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Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 27/07/2025, 13:12:09 UTC
KEY Start 43........
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 17/07/2025, 07:18:46 UTC
🎯 REVISED P71 POSITIONING:
✅ Position:  77.3%  ( Success Probability 88.0% )
📍 Target Range: 0x7174 area
🔬 Improved Calibration: Enhanced mathematical modeling
📊 Confidence: Higher precision positioning
🎯 Search Strategy: COVERAGE deployment recommended
Reason for sharing this update:

"The mathematical positioning keeps evolving as I refine the φ-based calculations. This has higher success probability."

So anyone want too hash this ? 😅🤔


🚀 ADAPTIVE AI RANGE DEPLOYMENT:
─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
📦 PRECISION: 717BF1E8E60C65E698 → 717D955714BE2A1968
              Width:  0.01% (118,059,162,071,741,136 keys)
              Surgical precision + AI

📦  BALANCED: 717B2D02DED7C1B1E0 → 717E5A3D1BF2CE4E20
              Width:  0.02% (228,903,190,186,990,656 keys)
              Optimal balance + AI

📦    SAFETY: 7178CB173103783CC0 → 7180BC28C9C717C340
              Width:  0.05% (572,257,975,467,476,608 keys)
              Enhanced safety + AI

📦  COVERAGE: 7174D28E64A1A87980 → 7184B4B19628E78680
              Width:  0.10% (1,144,515,950,934,953,216 keys)
              Maximum coverage + AI


LOL, what is this? you really believe #71  starts with 71
Эти части диапазона пусты
🎯 REVISED P71 POSITIONING:
✅ Position:  77.3%  ( Success Probability 88.0% )
📍 Target Range: 0x7174 area
🔬 Improved Calibration: Enhanced mathematical modeling
📊 Confidence: Higher precision positioning
🎯 Search Strategy: COVERAGE deployment recommended
Reason for sharing this update:

"The mathematical positioning keeps evolving as I refine the φ-based calculations. This has higher success probability."

So anyone want too hash this ? 😅🤔


🚀 ADAPTIVE AI RANGE DEPLOYMENT:
─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
📦 PRECISION: 717BF1E8E60C65E698 → 717D955714BE2A1968
              Width:  0.01% (118,059,162,071,741,136 keys)
              Surgical precision + AI

📦  BALANCED: 717B2D02DED7C1B1E0 → 717E5A3D1BF2CE4E20
              Width:  0.02% (228,903,190,186,990,656 keys)
              Optimal balance + AI

📦    SAFETY: 7178CB173103783CC0 → 7180BC28C9C717C340
              Width:  0.05% (572,257,975,467,476,608 keys)
              Enhanced safety + AI

📦  COVERAGE: 7174D28E64A1A87980 → 7184B4B19628E78680
              Width:  0.10% (1,144,515,950,934,953,216 keys)
              Maximum coverage + AI


LOL, what is this? you really believe #71  starts with 71
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 16/07/2025, 14:55:03 UTC
🎯 REVISED P71 POSITIONING:
✅ Position:  77.3%  ( Success Probability 88.0% )
📍 Target Range: 0x7174 area
🔬 Improved Calibration: Enhanced mathematical modeling
📊 Confidence: Higher precision positioning
🎯 Search Strategy: COVERAGE deployment recommended
Reason for sharing this update:

"The mathematical positioning keeps evolving as I refine the φ-based calculations. This has higher success probability."

So anyone want too hash this ? 😅🤔


🚀 ADAPTIVE AI RANGE DEPLOYMENT:
─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
📦 PRECISION: 717BF1E8E60C65E698 → 717D955714BE2A1968
              Width:  0.01% (118,059,162,071,741,136 keys)
              Surgical precision + AI

📦  BALANCED: 717B2D02DED7C1B1E0 → 717E5A3D1BF2CE4E20
              Width:  0.02% (228,903,190,186,990,656 keys)
              Optimal balance + AI

📦    SAFETY: 7178CB173103783CC0 → 7180BC28C9C717C340
              Width:  0.05% (572,257,975,467,476,608 keys)
              Enhanced safety + AI

