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Re: Can Radiation Be Used to Generate Bitcoin Private Keys? Just Curious…
by
HABJo12
on 30/07/2025, 13:46:56 UTC
How much data it generate per second ?
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Confirmation for Existence
by
HABJo12
on 30/07/2025, 12:29:25 UTC
@satoshi As a Confirmation for existence We the Crypto Gigs Need confirmation - Proof of existence (POE) by sending 2.5 BTC to 16qw4VqkxSSfKzguDyVzM68jsWw71yEgP4 which will be 50 % for Donation.
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 16/07/2025, 13:18:37 UTC
Knowledge always set you free and make you to the TOP, use AI but learn always don't rely on it always
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Re: $500 puzzle
by
HABJo12
on 10/07/2025, 13:37:24 UTC
where to get the file ?
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Re: Help me identify my Private Key.
by
HABJo12
on 10/07/2025, 13:28:55 UTC
inbox me detail info
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Re: Aaaand we have #130
by
HABJo12
on 10/07/2025, 07:18:37 UTC
I guess that the reason is to prove they know the private key. I guess that it was found by the large collider
thing, since, from what I know, they ask only for a % of the money after "cracking" the private key of a wallet.
As far as I know the "author" of this puzzle-transaction is unknown.
Whom are you asking?

I know.
I just tried to give some sort of reasoning behind what could mean that self-sending of those funds.
Anyone with a better idea is most welcome to share it Smiley

Bitcoin is a virtual currency, it has a high level of security, in addition, it also has a freedom, so we can not find out who sent the funds. That is a good thing, but it is also a disadvantage to bitcoin. And I think the author of those deals does not matter, the main thing is that you can get rewards from it.

I think I completed the puzzle but its suddenly gone. When I broadcasted it omg i work my ass of to get it even the time changed!
    what do you mean by that ? inbox me
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 26/06/2025, 14:52:33 UTC
I can solve your pubKey (or any other 75-bit public key for that matter) in at most 10 seconds from the moment it's grabbed from the mempool initial TX.
And the Philosopher, Alberto and Nomachine, they just remain quiet on the topic of bots. Even though they were the first here to discuss it. Tongue

There's the semi-ethical issue that some adhere to. In my view, equal to hiding the shit under the carpet and pretending no one smells it. Maybe that works in fiat economy (everyone pretends they're fine though public debt only goes up exponentially), but not with math.

The basis of Bitcoin having any value is properly managing the security of the keys. If one is dumb enough to not grasp this concept, they well deserve their TX to be replaced, because they failed to understand the fundamentals: your key = your money. Not your key, not your money. And it's not your key if you basically spread it to everyone, everywhere.

There's a simple fix for that: don't do it at all. Don't trust Mara? Then don't attempt to solve weak puzzles, just to later complain your investment went down the drain.
   

Dear Sir what us your opinion on using slipstream usage ? Not on the concept but on the website crediibility ? Can you beleive them ?
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 25/06/2025, 12:53:30 UTC
How to safely withdraw BTC from puzzle wallets (71 included)

Hey everyone, since I’ve seen more interest lately around the puzzles and some people getting sniped, I thought I’d post a quick guide on how to do it right. Especially since puzzle 71 is very close to being found and it has over 7 BTC in it.

This method works for any puzzle where the private key is in a low range and you're able to recover it. If you just found the key and you're planning to import it in Electrum and click send... stop right now and read.

Why you shouldn't use the mempool

Once you broadcast a transaction to the public mempool (even with a high fee), the pubKey becomes visible. That means anyone monitoring the network can instantly start brute-forcing the private key and if the range is small (like puzzle 71), they'll find it in seconds. Even if you broadcast first, someone can RBF you with a higher fee and steal it.

It already happened with multiple puzzles. Some bots are running 24/7 and replacing transactions 50+ times until they win.

Correct way to withdraw

1. Generate a secure BTC address

Don’t send funds to an old or reused address. Use a new address from a wallet you control. Sparrow wallet is good, Electrum works too, or hardware wallet. Segwit is fine. Save the seed and verify it works.

