Author

Topic: Bitcoin puzzle transaction ~32 BTC prize to who solves it - page 254. (Read 229433 times)

member
Activity: 174
Merit: 12
what the hell are you posting?
jr. member
Activity: 49
Merit: 1
Code:
# with status bar

import tqdm
import hashlib
import codecs
import random
from tqdm import tqdm

list = ["3ee4133d991f52fdf6a25c9834e0745ac74248a4","20d45a6a762535700ce9e0b216e31994335db8a5","739437bb3dd6d1983e66629c5f08c70e52769371","e0b8a2baee1b77fc703455f39d51477451fc8cfc","61eb8a50c86b0584bb727dd65bed8d2400d6d5aa","f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8","bf7413e8df4e7a34ce9dc13e2f2648783ec54adb","105b7f253f0ebd7843adaebbd805c944bfb863e4","9f1adb20baeacc38b3f49f3df6906a0e48f2df3d","86f9fea5cdecf033161dd2f8f8560768ae0a6d14","783c138ac81f6a52398564bb17455576e8525b29","35003c3ef8759c92092f8488fca59a042859018c","67671d5490c272e3ab7ddd34030d587738df33da","351e605fac813965951ba433b7c2956bf8ad95ce","20d28d4e87543947c7e4913bcdceaa16e2f8f061","24cef184714bbd030833904f5265c9c3e12a95a2","7c99ce73e19f9fbfcce4825ae88261e2b0b0b040"]

for x in tqdm(range(1000000000)):
 x0 = ("00","01","02","03","04","05","06","07","08","09","0a","0b","0c","0d","0e","0f","10","11","12","13","14","15","16","17","18","19","1a","1b","1c","1d","1e","1f",
       "20","21","22","23","24","25","26","27","28","29","2a","2b","2c","2d","2e","2f","30","31","32","33","34","35","36","37","38","39","3a","3b","3c","3d","3e","3f",
       "40","41","42","43","44","45","46","47","48","49","4a","4b","4c","4d","4e","4f","50","51","52","53","54","55","56","57","58","59","5a","5b","5c","5d","5e","5f",
       "60","61","62","63","64","65","66","67","68","69","6a","6b","6c","6d","6e","6f","70","71","72","73","74","75","76","77","78","79","7a","7b","7c","7d","7e","7f",
       "80","81","82","83","84","85","86","87","88","89","8a","8b","8c","8d","8e","8f","90","91","92","93","94","95","96","97","98","99","9a","9b","9c","9d","9e","9f",
       "a0","a1","a2","a3","a4","a5","a6","a7","a8","a9","aa","ab","ac","ad","ae","af","b0","b1","b2","b3","b4","b5","b6","b7","b8","b9","ba","bb","bc","bd","be","bf",
       "c0","c1","c2","c3","c4","c5","c6","c7","c8","c9","ca","cb","cc","cd","ce","cf","d0","d1","d2","d3","d4","d5","d6","d7","d8","d9","da","db","dc","dd","de","df",
       "e0","e1","e2","e3","e4","e5","e6","e7","e8","e9","ea","eb","ec","ed","ee","ef","f0","f1","f2","f3","f4","f5","f6","f7","f8","f9","fa","fb","fc","fd","fe","ff")
 z0 = random.choices(x0,k=8)
 z1 = random.choices(x0,k=8)
 z2 = random.choices(x0,k=8)
 z3 = random.choices(x0,k=8)
 c0 = ''.join(z0)
 c1 = ''.join(z1)
 c2 = ''.join(z2)
 c3 = ''.join(z3)
 xx = "02"+c0+c1+c2+c3
 cc = "03"+c0+c1+c2+c3
 publickey02 = codecs.decode(xx,"hex")
 publickey03 = codecs.decode(cc,"hex")
 s2 = hashlib.new('sha256', publickey02).digest()
 s3 = hashlib.new('sha256', publickey03).digest()
 r2 = hashlib.new('ripemd160', s2).digest()
 r3 = hashlib.new('ripemd160', s3).digest()
 z2 = codecs.encode(r2, 'hex').decode("utf-8")
 z3 = codecs.encode(r3, 'hex').decode("utf-8")
 if (str(z2)).endswith("248a4"):
  pass
 if (str(z3)).endswith("248a4"):
  print(z2,z3,xx,cc,x)
 if (str(z2)) in list:
  f=open("win.txt","a")
  f.write(str(z2)+"-"+(xx)+"\n")
  f.close()
 if (str(z3)) in list:
  f=open("win.txt","a")
  f.write(str(z3)+"-"+(cc)+"\n")
  f.close()
member
Activity: 272
Merit: 20
the right steps towerds the goal
Code:
import hashlib
import codecs
import random

