Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

Version 1

Last scraped

Edited on 17/06/2025, 11:43:09 UTC

Quote from: Andrey2225 on June 17, 2025, 03:31:14 AM

Пoчeмy 16 cлoв abandon ,1 cлoвo about, и 7 cлoв paндoмныx ?

Code:

import secp256k1 as ice
import mnemonic

def calculate_puzzle_parts():
   start = 1180591620717411303424
   end = 2361183241434822606847
   num_parts = 10

   part_size = (end - start) // num_parts
   for i in range(num_parts):
   start_dec = start + i * part_size
   end_dec = start + (i + 1) * part_size - 1

   start_hex = "%064x" % start_dec
   start_wif = ice.btc_pvk_to_wif(start_hex) # Compressed WIF
   private_key = bytes.fromhex(start_hex)
   language = 'english'
   mnemonic_words = mnemonic.Mnemonic(language).to_mnemonic(private_key)
   mnemonic_word_list = " ".join(mnemonic_words.split())
   print(f"Part {i + 1}", start_dec, start_wif, mnemonic_word_list)

calculate_puzzle_parts()

Quote

Part 1 1180591620717411303424 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qxGLkchTagWEWquHPtvw abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon able abandon abandon abandon abandon abandon abandon between
Part 2 1298650782789152433766 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qzcomN2tkj1Srntequ8e abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon able creek office smoke grid creek office ten
Part 3 1416709944860893564108 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3r2yGn7NKvmWfCjtNq4xH abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon able grid creek office smoke grid creek loyal
Part 4 1534769106932634694450 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3r5Kjnrhm6p1sYgtJyVwp abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon able office smoke grid creek office smoke excuse
Part 5 1652828269004375824792 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3r7gCoc3CGrX5tdmxEtfs abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon able smoke grid creek office smoke grid bottom
Part 6 1770887431076116955134 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rA2fpMNdSu2JEanwRKxh abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon able zoo zoo zoo zoo zoo zoo spike
Part 7 1888946593147858085476 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rCP8q6i4cwXWaXjXjtnt abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon about creek office smoke grid creek office normal
Part 8 2007005755219599215818 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rEjbqr3Vnz2ivUmns4h3 abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon about grid creek office smoke grid creek escape
Part 9 2125064917291340346160 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rH64rbNvy2XwGRg1SxL4 abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon about office smoke grid creek office smoke auction
Part 10 2243124079363081476502 KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3rKSXsLiN9539cNkQWuLv abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon abandon about smoke grid creek office smoke green spin

I assume that the private key is from 1888946593147858085476 to 2243124079363081476502.
So it is 16 abandon + about and 7 random words.
You can also go for 16 abandon and 8 random words. It's just madness to solve. The program 100% works on 1 or 2 missing words. I tested.

Maybe someone knows how to find a better method? I see that some words are repeated throughout the entire range. Tongue

Original archived Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it

Scraped on 17/06/2025, 11:13:28 UTC

Quote from: Andrey2225 on June 17, 2025, 03:31:14 AM

Пoчeмy 16 cлoв abandon ,1 cлoвo about, и 7 cлoв paндoмныx ?

Code:

import secp256k1 as ice
import mnemonic

def calculate_puzzle_parts():
start = 1180591620717411303424
end = 2361183241434822606847
num_parts = 10

part_size = (end - start) // num_parts
for i in range(num_parts):
start_dec = start + i * part_size
end_dec = start + (i + 1) * part_size - 1

start_hex = "%064x" % start_dec
start_wif = ice.btc_pvk_to_wif(start_hex) # Compressed WIF
private_key = bytes.fromhex(start_hex)
language = 'english'
mnemonic_words = mnemonic.Mnemonic(language).to_mnemonic(private_key)
mnemonic_word_list = " ".join(mnemonic_words.split())
print(f"Part {i + 1}", start_dec, start_wif, mnemonic_word_list)

calculate_puzzle_parts()

Quote