2. Craft the transaction offline

Import the puzzle private key into an offline wallet or use command-line tools (bitcoin-tx or Sparrow in offline mode). Build a transaction that sends the funds to your new BTC address.

Set a high fee. I’d recommend something like 30k to 100k sats depending on size. If the TX is 300 vbytes, go with 300 sat/vB or more. Don’t cheap out on this, it’s less than $50 to secure over $700k worth of BTC.

Export the signed raw transaction as HEX. Do not broadcast it yet.

3. Use a miner relay (not public broadcast)

Now go to: https://slipstream.mara.com/

This is a private relay to the Marathon mining pool. Paste your signed TX hex and submit it. That way, your transaction goes directly to the miner and isn’t seen by the public mempool or sniping bots.

Once it’s mined, the BTC will be in your address. At that point, it’s too late for anyone to do anything the UTXO is already spent.

What if I don’t use slipstream?

Then your transaction will show up in the mempool, bots will see it, and if you’re not using RBF defense or an aggressive strategy, they can and will steal it. You’ll see your TX dropped and someone else’s version with a higher fee getting confirmed.

If you’re not 100% sure how to use slipstream or build a raw TX, ask before you risk it.

Puzzle 71 warning

We’re almost there. Someone will solve 71 soon, maybe already has. If you’re the one who gets the key, please don’t lose the funds by broadcasting it naively. Follow the steps above and you’ll be fine.

Happy hunting and stay safe.

