Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 255. (Read 230409 times)

shishua random generator for fun
run faster use cython any codes
important all addresses are generated via x sequence you can't start from the beginning because they would be repeated it's too small save have fun
example I ended up at 372055 according to print you will know where you ended up and just rewrite it to your number and save the progress
you can use the words of the alphabet ascii etc,i used numbers

maybe it sounds funny ddd but this interests me
does not block writing and printing of the bit library if there is a collision of addresses with the balance or there is no protection against brute force dddd some programmer could check it and let me know

it seems to me that 64 is not in this range because if it had been there a long time ago, it would have been broken, so you better focus on 66 ddd this is just a theory that may be wrong
Bitcrack use
you use 8 rtx 3090, each gets 1 initial range 89abcdef and now you generate 3 + 3 hex + 1 = 7 hex one graphics card scans 9 hexadecimal 68719476735 which is 1 minute for rtx 3090 that's 1400 per day I tried it in python until the collision occurred at the beginning sometimes once every 1,000,000, etc.

This method is risky, it is better to mine ethereum or other profitable coins

New test code
you can customize it to suit your needs
puzle address 27 test
code :
from bit import Key
for xx in range(1,100000):
 for x in range(1,100000):
  y = 1
  p = 1
  p = (x - y) / xx # 100000  # 65536
  y += 1
  p *= p
  pp = 1 - p
  if pp >= 0.0: # 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
   c1 = ''.join(str(pp))[11:]
   for c in ("11",):
    ke = Key.from_int(int(c+c1))
    if (str(ke)).endswith("86k>"):
     print(c+c1,ke,x,xx)
    if (str(ke)) == "":
     f=open("win.txt","a")
     f.write(str(c+c1)+(str(ke))+"\n")
     f.close()
interesting for this collision is this number 1892x,,,3187xx I don't know what that means but it doesn't matter for fun

---

Hi all this easy matematic 1+1 code
code:
from bit import Key
import random
for x in range(1000000000):
 xx = random.randint(4611686018427387904,9223372036854775808)
 cc = xx+xx
 k1 = Key.from_int(int(cc))
 if (str(k1)).startswith("  print(cc,xx,k1,x)
 if (str(k1)) == "":
  f=open("win.txt","a")
  f.write(str(cc)+(str(k1))+"\n")
  f.close()
the simpler the faster
code compute 4611686018427387904 to 9223372036854775808 take one number maximum 9223372036854775808 keyspace and doubles it him final decimal is private key little reduces the key space

Random base58 happy hunting
code:
import base58
import secrets
from bit import *
for x in range(100000000):
 c0 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c1 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c2 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c3 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c4 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c5 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c6 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c7 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c8 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c9 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 cc = ("NPQRSTUVWXYZabcdefghij")
 for x0 in cc:
  k1 = Key.from_hex(base58.b58decode(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9).hex())
  ad = k1.address
  if ad == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
   f=open("xxx.txt","a")
   f.write(str(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9)+"-"+(ad)+"\n")
   f.close()
  print(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9,ad,x)

New version improve speed
code:
import base58
import secrets
from bit import *
for x in range(100000000):
 c0 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c1 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c2 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c3 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c4 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c5 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c6 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c7 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c8 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 c9 = secrets.choice("123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz")
 cc = ("NPQRSTUVWXYZabcdefghij")
 for x0 in cc:
  k1 = Key.from_hex(base58.b58decode(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9).hex())
  if (str(k1)).endswith("XQN>"):
   print(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9,k1,x)
  if (str(k1)) == "":
   f=open("win.txt","a")
   f.write(str(x0+c0+c1+c2+c3+c4+c5+c6+c7+c8+c9)+(str(k1))+"\n")
   f.close()

hello can you share your modified version of bitcrack please ?

bitCrack scan every single address

pro - scan every address
cons - use too much time for scan

reference from result
https://hashkeys.club/64/results/

Just an idea
may be need to modify new one bitCrack engine to JumpCrack or SkipCrack or PatternCrack
like to scan and jump to other number

but it is can possible to missing that address found 100%
but some one choose right jump maybe lucky
range 64 bit is 18446744073709551616

if jump every trillion it can be scan all easy (yes with missing address)

python code will be easy to code for scan by jump but will be very slow
CUDA GPU scan will be answer for jump scan

may be jump for 43 bit range and scan all on 32 bit range

I think this idea may be some one can found key #64

just idea and it possible missing address 100%

He actually found 9 characters (10 characters including the 1st "1"):

16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN - #64 puzzle
16jY7qLHKTofWA2DYS9ZaWyvZ5qnjfSXQN - found by Alpaste

But looks nice to have a vanity address with 10 characters and the same bit size.

Thanks i will follow the new one ;-)

willi9974
Have you read what is written above your post?
You are apparently not too smart for this puzzle because in your topic you include a link to an old topic when there is a new, relevant one.
First, learn to read, then think about the puzzle.

have at the moment time to try my luck.
what is the best way to start the searching for puzzle 64 and what for a script or program should i use for that?
a pc with 3 3070er GPU are here...

can anyone help to start?

Best regards
Willi

Update:
For the german Community i openenddiskussion a new german topic
—> https://bitcointalksearch.org/topic/suche-nach-dem-bitcoin-puzzle-64-fur-kostenlose-und-legale-bitcoins-5379045

Why bring up this old topic if there is a new, urgent one?
https://bitcointalk.org/index.php?topic=5218972.380

Everyone is brute forcing #64 since more than a two years now. You were very, very, optimistic that you will be the one with the stroke of unbelivable luck. Look at it this way: You uwere really lucky to understand after only few weeks how small chances are that this will happen and you can now move to more interesting things.

Each character has 58 possibilities, so if you have 1000 addresses, the chance of finding the same 3 characters at the end is about 0.5%. However, you need to find 24 more characters, and those aren't magically going to match once you find the last 3.

Hello guys, i've been trying to randomly bruteforce puzzle #64 for WEEKS in a row. I was using Bitcoincrack to randomly Bruteforce addresses that share the same 7 letters of 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN.
after bitcoincrack generated today 1000 addresses that share the same first 7 address characters, i have always looked for the last 3 characters of the generated addresses to go through the addresses quickly and see if #64 16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN is on the list.
and indeed, i saw an addresses that ends with the last 3 characters "XQN" that the original puzzle 64 address have. And the moment i saw XQN, i freaking screamed I WON 30K EUROS and smashed couple of glass bottles and went quickly to say and to celebrate with my friends.
after my friends came, we checked the address again and it was NOT the address i was looking for. it was so embarrassment, and i'm now broken and sad as hell.
What are the odds that i find the last 3 characters address, that i'm looking for in the range 64? ITS pretty much low, that's why i thought i found the 0.64 BTC. but unfortunately not. I should be the unluckiest guy ever...
Here is the Address, i thought i got the 0.64 bitcoin from it.
isn't that sad situation?
Pub Addr: 16jY7qLHKTofWA2DYS9ZaWyvZ5qnjfSXQN
Priv (WIF): p2pkh:KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZ3tw8CPZjZciVwEz8EV
Priv (HEX): 0x000000000000000000000000000000000000000000000000DAA7B7BDF277A7B7

xenon? its scamcoin.

Hello everyone, have you checked the open puzzles for the presence of coins in other networks? There is a token on the 115th puzzle on the BSC network, can it be taken away?

thank you for your kind reply
cheers

That's it, you just import the private key in WIF to core wallet and the BTC are yours.

Cheers.