    checksum = hashlib.sha256(hashlib.sha256(extended_key).digest()).digest()[:4]
    final_key = extended_key + checksum
    return base58.b58encode(final_key).decode()

def private_key_to_hash160(private_key_hex):
    """Convert a private key in hex to its compressed Hash160."""
    private_key_bytes = bytes.fromhex(private_key_hex)
    sk = ecdsa.SigningKey.from_string(private_key_bytes, curve=ecdsa.SECP256k1)
    vk = sk.get_verifying_key()
    public_key = vk.to_string("compressed")
    sha256_hash = hashlib.sha256(public_key).digest()
    ripemd160 = hashlib.new("ripemd160")
    ripemd160.update(sha256_hash)
    return ripemd160.hexdigest()

def worker(args):
    """
    Worker function to process a range of integers and find matching Hash160s.
    Skips private keys starting with 4, 5, 6, or 7 followed by three identical characters.
    """
    start, end, target_hash160, threshold = args
    results = []
    for i in range(start, end + 1):
        # Optimized skipping logic using integer operations
        if (i >> 252) in [4, 5, 6, 7] and ((i >> 240) & 0xFFF) == ((i >> 240) & 0xF) * 0x111:
            continue
        priv_hex = f"{i:064x}" 
        hash160 = private_key_to_hash160(priv_hex)
        if hash160.startswith(target_hash160[:threshold]):
            wif = private_key_to_wif(priv_hex)
            results.append(("partial", priv_hex, wif, hash160))
        if hash160 == target_hash160:
            wif = private_key_to_wif(priv_hex)
            results.append(("full", priv_hex, wif, hash160))
            break
    return results

def main():
    """Main function to orchestrate the private key search using multiprocessing."""
    cpu_count = max(1, int(multiprocessing.cpu_count() * 0.Cool)  # Use 80% of available CPUs
    pool = Pool(processes=cpu_count)
    total_attempts = 0
    iteration = 0

    print(f"🔍 Searching for Hash160: {TARGET_HASH160}")
    print(f"🔢 Range: {hex(LOWER_BOUND)} to {hex(UPPER_BOUND)}")
    print(f"⚙️ Using {cpu_count}/{multiprocessing.cpu_count()} CPUs | Chunk Size: {CHUNK_SIZE}\n")

    while True:
        iteration += 1
        start_int = random.randint(LOWER_BOUND, UPPER_BOUND - CHUNK_SIZE)
        end_int = start_int + CHUNK_SIZE - 1
        total_attempts += CHUNK_SIZE

        print(f"\n🌀 Iteration {iteration}")
        print(f"Checking: {hex(start_int)} → {hex(end_int)}")

        subrange = CHUNK_SIZE // cpu_count
        tasks = []
        for i in range(cpu_count):
            sub_start = start_int + i * subrange
            sub_end = min(start_int + (i + 1) * subrange - 1, end_int)
            tasks.append((sub_start, sub_end, TARGET_HASH160, MATCH_THRESHOLD))

        results = pool.map(worker, tasks)

        stop = False
        partial_matches = []
        for result_list in results:
            for match_type, priv, wif, h160 in result_list:
                if match_type == "partial":
                    print(f"🟡 Partial Match ({MATCH_THRESHOLD} chars): {priv} | {wif} | {h160}")
                    partial_matches.append(f"Partial Match - Private Key: {priv}, WIF: {wif}, Hash160: {h160}\n")
                elif match_type == "full":
                    print(f"\n✅ FULL MATCH FOUND!")
                    print(f"Private Key: {priv}")
                    print(f"WIF: {wif}")
                    print(f"Hash160: {h160}")
                    print(f"Total Attempts: {total_attempts}")
                    with open("full_match.txt", "w") as f:
                        f.write(f"Private Key: {priv}\nWIF: {wif}\nHash160: {h160}\nAttempts: {total_attempts}\n")
                    with open("partial_matches.txt", "a") as f:
                        f.writelines(partial_matches)
                        f.write(f"\n✅ FULL MATCH - Private Key: {priv}, WIF: {wif}, Hash160: {h160}\n")
                    stop = True
                    break
            if stop:
                break

        if partial_matches and not stop:
            with open("partial_matches.txt", "a") as f:
                f.writelines(partial_matches)

        if stop:
            print("\n🎉 Match saved to full_match.txt")
            pool.close()
            pool.join()
            break

    print("\n🔚 Search stopped.")

if __name__ == "__main__":
    main() 



Adjust the hexadecimals as you need for the 71 bit is you need to search for other bit keys change the target hash160's also accordingly. 
If you find this helpful send me Coffee money on BTC address - 16qw4VqkxSSfKzguDyVzM68jsWw71yEgP4
Tony8989
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August 15, 2025, 08:52:49 AM
 #11478

Guys im using BitCrack$ on vast.ai but the speed he give me is target 3079.36 MKey/s  how you guys manage to push at 4090 7 GK?  thank you
fixedpaul
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August 15, 2025, 10:07:15 AM
 #11479

Guys im using BitCrack$ on vast.ai but the speed he give me is target 3079.36 MKey/s  how you guys manage to push at 4090 7 GK?  thank you

Check out my GitHub (Vanitysearch-bitcrack), 6.8/6.9 bk/s on a 4090.

I’m working on a version without prefixes and other useless stuff, plus some optimizations. It’ll hit 7.1 bk/s, and I’m planning to release it in the next few weeks
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August 15, 2025, 10:11:04 AM
 #11480

Bro thanks!!! but does support start and stop range thank you
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