list = ["3ee4133d991f52fdf6a25c9834e0745ac74248a4","20d45a6a762535700ce9e0b216e31994335db8a5","739437bb3dd6d1983e66629c5f08c70e52769371","e0b8a2baee1b77fc703455f39d51477451fc8cfc","61eb8a50c86b0584bb727dd65bed8d2400d6d5aa","f6f5431d25bbf7b12e8add9af5e3475c44a0a5b8","bf7413e8df4e7a34ce9dc13e2f2648783ec54adb","105b7f253f0ebd7843adaebbd805c944bfb863e4","9f1adb20baeacc38b3f49f3df6906a0e48f2df3d","86f9fea5cdecf033161dd2f8f8560768ae0a6d14","783c138ac81f6a52398564bb17455576e8525b29","35003c3ef8759c92092f8488fca59a042859018c","67671d5490c272e3ab7ddd34030d587738df33da","351e605fac813965951ba433b7c2956bf8ad95ce","20d28d4e87543947c7e4913bcdceaa16e2f8f061","24cef184714bbd030833904f5265c9c3e12a95a2","7c99ce73e19f9fbfcce4825ae88261e2b0b0b040"]

for x in range(1000000000):
 x0 = ("00","01","02","03","04","05","06","07","08","09","0a","0b","0c","0d","0e","0f","10","11","12","13","14","15","16","17","18","19","1a","1b","1c","1d","1e","1f",
       "20","21","22","23","24","25","26","27","28","29","2a","2b","2c","2d","2e","2f","30","31","32","33","34","35","36","37","38","39","3a","3b","3c","3d","3e","3f",
       "40","41","42","43","44","45","46","47","48","49","4a","4b","4c","4d","4e","4f","50","51","52","53","54","55","56","57","58","59","5a","5b","5c","5d","5e","5f",
       "60","61","62","63","64","65","66","67","68","69","6a","6b","6c","6d","6e","6f","70","71","72","73","74","75","76","77","78","79","7a","7b","7c","7d","7e","7f",
       "80","81","82","83","84","85","86","87","88","89","8a","8b","8c","8d","8e","8f","90","91","92","93","94","95","96","97","98","99","9a","9b","9c","9d","9e","9f",
       "a0","a1","a2","a3","a4","a5","a6","a7","a8","a9","aa","ab","ac","ad","ae","af","b0","b1","b2","b3","b4","b5","b6","b7","b8","b9","ba","bb","bc","bd","be","bf",
       "c0","c1","c2","c3","c4","c5","c6","c7","c8","c9","ca","cb","cc","cd","ce","cf","d0","d1","d2","d3","d4","d5","d6","d7","d8","d9","da","db","dc","dd","de","df",
       "e0","e1","e2","e3","e4","e5","e6","e7","e8","e9","ea","eb","ec","ed","ee","ef","f0","f1","f2","f3","f4","f5","f6","f7","f8","f9","fa","fb","fc","fd","fe","ff")
 z0 = random.choices(x0,k=8)
 z1 = random.choices(x0,k=8)
 z2 = random.choices(x0,k=8)
 z3 = random.choices(x0,k=8)
 c0 = ''.join(z0)
 c1 = ''.join(z1)
 c2 = ''.join(z2)
 c3 = ''.join(z3)
 xx = "02"+c0+c1+c2+c3
 cc = "03"+c0+c1+c2+c3
 publickey02 = codecs.decode(xx,"hex")
 publickey03 = codecs.decode(cc,"hex")
 s2 = hashlib.new('sha256', publickey02).digest()
 s3 = hashlib.new('sha256', publickey03).digest()
 r2 = hashlib.new('ripemd160', s2).digest()
 r3 = hashlib.new('ripemd160', s3).digest()
 z2 = codecs.encode(r2, 'hex').decode("utf-8")
 z3 = codecs.encode(r3, 'hex').decode("utf-8")
 if (str(z2)).endswith("248a4"):
  pass
 if (str(z3)).endswith("248a4"):
  print(z2,z3,xx,cc,x)
 if (str(z2)) in list:
  f=open("win.txt","a")
  f.write(str(z2)+"-"+(xx)+"\n")
  f.close()
 if (str(z3)) in list:
  f=open("win.txt","a")
  f.write(str(z3)+"-"+(cc)+"\n")
  f.close()
jr. member
