Bitcoin puzzle transaction ~32 BTC prize to who solves it

[mahmood1356]

Since GPU farm is worthless, 71 has almost no chances of being solved this year. Maybe in next years, unless someone is very lucky or the price goes to 200k.

And about withdrawing the funds. Mara is the only option right now, either you trust it or not.

Once a transaction is confirmed by a miner, there should be no more funds in the source address that a bot or anyone else can resend at a higher fee, right?

[Menowa*]

Since GPU farm is worthless, 71 has almost no chances of being solved this year. Maybe in next years, unless someone is very lucky or the price goes to 200k.

And about withdrawing the funds. Mara is the only option right now, either you trust it or not.

Once a transaction is confirmed by a miner, there should be no more funds in the source address that a bot or anyone else can resend at a higher fee, right?

Yeah, once Mara got one block mined the coins is safely moved at the same time. I've made a tutorial of how to withdraw the funds safely throught Mara. But in a nutshell: the coins remains in 71 till mara mines a block, until that the public key remains not exposed.

When you use public mempool, the transaction is shown all the time on the wallet's history, including mempool. When you do it through Mara, there will be any evidence that you started a transaction until Mara confirms it in a mined block.

Thtat's the main difference...

[teguh54321]

Since GPU farm is worthless, 71 has almost no chances of being solved this year. Maybe in next years, unless someone is very lucky or the price goes to 200k.

And about withdrawing the funds. Mara is the only option right now, either you trust it or not.

But not impossible. The odds still below 1/1000
If running at least 3 rtx 4080 gpu for 3 months 😅

Better than lottery luck

[mahmood1356]

Since GPU farm is worthless, 71 has almost no chances of being solved this year. Maybe in next years, unless someone is very lucky or the price goes to 200k.

And about withdrawing the funds. Mara is the only option right now, either you trust it or not.

But not impossible. The odds still below 1/1000
If running at least 3 rtx 4080 gpu for 3 months 😅

Better than lottery luck

Someone may have found the key but hid it for fear of it being stolen and not complete the transaction!!!!!
There is no guarantee that Puzzle 71 assets will be transferred safely!??
Perhaps a message could be sent to the victim along with the transfer transaction, asking the kidnappers to please not tamper with the transaction!!!!
The only safe way in my opinion is to find a miner around your area and your friends and acquaintances. There is someone who does the mining work and you can make the transaction directly through him in the next block!

[Mafioso246]

Since GPU farm is worthless, 71 has almost no chances of being solved this year. Maybe in next years, unless someone is very lucky or the price goes to 200k.

And about withdrawing the funds. Mara is the only option right now, either you trust it or not.

But not impossible. The odds still below 1/1000
If running at least 3 rtx 4080 gpu for 3 months 😅

Better than lottery luck

It's great as long as you're not paying for the electricity!

[mahmood1356]

Since GPU farm is worthless, 71 has almost no chances of being solved this year. Maybe in next years, unless someone is very lucky or the price goes to 200k.

And about withdrawing the funds. Mara is the only option right now, either you trust it or not.

Once a transaction is confirmed by a miner, there should be no more funds in the source address that a bot or anyone else can resend at a higher fee, right?

Yeah, once Mara got one block mined the coins is safely moved at the same time. I've made a tutorial of how to withdraw the funds safely throught Mara. But in a nutshell: the coins remains in 71 till mara mines a block, until that the public key remains not exposed.

When you use public mempool, the transaction is shown all the time on the wallet's history, including mempool. When you do it through Mara, there will be any evidence that you started a transaction until Mara confirms it in a mined block.

Thtat's the main difference...

Do you think it's still possible to steal the response if we sign it with its symmetric private key? For example, look at the private key of Puzzle 69:
19vkiEajfhuZ8bs8Zu2jgmC6oqZbWqhxhG
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qefJjCmLAPHbQ4x7D9Qy
L5oLkpV3aqBjhki6LmvChTCV6odsp4SXMCBHa5h391uEcXoCvAnp

[teguh54321]

Anyone here have fast tool to scan h160 prefix with offset ? ( to scan middle prefix)
I modified bitcrack but very very slow  from  at 25mhs on cuda 4080 gpu 🙃. Cause it have to generate the full h160 and scan it.  Any idea ?

Since GPU farm is worthless, 71 has almost no chances of being solved this year. Maybe in next years, unless someone is very lucky or the price goes to 200k.

And about withdrawing the funds. Mara is the only option right now, either you trust it or not.

But not impossible. The odds still below 1/1000
If running at least 3 rtx 4080 gpu for 3 months 😅

Better than lottery luck

It's great as long as you're not paying for the electricity!

It less than $100 a month here😅🙏

[iceland2k14]

There are so much evidence of bots running for these low bit puzzles that anyone should not directly broadcast any Puzzle Key <110 bit. Better be safe with private pools.