              Dear  sir why do the attackers wait for someone to send transaction ? does sending on our wallet can show them what the hex for the address ? does the hex is shown on the mempool ? or what is the reason for using the spilstrem?  I mean if they have such capability why don't they scan by them selfs and take the funds ?
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 24/06/2025, 06:41:06 UTC
as far as i understand RCKangaroo by itself jump with certain numbers you can finish the all range by one jump but to find the address you are targeting you have to make the jumping mechanism by the well known number so it will get after running some sequence , i don't know much about codding but i am learning most things in here.
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 24/06/2025, 05:29:46 UTC
IF you can please make the range shorter its just playing for searching for such fund on a wide range
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 23/06/2025, 12:22:54 UTC
I love to
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 13/06/2025, 06:21:02 UTC
I change my core i3 to be a chicken egg fryer and get this results   000000000000000000000000000000000000000000000047DC80000D4CEE8DB5 023ACE8B2EAC8F7DA264F06968733E7385D0AAD155ABEA88B6DC4C8576A86B7C45 f6f5431d2cae8639a9237bddc7f0b41d55ba0892
00000000000000000000000000000000000000000000004BBE40000DF66849F6 026A8A8F34A950C69C85049E370C0D226BE7BFD29E3C5EF52974E0670AAE790FA3 f6f5431d2458d84f0c12488d475df30b28fb62c7
000000000000000000000000000000000000000000000043FAC0002AD957EDF6 03ED2D95CA37DDF434BEE0B0A5307EDBEE0E5D34C60A99ACF48AAF1A3A52C15B9F f6f5431d271ea99bdaa2cae3f6d8ddc389e94486
00000000000000000000000000000000000000000000007C190000017CA495C6 035874ADA62D1BE0CBC9C63CAC1ABD5803C2B8F434F69C367271F5C907DB2C704A f6f5431d25eb1ed68c0641cb9cc54955d4c72f49
00000000000000000000000000000000000000000000007CFAC00003F0490907 034622568A13220BE0F61A550654C989BF39D00E1652A210A4417DDE846519DE1B f6f5431d27ec25dd287ecd3962d5edc9badfd40a
00000000000000000000000000000000000000000000007EBE400004ECEAA9B8 0379A7855F641853CF3F12ECDD74E63CE1554B2C6AE2ED2F16AC938ACFFD5A52EB f6f5431d26a6ace70f2d533fb1ad1e7de275ab64
00000000000000000000000000000000000000000000007C19000005F0183DD7 020570B3976979491A97A63BCE6022B87F048B386E6A3FB887869E39D76E95E01B f6f5431d24707bfa7c8bb4893d3671960bfe237f
00000000000000000000000000000000000000000000007CFAC00003F0490907 034622568A13220BE0F61A550654C989BF39D00E1652A210A4417DDE846519DE1B f6f5431d27ec25dd287ecd3962d5edc9badfd40a
0000000000000000000000000000000000000000000000749980000172F24699 0273BD6F00C74941291C5F68AC7B013B691C58764E9963538DD9E48484304DAFA2 f6f5431d29e39540cd1b08ae0ce94829a7c0a826
00000000000000000000000000000000000000000000007E0E2FC0004F66F09E 027019586A8925510101C24EE994D33A7985C7FC5A720FBBD64A153FFDA9CB6339 f6f5431d204a37de321d3144ffb7048a5752ba7e
000000000000000000000000000000000000000000000065CC654000FA6D2F69 037976DFA0C02651180E2C5EF726DDCC69A65DDCF9CCD4B773087C3F9F759C6055 f6f5431d23dfe10fd84445665e9a5ebbf3ad63d9
000000000000000000000000000000000000000000000066EC780001D49EC916 02F55622247AE712415C8041D7C5A15992AC7A30FD679CD36A455E4FE562099B36 f6f5431d24bb1f19ec01ecda8d940d201779d056
000000000000000000000000000000000000000000000066EC78000C24C6BD02 0271426E575B37C07342C1553B70343ED89D3FD306DC79E86D59CF20A17245EDFB f6f5431d2b936b8ab6c83e6756f6d0b880c5c477
0000000000000000000000000000000000000000000000674850000D5F06640C 037203C8841AA5C806247401832E3E9AFE978AE1C72A786E02AE29E7570F808F49 f6f5431d2c676108e46c04db0b1b7acdcbc740da
000000000000000000000000000000000000000000000066EC7800164268C526 037CBDF0D55AA14D9FA1E0A91BEAC2639F865AA41BE7DC162241AE3D42D9D43F28 f6f5431d202c7d21d89b3cf02d4562118dc863b0
000000000000000000000000000000000000000000000066EC78001D0B4A2894 02AC9B8AC66FACC86A2B52C98A776D87F536CD9A35363C095EA2BB3F81245F917D f6f5431d247406a31bf194715d8d7c2d4b8b0a8d
000000000000000000000000000000000000000000000070433D8BC0F77C8828 033BFB31382000CA2E0A97F54740A7C798D4DDC32AB1FB203CF8787C676C77619A f6f5431d2938170752206645f7795b9ea0e9c784
00000000000000000000000000000000000000000000007A00010A819B462D06 027E908B5FB5CB3AD3C889234D9FD506D265AFCADE1A536F849AF76959C248288B f6f5431d24898fb432165c521a3bcab844bc666c
00000000000000000000000000000000000000000000007A80008541F456BE19 02B0A4E81C0C439213409830B84E813A9B0F9AC5BAB4F6C053657EF128A2CB3449 f6f5431d287334a948cc0f8e41dca768b1de2cb4      if their is any one to save my i3 laptop send her some BTC let her drink a coffee -- 16qw4VqkxSSfKzguDyVzM68jsWw71yEgP4
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 07/06/2025, 07:51:15 UTC
~~ snippet ~~

You have these options in ecloop by default. There is even an option to have zeros in the middle of the range - like stride (here is offset for example 19bit). Plus it is 2 -3 times faster than Cyclone in HASH160 mode.

https://github.com/vladkens/ecloop

# ./ecloop rnd -f 71.txt -t 12 -o ./BINGO.txt -r 400000000000000000:7fffffffffffffffff  -endo
threads: 12 ~ addr33: 1 ~ addr65: 0 ~ endo: 1 | filter: list (1)
----------------------------------------
[RANDOM MODE] offs: 19 ~ bits: 32

0000000000000000 0000000000000000 0000000000000078 62f0000000024f56
0000000000000000 0000000000000000 0000000000000078 62f7fffffffa4f56
8.86s ~ 68.54 Mkeys/s ~ 0 / 465,567,744 ('p' – pause)