Activity: 67
Merit: 1
Hi all this is 64 public key hunters work correct
She small chance to hit the 64 public key, but that's all for fun
it doesn't use a bitcoin library, so the running code quickly converts to c+ and runs at full power
now it's faster that one print was holding it back
it generates a public key not a public address it's even worse than brute force because you have to break the whole 64 hexadecimal
and generate equivalent hash160 public address easy peasy
if I added base58 there I could convert hash 160 to bitcoin addresses but that would slow down the speed
Code:
import hashlib
import codecs
import random
for x in range(1000000000):
 x0 = ("00","01","02","03","04","05","06","07","08","09","0a","0b","0c","0d","0e","0f","10","11","12","13","14","15","16","17","18","19","1a","1b","1c","1d","1e","1f",
       "20","21","22","23","24","25","26","27","28","29","2a","2b","2c","2d","2e","2f","30","31","32","33","34","35","36","37","38","39","3a","3b","3c","3d","3e","3f",
       "40","41","42","43","44","45","46","47","48","49","4a","4b","4c","4d","4e","4f","50","51","52","53","54","55","56","57","58","59","5a","5b","5c","5d","5e","5f",
       "60","61","62","63","64","65","66","67","68","69","6a","6b","6c","6d","6e","6f","70","71","72","73","74","75","76","77","78","79","7a","7b","7c","7d","7e","7f",
       "80","81","82","83","84","85","86","87","88","89","8a","8b","8c","8d","8e","8f","90","91","92","93","94","95","96","97","98","99","9a","9b","9c","9d","9e","9f",
       "a0","a1","a2","a3","a4","a5","a6","a7","a8","a9","aa","ab","ac","ad","ae","af","b0","b1","b2","b3","b4","b5","b6","b7","b8","b9","ba","bb","bc","bd","be","bf",
       "c0","c1","c2","c3","c4","c5","c6","c7","c8","c9","ca","cb","cc","cd","ce","cf","d0","d1","d2","d3","d4","d5","d6","d7","d8","d9","da","db","dc","dd","de","df",
       "e0","e1","e2","e3","e4","e5","e6","e7","e8","e9","ea","eb","ec","ed","ee","ef","f0","f1","f2","f3","f4","f5","f6","f7","f8","f9","fa","fb","fc","fd","fe","ff")
 z0 = random.choices(x0,k=8)
 z1 = random.choices(x0,k=8)
 z2 = random.choices(x0,k=8)
 z3 = random.choices(x0,k=8)
 c0 = ''.join(z0)
 c1 = ''.join(z1)
 c2 = ''.join(z2)
 c3 = ''.join(z3)
 xx = "02"+c0+c1+c2+c3
 cc = "03"+c0+c1+c2+c3
 publickey02 = codecs.decode(xx,"hex")
 publickey03 = codecs.decode(cc,"hex")
 s2 = hashlib.new('sha256', publickey02).digest()
 s3 = hashlib.new('sha256', publickey03).digest()
 r2 = hashlib.new('ripemd160', s2).digest()
 r3 = hashlib.new('ripemd160', s3).digest()
 z2 = codecs.encode(r2, 'hex').decode("utf-8")
 z3 = codecs.encode(r3, 'hex').decode("utf-8")
 if (str(z2)).endswith("248a4"):
  pass
 if (str(z3)).endswith("248a4"):
  print(z2,z3,xx,cc,x)
 if (str(z2)) == "3ee4133d991f52fdf6a25c9834e0745ac74248a4":
  f=open("win2.txt","a")
  f.write(str(z2)+"-"+(xx)+"\n")
  f.close()
 if (str(z3)) == "3ee4133d991f52fdf6a25c9834e0745ac74248a4":
  f=open("win3.txt","a")
  f.write(str(z3)+"-"+(cc)+"\n")
  f.close()
full member
Activity: 980
Merit: 152