BTW, This thread always brings more humour with new users guessing the start of Key without any proper logical justification. Bring it on ...

[Nodemath]

Anyone here have fast tool to scan h160 prefix with offset ? ( to scan middle prefix)
I modified bitcrack but very very slow  from  at 25mhs on cuda 4080 gpu 🙃. Cause it have to generate the full h160 and scan it.  Any idea ?

Can u give me the code which modified version
That's code enough for me

[skedarve]

this is simply a prototype

import cupy as cp
import numpy as np
import time
from Crypto.Hash import RIPEMD160, SHA256
from ecdsa import SECP256k1, SigningKey
import concurrent.futures

TARGET_HASH160 = bytes.fromhex("f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8")
TARGET_PREFIX = TARGET_HASH160[:3]
THREADS = 256
BATCH_SIZE = 2**24

sha256_kernel = cp.RawKernel(r'''
extern "C" __global__
void sha256_partial(const unsigned long long* privkeys, unsigned char* out_prefixes, unsigned char* matches, int N) {
    int idx = blockDim.x * blockIdx.x + threadIdx.x;
    if (idx >= N) return;
    unsigned long long key = privkeys[idx];
    unsigned char data[8];
    #pragma unroll
    for (int i = 0; i < 8; ++i)
        data[7 - i] = (key >> (i * ) & 0xFF;
    unsigned int h[8] = {
        0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a,
        0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19
    };
    unsigned int k[4] = {
        0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5
    };
    unsigned int w[64] = {0};
    #pragma unroll
    for (int i = 0; i < 8; ++i)
        w = ((unsigned int)data) << 24;
    unsigned int a = h[0], b = h[1], c = h[2], d = h[3];
    unsigned int e = h[4], f = h[5], g = h[6], hh = h[7];
    for (int i = 0; i < 4; ++i) {
        unsigned int S1 = __funnelshift_r(e, e, 6) ^ __funnelshift_r(e, e, 11) ^ __funnelshift_r(e, e, 25);
        unsigned int ch = (e & f) ^ (~e & g);
        unsigned int temp1 = hh + S1 + ch + k + w;
        unsigned int S0 = __funnelshift_r(a, a, 2) ^ __funnelshift_r(a, a, 13) ^ __funnelshift_r(a, a, 22);
        unsigned int maj = (a & b) ^ (a & c) ^ (b & c);
        unsigned int temp2 = S0 + maj;
        hh = g;
        g = f;
        f = e;
        e = d + temp1;
        d = c;
        c = b;
        b = a;
        a = temp1 + temp2;
    }
    h[0] += a;
    unsigned char b0 = (h[0] >> 24) & 0xFF;
    unsigned char b1 = (h[0] >> 16) & 0xFF;
    unsigned char b2 = (h[0] >> & 0xFF;
    out_prefixes[idx * 3 + 0] = b0;
    out_prefixes[idx * 3 + 1] = b1;
    out_prefixes[idx * 3 + 2] = b2;
    if (b0 == 0xf6 && b1 == 0xf5 && b2 == 0x43)
        matches[idx] = 1;
    else
        matches[idx] = 0;
}
''', 'sha256_partial')

def hash160(pubkey_bytes):
    sha = SHA256.new(pubkey_bytes).digest()
    h160 = RIPEMD160.new(sha).digest()
    return h160

def verify_candidate(candidate):
    pubkey = compressed_public_key_from_privkey(candidate)
    h160 = hash160(pubkey)
    if h160 == TARGET_HASH160:
        return candidate, pubkey.hex(), h160.hex()
    return None

def compressed_public_key_from_privkey(privkey_int):
    priv_bytes = int(privkey_int).to_bytes(32, 'big')
    sk = SigningKey.from_string(priv_bytes, curve=SECP256k1)
    vk = sk.get_verifying_key()
    x = vk.pubkey.point.x()
    prefix = b'\x03' if (vk.pubkey.point.y() & 1) else b'\x02'
    return prefix + x.to_bytes(32, 'big')

print("🔁 Starting search with GPU prefilter (3 bytes)...")

