ARG Puzzle with 3.5 BTC Private Key Prize, Game Over

[arg]

There's an l in the given string so that can't be used to mutate the header key.

[1715184752]

1715184752

[1715184752]

1715184752

[1715184752]

1715184752

[pongo]

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

[arg]

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

Tried that a day ago.

61 6 9 19 35 52 47 19 49 8 6 2 13 1 49 39 59 7 50 21 31 14 22 48 31 24 47 25 1 61 40 58 14 40 30 12 53 54 18 7 27 50 36 43 61 51 40 49 60 7 25

[pongo]

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.

[rgm108]

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.

Thanks but we are not looking for a Doge priv key.
We have an existing Bitcoin address which is funded so that does not make much sense.

[arg]

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

Tried something like that too.

you'd get a dogecoin private key.

What? We are looking for a Bitcoin key, not dogecoin.

[pongo]

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)))

[itod]

The code from Wow.jpg (z69JZqlJn862D1ndx7oLVEMmVolP1zewEeUCrsI7Roahzpeny7P) maps to the following numbers using the above character mapping:

Code:

61 6 9 19 35 52 47 19 49 8 6 2 13 1 49 39 59 7 50 21 31 14 22 48 31 50 47 25 1 61 40 58 14 40 30 12 53 54 18 7 27 50 36 43 61 51 40 49 60 7 25

Maybe to shift this by 64 to get ASCII characters?

[arg]

The code from Wow.jpg (z69JZqlJn862D1ndx7oLVEMmVolP1zewEeUCrsI7Roahzpeny7P) maps to the following numbers using the above character mapping:

Code:

61 6 9 19 35 52 47 19 49 8 6 2 13 1 49 39 59 7 50 21 31 14 22 48 31 50 47 25 1 61 40 58 14 40 30 12 53 54 18 7 27 50 36 43 61 51 40 49 60 7 25

Maybe to shift this by 64 to get ASCII characters?

No, that gives you a newline char which I don't think the OP would have used.

[pongo]

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.

[rgm108]

@ pongo

The Doge pic was a joke m8.

[arg]

Er, maybe I'm on the wrong track.

All of this seems too.. complex.

The solution probably needs to involve a dual sha256 too. That hint is still on the front post.

My pet theory is that the line on the right hand side of Wow.jpeg is somehow the answer, and the 51 character is just a red herring somehow. If it was just a transformation on the "private key" what was the point of the hidden swirls and lines on the image?

[hoba]

The solution probably needs to involve a dual sha256 too. That hint is still on the front post.

Hint is gone now.

[arg]

The solution probably needs to involve a dual sha256 too. That hint is still on the front post.

Hint is gone now.

Well that nukes that theory then. 

[Mitchell]

The solution probably needs to involve a dual sha256 too. That hint is still on the front post.

Hint is gone now.

Well that nukes that theory then. 

Not really. OP has been placing and removing hints for a while. The hints themselves are valid.

[xELx]

The solution probably needs to involve a dual sha256 too. That hint is still on the front post.

Hint is gone now.

Well that nukes that theory then.  

Not really. OP has been placing and removing hints for a while. The hints themselves are valid.

I wouldn't think that "The Final Countdown" is a hint. Seems more like hyping that it's the end. The only thing that has really been revealed is that we have to use the Crops picture with the Wow one.

[arg]

The solution probably needs to involve a dual sha256 too. That hint is still on the front post.

Hint is gone now.

Well that nukes that theory then. 

Not really. OP has been placing and removing hints for a while. The hints themselves are valid.

Right, she removes them once they've been mentioned. So it must involve a double sha256 hash somehow, and the 51 character string is a red herring. arg-sidekick thinks that's related to the bit in Crops.jpg about deception.

[shorena]

-snip-
Please explain, because I saw a couple of versions of the code translations and I will like to try myself.
Thank you

Found my notes

Code:

grn = 1
red = 0
oth = brk

11111 
10001 
11111 
01 
10001 
01111 
11111 
1000 
01 
1000 
1011 
01 
10001 
00011 
10000 
011 
111 
011 
00110
111
8-bit code afterwards

in morse

grn = -
red = .
----- 
-...- 
----- 
.- 
-...- 
.---- 
----- 
-... 
.- 
-... 
-.-- 
.- 
-...- 
...-- 
-.... 
.-- 
--- 
.-- 
..--.
---

->
0=0A=10BABYA=36WOW?O

[itod]

My pet theory is that the line on the right hand side of Wow.jpeg is somehow the answer, and the 51 character is just a red herring somehow. If it was just a transformation on the "private key" what was the point of the hidden swirls and lines on the image?

I've been saying this since yesterday. Vertical number sequence is:

Code:

27913579135791357913579135791357913579024

Notice that we have 'WRITTEN VERTICALLY' added to the image, although it is for 'TWO DIGITS'.

In the number sequence above series '13579' is repeated 7x, with '279' before and '024' after. I could not make anything out of it, but it's clearly underlined with blue stroke from image metadata.

[arg]

Notice that we have 'WRITTEN VERTICALLY' added to the image

That's from the original image, the OP just rewrote it in the same font as the key.

It's too short to use as a mask against the top key in any sane way. Shifting every character by the number in the timestamp does nothing. I too, did all of this yesterday in endless permutations. We're missing something.

https://i.imgur.com/rH3DX2I.gif

Hey OP, bout those hints?