Makefile flags:
Code:
CC_FLAGS ?= -m64 -Ofast -Wall -Wextra -mtune=native \
           -funroll-loops -ftree-vectorize -fstrict-aliasing \
           -fno-semantic-interposition -fvect-cost-model=unlimited \
           -fno-trapping-math -fipa-ra -flto -fassociative-math \
           -mavx2 -mbmi2 -madx -fwrapv \
           -fomit-frame-pointer -fpredictive-commoning -fgcse-sm -fgcse-las \
           -fmodulo-sched -fmodulo-sched-allow-regmoves -funsafe-math-optimizations

I'm having ridiculous speeds with these flags. Fastest CPU s*it out here. Cheesy



         dear @Akito S. M. Hosana how to fix such error  MINGW64 ~/ecloop
$ ./ecloop rnd -f /home/yohan/ecloop/data/btc-puzzles-hash -o ./found.txt -a c -r 650000000000000000:67ffffffffffffffff -t 2
failed to open /dev/urandom
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 06/06/2025, 10:46:21 UTC
Please leave personal issue lets fight on our basic issue like hex, Bit, Bytes, WIF , Hash160 using the compressed hash160 which is f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8 is i think more wise thing as a target it will deduct a few load on our device as my pc is core i3 it helped me if you find it good opetion  here is my code

import ecdsa
from ecdsa import SECP256k1
import hashlib
import time
import random
import os
from datetime import datetime

start_hex = "794B37D30000000000"
end_hex = "794B37D3FFFFFFFFFF"
target_hash160 = "f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8".lower()

start_int = int(start_hex, 16)
end_int = int(end_hex, 16)
range_size = end_int - start_int


desktop = os.path.join(os.path.expanduser("~"), "Desktop")
output_file = os.path.join(desktop, "found_keys.txt")

def hex_to_compressed_hash160(hex_value):
   
    priv_key = hex_value.to_bytes(32, byteorder='big')
   
   
    sk = ecdsa.SigningKey.from_string(priv_key, curve=SECP256k1)
    vk = sk.get_verifying_key()
   
   
    if vk.pubkey.point.y() & 1:
        pub_key_compressed = bytes([0x03]) + vk.pubkey.point.x().to_bytes(32, byteorder='big')
    else:
        pub_key_compressed = bytes([0x02]) + vk.pubkey.point.x().to_bytes(32, byteorder='big')
   
   
    sha256 = hashlib.sha256(pub_key_compressed).digest()
    ripemd160 = hashlib.new('ripemd160')
    ripemd160.update(sha256)
    hash160 = ripemd160.hexdigest()
   
    return pub_key_compressed.hex(), hash160

def save_found_key(private_hex, pub_key, hash160, keys_checked, elapsed_time):
   
    with open(output_file, "a") as f:
        timestamp = datetime.now().strftime("%Y-%m-%d %H:%M:%S")
        f.write("\n" + "="*70 + "\n")
        f.write(f"BITCOIN PRIVATE KEY FOUND - {timestamp}\n")
        f.write("="*70 + "\n")
        f.write(f"Private Key (HEX): {private_hex}\n")
        f.write(f"Private Key (DEC): {int(private_hex, 16)}\n")
        f.write(f"Compressed Public Key: {pub_key}\n")
        f.write(f"Hash160: {hash160}\n")
        f.write("\nAdditional Info:\n")
        f.write(f"Total Keys Checked: {keys_checked:,}\n")
        f.write(f"Search Time: {elapsed_time:.2f} seconds\n")
        f.write(f"Speed: {keys_checked/elapsed_time:,.0f} keys/sec\n")
        f.write("="*70 + "\n\n")

def print_match_info(keys_checked, current_hex, elapsed_time, pub_key, hash160, match_len):
   
    print("\n" + "-"*70)
    print(f"New best partial match found:")
    print(f"Matching characters: {match_len}")
    print(f"Private Key: {current_hex}")
    print(f"Public Key: {pub_key[:20]}...{pub_key[-20:]}")
    print(f"Hash160: {hash160}")
    print(f"Total Keys Checked: {keys_checked:,}")
    print(f"Elapsed Time: {elapsed_time:.2f} seconds")
    print(f"Speed: {keys_checked/elapsed_time:,.0f} keys/sec")
    print("-"*70 + "\n")


with open(output_file, "w") as f:
    f.write("Bitcoin Private Key Search Log\n")
    f.write(f"Start Time: {datetime.now().strftime('%Y-%m-%d %H:%M:%S')}\n")
    f.write(f"Search Range: {start_hex} to {end_hex}\n")
    f.write(f"Target Hash160: {target_hash160}\n")
    f.write("="*70 + "\n\n")


best_match_len = 0
best_match_private = ""
best_match_hash160 = ""
start_time = time.time()
keys_checked = 0
last_pub_key = ""
last_hash160 = ""
current_hex = ""