📦  COVERAGE: 7174D28E64A1A87980 → 7184B4B19628E78680
              Width:  0.10% (1,144,515,950,934,953,216 keys)
              Maximum coverage + AI
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 18/06/2025, 16:51:52 UTC
There's no way this can be solved this way.  .. go away now.  Grin

Here is the whole code

https://github.com/AkitoHosana/mnemonic-puzzle-solver

Unfortunately this link does not exist you made a mistake

Try again.  Tongue
. Так как все пазлы это адресса одного детерминированного кошелька,о чем говорил создатель ,то в случае использования сид фразы нужен путь создания каждого адресса пазла.Какой путь вы используете для 71 пазла ?
There's no way this can be solved this way.  .. go away now.  Grin

Here is the whole code

https://github.com/AkitoHosana/mnemonic-puzzle-solver

Unfortunately this link does not exist you made a mistake

Try again.  Tongue
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 17/06/2025, 03:31:14 UTC
Hello, what do you think about this idea?
I hope it is useful and I am waiting for your comments

Good idea. It will only take 100,000 years to find puzzle 71 private key.  Embarrassed

Here is the AI script

Code:
import mnemonic
import secp256k1 as ice
import multiprocessing
import time
import random

# Configuration
TARGET_HASH160 = "f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8"  # target hash160
FIXED_WORDS = ["abandon"] * 16 + ["about"]  # Fixed part of mnemonic
LANGUAGE = 'english'
BATCH_SIZE = 10000  # Print progress every X valid attempts

# Initialize components
mnemo = mnemonic.Mnemonic(LANGUAGE)
wordlist = mnemo.wordlist
target_binary = bytes.fromhex(TARGET_HASH160)

# Shared counters
attempts = multiprocessing.Value('i', 0)        # Total tries
valid_attempts = multiprocessing.Value('i', 0)  # Valid mnemonics only
start_time = time.time()
lock = multiprocessing.Lock()

def worker(result_queue):
    while True:
        try:
            # Generate 7 random words
            random_words = random.choices(wordlist, k=7)

            # Build full mnemonic phrase
            candidate_words = FIXED_WORDS + random_words
            mnemonic_phrase = ' '.join(candidate_words)

            #Skip if mnemonic is not valid
            if not mnemo.check(mnemonic_phrase):
                with lock:
                    attempts.value += 1  # Count total attempts
                continue

            # Valid mnemonic found
            with lock:
                attempts.value += 1
                valid_attempts.value += 1

                # Print progress based on valid attempts
                if valid_attempts.value % BATCH_SIZE == 0:
                    elapsed = time.time() - start_time
                    rate = valid_attempts.value / max(1, elapsed)
                    print(f"[Valid] {valid_attempts.value:,} | {rate:,.0f} valid/sec | Current: {' '.join(random_words)}")

            # Convert to private key
            entropy = mnemo.to_entropy(mnemonic_phrase.split())
            private_key_int = int.from_bytes(entropy, 'big')

            # Generate hash160
            h160 = ice.privatekey_to_h160(0, True, private_key_int)

            # Check for match
            if h160 == target_binary:
                result_queue.put((
                    True,
                    mnemonic_phrase,
                    private_key_int.to_bytes(32, 'big').hex(),
                    h160.hex()
                ))
                return

        except Exception as e:
            continue  # Skip on error

if __name__ == '__main__':
    print(f"Starting random mnemonic search with {multiprocessing.cpu_count()} workers...")

    result_queue = multiprocessing.Queue()
    processes = []

    # Start workers
    for _ in range(multiprocessing.cpu_count()):
        p = multiprocessing.Process(target=worker, args=(result_queue,))
        processes.append(p)
        p.start()

    # Wait for results
    while True:
        success, mnemonic_phrase, privkey_hex, found_hash160 = result_queue.get()
        if success:
            print("\nSUCCESS! Found matching mnemonic:")
            print(f"Full Mnemonic: {mnemonic_phrase}")
            print(f"Private Key: {privkey_hex}")
            print(f"Hash160: {found_hash160}")

