Napstercrew
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Super Carder

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- Apr 2020
Hie guys
i am back, as you all know i am a software developer and a Hacker so a few weeks ago i found a post on bitcoin talk forum with puzzle addresses and some of them are cracked already but here is the interesting part.
the private keys for those addresses are in bit ranges from 1 to 256
so a random from each bitrange is a private key
some ranges are too big so i developed a program to narrow it down leaving only 10% using some algorithms including polynormial math pseudo random and birthday paradox attack
so i know where the private key for puzzle 66 is but my resources are limited and too much electricity blackouts here, i need brute force 382 764 987 624 keys and within those keys there is a key for 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so which has 0.66 BTC.
my i3 laptop can crack 2 million keys/s
So i am calling anyone with few more resources for a deal of 60/40%
all i need is remote or physical(if you are in South Africa) access to a pc with a little more power or a VPS with a few more cores or GPU(Graphic card)
if you are in South Africa we can deal on face to face cause this is serious cash
If you have funds you buy a Few VPS on ionos with paypal or card and give me a monitored access to them 10 Vps with 4 cores each cost $2
or if you have a gaming pc or laptop with nvidia, you can give me remote access.
here is sample python code which tells you how many characters match you get before getting into the range where private key is, for bit range 12 its 3
4 characters is the key
i used it to find 11 characters for bit range 66 which means 12 is the key
Comments and PM
i am back, as you all know i am a software developer and a Hacker so a few weeks ago i found a post on bitcoin talk forum with puzzle addresses and some of them are cracked already but here is the interesting part.
the private keys for those addresses are in bit ranges from 1 to 256
so a random from each bitrange is a private key
some ranges are too big so i developed a program to narrow it down leaving only 10% using some algorithms including polynormial math pseudo random and birthday paradox attack
so i know where the private key for puzzle 66 is but my resources are limited and too much electricity blackouts here, i need brute force 382 764 987 624 keys and within those keys there is a key for 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so which has 0.66 BTC.
my i3 laptop can crack 2 million keys/s
So i am calling anyone with few more resources for a deal of 60/40%
all i need is remote or physical(if you are in South Africa) access to a pc with a little more power or a VPS with a few more cores or GPU(Graphic card)
if you are in South Africa we can deal on face to face cause this is serious cash
If you have funds you buy a Few VPS on ionos with paypal or card and give me a monitored access to them 10 Vps with 4 cores each cost $2
or if you have a gaming pc or laptop with nvidia, you can give me remote access.
here is sample python code which tells you how many characters match you get before getting into the range where private key is, for bit range 12 its 3
4 characters is the key
i used it to find 11 characters for bit range 66 which means 12 is the key
Python:
import random
import bitcoin
import sys
def hex2addr(hx,ds):
decoded_private_key = ds
private_key=hx
compressed_private_key = private_key + '01'
# Multiply the EC generator point G with the private key to get a public key point
public_key = bitcoin.fast_multiply(bitcoin.G, decoded_private_key)
# Encode as hex, prefix 04
hex_encoded_public_key = bitcoin.encode_pubkey(public_key,'hex')
# Compress public key, adjust prefix depending on whether y is even or odd
(public_key_x, public_key_y) = public_key
if (public_key_y % 2) == 0:
compressed_prefix = '02'
else:
compressed_prefix = '03'
hex_compressed_public_key = compressed_prefix + bitcoin.encode(public_key_x, 16)
ad =bitcoin.pubkey_to_address(public_key)
# Generate compressed bitcoin address from compressed public key
adr= bitcoin.pubkey_to_address(hex_compressed_public_key)
return adr
def check(ad):
if "1DBau" in ad:
print(ad)
st = "0"
n = 1000000
while n>0:
ps = random.getrandbits(12)
pt = str(ps).rstrip("L")
ps = int(pt)
hx = hex(ps).lstrip("0x").rstrip("L")
tp = 64-len(hx)
hxs = st*tp+hx
ads = hex2addr(hxs,ps)
check(ads)
n =n-1
print("done")