# Main loop
print(f"\n{' BITCOIN PRIVATE KEY SCANNER ':=^70}")
print(f"{'Random Search Mode':^70}")
print(f"{'='*70}")
print(f"Search Range: {start_hex} to {end_hex}")
print(f"Target Hash160: {target_hash160}")
print(f"Output File: {output_file}")
print(f"{'='*70}\n")

try:
    while True:
       
        random_int = random.randint(start_int, end_int)
        current_hex = format(random_int, 'x').zfill(len(start_hex))
       
       
        pub_key_compressed, hash160 = hex_to_compressed_hash160(random_int)
        keys_checked += 1
        last_pub_key = pub_key_compressed
        last_hash160 = hash160
       
       
        match_len = len(os.path.commonprefix([hash160, target_hash160]))
       
        if match_len == 40: 
            elapsed = time.time() - start_time
            print("\n" + "!"*70)
            print(f"!!! MATCH FOUND AFTER {keys_checked:,} KEYS !!!")
            print(f"Private Key: {current_hex}")
            print(f"Public Key: {pub_key_compressed}")
            print(f"Hash160: {hash160}")
            print(f"Elapsed Time: {elapsed:.2f} seconds")
            print(f"Speed: {keys_checked/elapsed:,.0f} keys/sec")
            print("!"*70)
            save_found_key(current_hex, pub_key_compressed, hash160, keys_checked, elapsed)
            break
        elif match_len >= 5 and match_len > best_match_len:
           
            best_match_len = match_len
            best_match_private = current_hex
            best_match_hash160 = hash160
            elapsed = time.time() - start_time
            print_match_info(keys_checked, current_hex, elapsed, pub_key_compressed, hash160, match_len)

except KeyboardInterrupt:
    elapsed = time.time() - start_time
    print("\n\n" + "-"*70)
    print(f"SCAN INTERRUPTED BY USER")
    print(f"Total Keys Checked: {keys_checked:,}")
    print(f"Elapsed Time: {elapsed:.2f} seconds")
    print(f"Last Private Key: {current_hex}")
    if best_match_len >= 5:
        print(f"Best Partial Match Found:")
        print(f"Matching Characters: {best_match_len}")
        print(f"Private Key: {best_match_private}")
        print(f"Hash160: {best_match_hash160}")
    else:
        print("No partial match of 5 or more characters found.")
    print("-"*70)


elapsed = time.time() - start_time
print("\n\n" + "="*70)
print("SCAN COMPLETE - FINAL STATISTICS")
print("="*70)
print(f"Total Keys Checked: {keys_checked:,}")
print(f"Total Time: {elapsed:.2f} seconds")
print(f"Average Speed: {keys_checked/elapsed:,.0f} keys/sec")
print(f"Last Private Key: {current_hex}")
print(f"Last Public Key: {last_pub_key[:20]}...{last_pub_key[-20:]}")
print(f"Last Hash160: {last_hash160}")
print("="*70)      use the Hex you need on start_hex and  end_hex  also you can modify the target hash160

If this helps buy me a coffee by sending some $ on my BTC address 16qw4VqkxSSfKzguDyVzM68jsWw71yEgP4
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 02/05/2025, 08:51:10 UTC
Guys Please try to make your own creative code whether C++ or python or if you are using even others code try to analyze it before running it on your device , being eager to get the prize may lead you to not getting after even getting the hex .
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 07/04/2025, 11:05:03 UTC
i have a good Coding skill and also time but don't have at list core i7 4060 graphics and also so regretting for not having it my past business failure lead me to this , but i am working and love to see all the coding progress, in here fighting to find the best matched hash160.
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Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 25/03/2025, 12:54:18 UTC
uSE THIS FOR PYTHON import hashlib
import ecdsa
import random
from multiprocessing import Pool, cpu_count, Manager
import queue

