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Showing 5 of 5 results by pongo
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Board Bitcoin Discussion
Re: ARG Puzzle with 3.5 BTC Private Key Prize **Game Over**
by
pongo
on 07/07/2014, 03:18:16 UTC
I'm not taking part because the math lingo is too intimidating.

address: 187dAwrLyuaoVHG7aTk6zjNRi3f8xVMH2j
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Re: ARG Puzzle with 3.5 BTC Private Key Prize
by
pongo
on 05/07/2014, 17:20:42 UTC
Er, maybe I'm on the wrong track.  In that format, what I got is this:

0x3067c24761f46c1f411ee50a802f9451d04ce566e6c6bca2d0c0d42ba448e5969ca5c2298fc8a 4

which is 312 bits, a bit overqualified to be a private key.
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Re: ARG Puzzle with 3.5 BTC Private Key Prize
by
pongo
on 05/07/2014, 17:13:28 UTC
Heh, not OP.

Yeah, though isn't it suspicious that the dogecoin symbol is there?  Maybe we're not at the last step and there's something to do with dogecoin?

Here's some quick code I was trying out:



import hashlib, binascii
import sys

__b62chars = '0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'
__b62base = len(__b62chars)



def b62decode(v, length):
  """ decode v into a string of len bytes
  """
  long_value = 0
  for (i, c) in enumerate(v[::-1]):
    long_value += __b62chars.find(c) * (__b62base**i)

  result = bytes()
  while long_value >= 256:
    div, mod = divmod(long_value, 256)
    result = chr(mod) + result
    long_value = div
  result = chr(long_value) + result

  nPad = 0
  for c in v:
    if c == __b62chars[0]: nPad += 1
    else: break

  result = chr(0)*nPad + result
  if length is not None and len(result) != length:
    return None

  return result


bytes = b62decode('z69JZqlJn862D1ndx7oLVEMmVOlP1zewEeUCrsI7Roahzpeny7PA', None)

for byte in bytes:
   # sys.stdout.write("{:02x} ".format(ord(byte)))
   sys.stdout.write("{:08b} ".format(ord(byte)))

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Board Bitcoin Discussion
Re: ARG Puzzle with 3.5 BTC Private Key Prize
by
pongo
on 05/07/2014, 17:07:16 UTC
That's just using each character as a number.  Did you try converting it to the binary representation?  for example, 11 in base-62 equals the number 63

I unfortunately have to leave right now, but my prediction is that if you took z69JZqlJn862D1ndx7oLVEMmV0lP1zewEeUCrsI7Roahzpeny7P and converted it to the binary representation, assuming that's base62, you'd get a dogecoin private key.
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Board Bitcoin Discussion
Re: ARG Puzzle with 3.5 BTC Private Key Prize
by
pongo
on 05/07/2014, 16:53:23 UTC
Maybe it's a form of base-62 encoding like this:

new encoding: 0  1  2  3  4  5  6  7  8  9  A  B  C  D  E  F  G  H  I  J  K  L  M  N  O  P  Q  R  S  T  U  V  W  X  Y  Z  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o  p  q  r  s  t  u  v  w  x  y  z
decimal:      0  1  2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61