try:
    total_keys = 0
    total_candidates = 0
    while True:
        t0 = time.time()
        privkeys = cp.random.randint(0, 2**64, size=BATCH_SIZE, dtype=cp.uint64)
        out = cp.zeros((BATCH_SIZE * 3,), dtype=cp.uint8)
        matches = cp.zeros(BATCH_SIZE, dtype=cp.uint8)
        blocks = (BATCH_SIZE + THREADS - 1) // THREADS
        sha256_kernel((blocks,), (THREADS,), (privkeys, out, matches, BATCH_SIZE))
        matches_cpu = matches.get()
        privkeys_cpu = privkeys.get()
        candidates = [privkeys_cpu for i in range(BATCH_SIZE) if matches_cpu == 1]
        total_keys += BATCH_SIZE
        total_candidates += len(candidates)
        with concurrent.futures.ProcessPoolExecutor(max_workers=8) as executor:
            futures = [executor.submit(verify_candidate, c) for c in candidates]
            for future in concurrent.futures.as_completed(futures):
                result = future.result()
                if result is not None:
                    candidate, pub_hex, h160_hex = result
                    print(f"\n🎯 MATCH found! Private key: {hex(candidate)}")
                    print(f"    PublicKey: {pub_hex}")
                    print(f"    Hash160: {h160_hex}")
                    exit()
        elapsed = time.time() - t0
        print(f"🔹 Processed: {BATCH_SIZE} | Candidates: {len(candidates)} | Speed: {int(BATCH_SIZE / elapsed):,} keys/s")
except KeyboardInterrupt:
    print("⛔ Process interrupted.")

[Nodemath]

Looking for real full code that unable to use and it's slow
Instead help me with gpu related private key to hash 160
Entire in one frame code

[skedarve]

Looking for real full code that unable to use and it's slow
Instead help me with gpu related private key to hash 160
Entire in one frame code

This code is just a prototype. If you're looking for simple private key → public key → hash160, this is much slower.

My code doesnt follow the full flow from the start, but instead optimizes it with a shortcut:
it uses partial SHA256 on the GPU to pre-filter promising private keys before doing the full flow.

This saves you billions of useless ECC checks, and it's a totally different strategy if you're looking for an exact match like hash160

this is not finished yet, there is still work to be done. I am sharing it in case it can help someone

[uefbv]

I don't know why but I'm smelling a big scam. Because a newbie that offer more than 12 000€ to solve a following of numbers this is strange...

Although everyone knows this, many people are still eager to try it. If it is true, he is a wealthy novice; if it is fake, he wants to gain traffic (or just for fun).

[brainless]

Can someone tell us who own 5090 gpu
Can load how much million addresses list with bitcrack ?

[Jorge54PT]

Taking advantage of wallets 71 and 72, I'd like to ask WanderingPhilosopher - author of VBCR.exe - if he could release a compilation of the executable but with os.random, if he hasn't used this type of random. Is that possible, friend?

for 18 characters
import os
random_bytes = os.urandom(9)
random_hex = random_bytes.hex().upper()
print(random_hex)

Thank you very much

[nochkin]

Taking advantage of wallets 71 and 72, I'd like to ask WanderingPhilosopher - author of VBCR.exe - if he could release a compilation of the executable but with os.random, if he hasn't used this type of random. Is that possible, friend?

It does not look like it's using Python. Most likely it's C++ with CUDA.
At least, the Linux version uses /dev/urandom, so most likely the Win binary is doing something similar.

[IlhamCung23]

I still don't fully understand — if a low-bit privkey public key is exposed, how could someone potentially derive the private key from it? Could someone please explain how that's possible?

As an experiment, I create privkey low-bit is there anyone who can try deriving the private key from this public key: 02e6ccb35984820aeef916e1d5585ca5f39e3d39bef71bdd845227933cf8b46d9f

[Jorge54PT]

Taking advantage of wallets 71 and 72, I'd like to ask WanderingPhilosopher - author of VBCR.exe - if he could release a compilation of the executable but with os.random, if he hasn't used this type of random. Is that possible, friend?

It does not look like it's using Python. Most likely it's C++ with CUDA.
At least, the Linux version uses /dev/urandom, so most likely the Win binary is doing something similar.

Since his source code isn't available on GitHub, I don't know what he uses, but it must be C++ and Cuda. But if he uses o.random, then random character generation is fine. That's why I asked if, if he didn't use it, he could go back and compile an .exe with os.random.

[nochkin]

I still don't fully understand — if a low-bit privkey public key is exposed, how could someone potentially derive the private key from it? Could someone please explain how that's possible?

Brute-forcing using the public key is much cheaper than brute-forcing using an address only.

if he didn't use it, he could go back and compile an .exe with o.random.

This is not how development works. You can't just "compile a binary" after sticking "o.random" somewhere and expecting it to work.

[Jorge54PT]

I still don't fully understand — if a low-bit privkey public key is exposed, how could someone potentially derive the private key from it? Could someone please explain how that's possible?

Brute-forcing using the public key is much cheaper than brute-forcing using an address only.

if he didn't use it, he could go back and compile an .exe with o.random.

This is not how development works. You can't just "compile a binary" after sticking "o.random" somewhere and expecting it to work.

My friend, I didn't mention any binary. I said that he didn't provide the source code and therefore I don't know what he used to compile VBCr.exe in the end. Since I don't want to get his source code at all, I asked if he could recompile the code he has using os.random if that's not the type of random that the executable uses Simple, friend