# Define constants
TARGET_HASH160 = "e0b8a2baee1b77fc703455f39d51477451fc8cfc".lower()
LOWER_BOUND = int("80000000000000000", 16)
UPPER_BOUND = int("ffFFFFFFFFFFFFFFF", 16)
KEYS_PER_ITERATION = 1_000_000  # Number of keys to generate per iteration

def private_key_to_hash160(private_key_hex):
    """Generate the Hash160 from a private key using compressed public key."""
    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)
    hash160 = ripemd160.hexdigest()
    return hash160

def check_key_range(args):
    """Check a set of random private keys and queue their Hash160 values."""
    num_keys, target_hash160, print_queue, lower_bound, upper_bound = args
    for _ in range(num_keys):
        current_int = random.randint(lower_bound, upper_bound)
        private_key_hex = f"{current_int:064x}"
        hash160 = private_key_to_hash160(private_key_hex)
        print_queue.put((private_key_hex, hash160))
        if hash160.lower() == target_hash160:
            return (private_key_hex, hash160)
    return None

def process_iteration(keys_per_iteration, target_hash160, num_processes, print_queue):
    """Process a set of random keys, printing private keys and Hash160 values."""
    keys_per_process = keys_per_iteration // num_processes
    # Adjust for any remainder
    ranges = [
        (
            keys_per_process if i < num_processes - 1 else keys_per_iteration - (keys_per_process * (num_processes - 1)),
            target_hash160,
            print_queue,
            LOWER_BOUND,
            UPPER_BOUND
        )
        for i in range(num_processes)
    ]

    with Pool(processes=num_processes) as pool:
        result = pool.map_async(check_key_range, ranges)
        while True:
            try:
                priv, h160 = print_queue.get(timeout=0.1)
                print(f"Private Key: {priv}, Hash160: {h160}")
            except queue.Empty:
                if result.ready():
                    break
        results = result.get()

    for res in results:
        if res:
            return res
    return None

def main():
    """Main function to search for a matching Hash160 with random key generation."""
    total_attempts = 0
    iteration = 0
    num_processes = cpu_count()
    manager = Manager()
    print_queue = manager.Queue()

    print(f"Searching for Hash160: {TARGET_HASH160}")
    print(f"Range: {LOWER_BOUND:X} to {UPPER_BOUND:X}")
    print(f"Using {num_processes} processes with {KEYS_PER_ITERATION} keys per iteration\n")

    while True:
        iteration += 1
        total_attempts += KEYS_PER_ITERATION

        print(f"Iteration {iteration}:")
        print(f"Generating {KEYS_PER_ITERATION} random keys...")

        result = process_iteration(KEYS_PER_ITERATION, TARGET_HASH160, num_processes, print_queue)

        if result:
            private_key_hex, hash160 = result
            print(f"\nMatch found after {total_attempts} attempts!")
            print(f"Private Key: {private_key_hex}")
            print(f"Hash160: {hash160}")
            with open("found_match.txt", "w") as f:
                f.write(f"Private Key: {private_key_hex}\nHash160: {hash160}")
            break

        print(f"Completed iteration {iteration}. Total attempts: {total_attempts}\n")

if __name__ == "__main__":
    main()             IF YOU ARE SATISFIED MAKE DROP A TIP HERE ON  16qw4VqkxSSfKzguDyVzM68jsWw71yEgP4 
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 17/03/2025, 13:03:39 UTC
Most of the application posted here are whether they are not full or they are samples no one is going to pass the probabilities chance of getting the prize for anyone for example the last applications are meant to solve puzzle 68 but the bit it can solve is limited to 1 up to 20 (Puzzle 68 is 68 bits) so most of the coders here is playing on you work on your own Guys
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 17/03/2025, 08:47:50 UTC
 Cheesy Cheesy Cheesy Cheesy Cheesy Cheesy Cheesy Love your expression
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
HABJo12
on 17/03/2025, 07:57:36 UTC
My pray and working hard is to get the 5000 $ then i will post here the HEX key of the Address after collecting the funds. 100% sure