            # Terminate all workers
            for p in processes:
                p.terminate()
            break

    # Cleanup
    for p in processes:
        p.join()

    elapsed = time.time() - start_time
    print(f"\nTotal attempts: {attempts.value:,}")
    print(f"Valid mnemonics tested: {valid_attempts.value:,}")
    print(f"Time elapsed: {elapsed:.2f} seconds")
    print(f"Speed: {valid_attempts.value / max(1, elapsed):,.0f} valid mnemonics/sec")
Почему 16 слов abandon ,1 слово about, и 7 слов рандомных ?
Hello, what do you think about this idea?
I hope it is useful and I am waiting for your comments

Good idea. It will only take 100,000 years to find puzzle 71 private key.  Embarrassed

Here is the AI script

Code:
import mnemonic
import secp256k1 as ice
import multiprocessing
import time
import random

# Configuration
TARGET_HASH160 = "f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8"  # target hash160
FIXED_WORDS = ["abandon"] * 16 + ["about"]  # Fixed part of mnemonic
LANGUAGE = 'english'
BATCH_SIZE = 10000  # Print progress every X valid attempts

# Initialize components
mnemo = mnemonic.Mnemonic(LANGUAGE)
wordlist = mnemo.wordlist
target_binary = bytes.fromhex(TARGET_HASH160)

# Shared counters
attempts = multiprocessing.Value('i', 0)        # Total tries
valid_attempts = multiprocessing.Value('i', 0)  # Valid mnemonics only
start_time = time.time()
lock = multiprocessing.Lock()

def worker(result_queue):
    while True:
        try:
            # Generate 7 random words
            random_words = random.choices(wordlist, k=7)

            # Build full mnemonic phrase
            candidate_words = FIXED_WORDS + random_words
            mnemonic_phrase = ' '.join(candidate_words)

            #Skip if mnemonic is not valid
            if not mnemo.check(mnemonic_phrase):
                with lock:
                    attempts.value += 1  # Count total attempts
                continue

            # Valid mnemonic found
            with lock:
                attempts.value += 1
                valid_attempts.value += 1

                # Print progress based on valid attempts
                if valid_attempts.value % BATCH_SIZE == 0:
                    elapsed = time.time() - start_time
                    rate = valid_attempts.value / max(1, elapsed)
                    print(f"[Valid] {valid_attempts.value:,} | {rate:,.0f} valid/sec | Current: {' '.join(random_words)}")

            # Convert to private key
            entropy = mnemo.to_entropy(mnemonic_phrase.split())
            private_key_int = int.from_bytes(entropy, 'big')

            # Generate hash160
            h160 = ice.privatekey_to_h160(0, True, private_key_int)

            # Check for match
            if h160 == target_binary:
                result_queue.put((
                    True,
                    mnemonic_phrase,
                    private_key_int.to_bytes(32, 'big').hex(),
                    h160.hex()
                ))
                return

        except Exception as e:
            continue  # Skip on error

if __name__ == '__main__':
    print(f"Starting random mnemonic search with {multiprocessing.cpu_count()} workers...")

    result_queue = multiprocessing.Queue()
    processes = []

    # Start workers
    for _ in range(multiprocessing.cpu_count()):
        p = multiprocessing.Process(target=worker, args=(result_queue,))
        processes.append(p)
        p.start()

    # Wait for results
    while True:
        success, mnemonic_phrase, privkey_hex, found_hash160 = result_queue.get()
        if success:
            print("\nSUCCESS! Found matching mnemonic:")
            print(f"Full Mnemonic: {mnemonic_phrase}")
            print(f"Private Key: {privkey_hex}")
            print(f"Hash160: {found_hash160}")

            # Terminate all workers
            for p in processes:
                p.terminate()
            break

    # Cleanup
    for p in processes:
        p.join()

    elapsed = time.time() - start_time
    print(f"\nTotal attempts: {attempts.value:,}")
    print(f"Valid mnemonics tested: {valid_attempts.value:,}")
    print(f"Time elapsed: {elapsed:.2f} seconds")
    print(f"Speed: {valid_attempts.value / max(1, elapsed):,.0f} valid mnemonics/sec")
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 17/06/2025, 03:22:29 UTC
Hello, what do you think about this idea?
I hope it is useful and I am waiting for your comments

Good idea. It will only take 100,000 years to find puzzle 71 private key.  Embarrassed

Here is the AI script

Code:
import mnemonic
import secp256k1 as ice
import multiprocessing
import time
import random