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

Nice work. Thank  you for share.
jr. member
Activity: 67
Merit: 1
to be honest, I don't use these codes myself  Grin Grin Grin
simply, if someone has the upper hand over others, then the code would not publish the logically they will always keep it for themselves
although I'm of the opinion that luck there plays a big role or that start range you can type as in the lottery
or let it go sequentially and wait 300 years Grin
someone will write the initial range and empty the address now
those 300 years it's just figurative it can break now or never
+ you are not sure yet that the address is in the 16 hexadecimal range, what if this address was generated incorrectly and the creator did not check all the addresses
maybe I'm wrong, so all the first addresses have the right range, but you never know
newbie
Activity: 24
Merit: 9
shishua random generator for fun
run faster use cython any codes
Code:
from shishua import SHISHUA
from bit import Key
for x in range(372055,10000000):
 rng = SHISHUA(x).random_raw(7).hex()
 zzz = ["80","81","82","83","84","85","86","87","88","89","8a","8b","8c","8d","8e","8f","90","91","92","93","94","95","96","97","98","99","9a","9b","9c","9d","9e","9f","a0","a1","a2","a3","a4","a5","a6","a7","a8","a9","aa","ab","ac","ad","ae","af","b0","b1","b2","b3","b4","b5","b6","b7","b8","b9","ba","bb","bc","bd","be","bf","c0","c1","c2","c3","c4","c5","c6","c7","c8","c9","ca","cb","cc","cd","ce","cf","d0","d1","d2","d3","d4","d5","d6","d7","d8","d9","da","db","dc","dd","de","df","e0","e1","e2","e3","e4","e5","e6","e7","e8","e9","ea","eb","ec","ed","ee","ef","f0","f1","f2","f3","f4","f5","f6","f7","f8","f9","fa","fb","fc","fd","fe","ff"]
 for c in zzz:
  ke1 = Key.from_hex(c+rng)
  ad1 = ke1.address
  if ad1.endswith("XQN"):
   print(c+rng,ad1,x)
  if ad1 == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
   f=open("win.txt","a")
   f.write(str(c+rng)+"-"+(ad1)+"\n")
   f.close()

@Feron

How can I run this in Ubuntu?

Can I use pip to install the shishua python package?

jr. member
Activity: 67
Merit: 1
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
Code:
from shishua import SHISHUA
from bit import Key
for x in range(372055,10000000):
 rng = SHISHUA(x).random_raw(7).hex()
 zzz = ["80","81","82","83","84","85","86","87","88","89","8a","8b","8c","8d","8e","8f","90","91","92","93","94","95","96","97","98","99","9a","9b","9c","9d","9e","9f","a0","a1","a2","a3","a4","a5","a6","a7","a8","a9","aa","ab","ac","ad","ae","af","b0","b1","b2","b3","b4","b5","b6","b7","b8","b9","ba","bb","bc","bd","be","bf","c0","c1","c2","c3","c4","c5","c6","c7","c8","c9","ca","cb","cc","cd","ce","cf","d0","d1","d2","d3","d4","d5","d6","d7","d8","d9","da","db","dc","dd","de","df","e0","e1","e2","e3","e4","e5","e6","e7","e8","e9","ea","eb","ec","ed","ee","ef","f0","f1","f2","f3","f4","f5","f6","f7","f8","f9","fa","fb","fc","fd","fe","ff"]
 for c in zzz:
  ke1 = Key.from_hex(c+rng)
  ad1 = ke1.address
  if ad1.endswith("XQN"):
   print(c+rng,ad1,x)
  if ad1 == "16jY7qLJnxb7CHZyqBP8qca9d51gAjyXQN":
   f=open("win.txt","a")
   f.write(str(c+rng)+"-"+(ad1)+"\n")
   f.close()
jr. member
Activity: 67
Merit: 1
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 Grin dddd Grin some programmer could check it and let me know
jr. member
Activity: 67
Merit: 1
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
jr. member
Activity: 67
Merit: 1
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
jr. member
Activity: 67
Merit: 1
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()
full member
Activity: 431
Merit: 105
something this damn shit doesn't want to give up  Grin maybe we can try to get in through ripemd160...

Quote
These stories abour lot's of BTC being lost are just legends


"from Russia with love"


Hi there Andzhig, you have a complete package of these things you are programming to start searching somehow.
thanks man,
member
Activity: 206
Merit: 16

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%


hello can you share your modified version of bitcrack please ?
member
Activity: 406
Merit: 47

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%
sr. member
Activity: 443
Merit: 350
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 ~ and indeed, i saw an addresses that ends with the last 3 characters "XQN" that the original puzzle 64 address have.
~
What are the odds that i find the last 3 characters address
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.

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.
legendary
Activity: 3500
Merit: 2792
Enjoy 500% bonus + 70 FS
Thanks i will follow the new one ;-)
member
Activity: 174
Merit: 12
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.
legendary
Activity: 3500
Merit: 2792
Enjoy 500% bonus + 70 FS
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
member
Activity: 174
Merit: 12
Why bring up this old topic if there is a new, urgent one?
https://bitcointalk.org/index.php?topic=5218972.380
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