# Configuration
TARGET_HASH160 = "f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8"  # target hash160
FIXED_WORDS = ["abandon"] * 16 + ["about"]  # Fixed part of mnemonic
LANGUAGE = 'english'
BATCH_SIZE = 10000  # Print progress every X valid attempts

# Initialize components
mnemo = mnemonic.Mnemonic(LANGUAGE)
wordlist = mnemo.wordlist
target_binary = bytes.fromhex(TARGET_HASH160)

# Shared counters
attempts = multiprocessing.Value('i', 0)        # Total tries
valid_attempts = multiprocessing.Value('i', 0)  # Valid mnemonics only
start_time = time.time()
lock = multiprocessing.Lock()

def worker(result_queue):
    while True:
        try:
            # Generate 7 random words
            random_words = random.choices(wordlist, k=7)

            # Build full mnemonic phrase
            candidate_words = FIXED_WORDS + random_words
            mnemonic_phrase = ' '.join(candidate_words)

            #Skip if mnemonic is not valid
            if not mnemo.check(mnemonic_phrase):
                with lock:
                    attempts.value += 1  # Count total attempts
                continue

            # Valid mnemonic found
            with lock:
                attempts.value += 1
                valid_attempts.value += 1

                # Print progress based on valid attempts
                if valid_attempts.value % BATCH_SIZE == 0:
                    elapsed = time.time() - start_time
                    rate = valid_attempts.value / max(1, elapsed)
                    print(f"[Valid] {valid_attempts.value:,} | {rate:,.0f} valid/sec | Current: {' '.join(random_words)}")

            # Convert to private key
            entropy = mnemo.to_entropy(mnemonic_phrase.split())
            private_key_int = int.from_bytes(entropy, 'big')

            # Generate hash160
            h160 = ice.privatekey_to_h160(0, True, private_key_int)

            # Check for match
            if h160 == target_binary:
                result_queue.put((
                    True,
                    mnemonic_phrase,
                    private_key_int.to_bytes(32, 'big').hex(),
                    h160.hex()
                ))
                return

        except Exception as e:
            continue  # Skip on error

if __name__ == '__main__':
    print(f"Starting random mnemonic search with {multiprocessing.cpu_count()} workers...")

    result_queue = multiprocessing.Queue()
    processes = []

    # Start workers
    for _ in range(multiprocessing.cpu_count()):
        p = multiprocessing.Process(target=worker, args=(result_queue,))
        processes.append(p)
        p.start()

    # Wait for results
    while True:
        success, mnemonic_phrase, privkey_hex, found_hash160 = result_queue.get()
        if success:
            print("\nSUCCESS! Found matching mnemonic:")
            print(f"Full Mnemonic: {mnemonic_phrase}")
            print(f"Private Key: {privkey_hex}")
            print(f"Hash160: {found_hash160}")

            # Terminate all workers
            for p in processes:
                p.terminate()
            break

    # Cleanup
    for p in processes:
        p.join()

    elapsed = time.time() - start_time
    print(f"\nTotal attempts: {attempts.value:,}")
    print(f"Valid mnemonics tested: {valid_attempts.value:,}")
    print(f"Time elapsed: {elapsed:.2f} seconds")
    print(f"Speed: {valid_attempts.value / max(1, elapsed):,.0f} valid mnemonics/sec")
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 23/04/2025, 12:20:00 UTC
Why people keep sending sats to puzzle wallets?
Coздaтeль дocылaeт нa aдpecc биткoинa,этo пoдcкaзкa,гдe в диaпaзoнe нaxoдитcя ключ

This is your theory, but I don't see any proof
Первая подсказка создателя была о том ,что это адресса одного и того же детерминированного кошелька,значит сид фраза ко всем адресам одна и та же,только пути деривации разные,вторая подсказка спрятана в количестве транзакций на адресс и количество биткоинов в каждой транзакции ,возможно и плата за транзакции.Да ,к сожалению,это моя теория.Но ведь это пазлы и решение производится по подсказкам.Smiley
Post
Topic
Board Bitcoin Discussion
Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
by
Andrey2225
on 23/04/2025, 02:24:06 UTC
Why people keep sending sats to puzzle wallets?
Создатель досылает на адресс биткоина,это подсказка,где в диапазоне находится ключ