Topic: The Legend of Satoshi Nakamato, FINAL STEP PUBLISHED.... 4.87 BTC GRAND PRIZE! - page 50. (Read 108519 times)

Can it be solved, yes, will it, no.

Why? Unless you are willing to spend the time and money to buy out dated computers, software and travel the world you will not solve it. Oh and be a genius.

When you think about it the original value was never worth it, still isn't.

Unless I'm wrong, something has changed here.

I remember that first 2 comments on page 28 were at the bottom of the page/thread, not the first of a new page.

Anyone edited/deleted a post ?

Would be amazing to get 4.87 BTC for free lol , I cant begin to fathom how this is even possible.

I think we are looking for a hex code. 

8 variations of the flames. $a,$b,$c,$d,$e,$f,$g,$h

Also I believe the heights of the non pattern flames are the second part of the hex.  They can be either (0,1,2,3,4,5,6,7) short or (8,9,A,B,C,D,E,F) tall

$i = tall $a
$j = tall $b
$k = tall $c
etc...

Here are the flames in the order of the pattern.  inner top, inner right, inner bottom, etc...

$total = $b.$k.$q.$d.$h.$q.$a.$m.$q.$d.$n.$q.$d.$d.$q.$e.$c.$q.$f.$h.$q.$b.$e.$q.$e.$m.$q.$d.$k.$q.$h.$c.$q.$e.$m.$q.$b.$d.$q.$b.$i.
$q.$c.$i.$q.$b.$j.$q.$f.$j.$q.$g.$m.$q.$e.$l.$q.$h.$f.$q.$g.$k.$q.$d.$k.$q.$f.$o.$q.$e.$m.$q.$d.$b.$q.$d.$c.$q.$h.$d.
$q.$h.$d.$q.$e.$m.$q.$g.$c.$q.$a.$a.$q.$d.$f.$q.$f.$m.$q.$c.$l.$q.$d.$m.$q.$a.$p.$q.$h.$p.$q.$e.$p.$q.$f.$p.$q.$a.$o.$q.$d.$m.$q.$f.$n.$q.$b.$d.$q.$h.$m.$q.$g.$h.$q.$d.$o.$q.$d.$d.$q.$e.$m.$q.$b.$k.$q.$f.$d.$q.$g.$i.$q.$d.
$p.$q.$h.$a.$q.$g.$f.$q.$b.$b.$q.$e.$f.$q.$a.$j.$q.$b.$m.$q.$h.$m.$q.$e.$k.$q.$d.$d.$q.$h.$e.$q.$a.$a.$q.$a.$a.$q.$e.$k.$q.$f.$o.$q.$h.$b.$q.$g.$a.$q.$e.$p.$q.$a.$a.$q.$d.$m.$q.$g.$h.$q.$b.$p.$q.$e.$g.$q.$a.$k.$q.$h.$d;

The $q is just a space.

I also think we are looking for 6PR which are the first 3 letters in a password encrypted private key.  There also must be a primer in the flames as you have 36 flames not needed for the encrypted key.  I think these 18 letters (36 flames) either at the front or the back have written text to let you know you have solved the code or are at least getting close.

Good question. It' very likely some https://en.wikipedia.org/wiki/Steganography is applied. In previous puzzle we even had files hidden within files, with different kinds of compression algorithms.

You shouldn't assume you can solve the riddle without looking into the file, but it is possible.

There is no trace about clues given back in 2015 by Coin_artist.

Looks like there was retweet about Base6 and Baconian Cipher but can't find the tweets from 2015.

Does anyone know if we have to look at the picture as a painting (no digital editing) or if we should look at it like a photo of the painting (could have been edited).

Could we solve the puzzle by only looking at the painting in a gallery for example ? (still having a computer to compute the base conversion, sha256 etc...)

Thx !

Wow.
Is this still going on? That is a pretty big amount if you ask me. Imagine getting 4.87 bitcoins for free? Damn! That would be a hell lot of a fortune!ake me wonder who is really behind this.
But whoever he is, he's a damn generous guy. Lol.

But even if don't find it. I would really love to see someone finding it and exolaining how he did it. That would be so much fun to know.

Hello guys, let's do this, LEROYYYYYYYYY

I'm on the IRC channel (#barg on freenode), but I'm all alone, see you there !

I'm probably way off on this, but the FO and LD found might actually be F0 and 1D.
These might be positions of pieces on a chess board, if we assume non-standard position naming conventions (Usually, positions start from 1).
If we reverse the 1D to D1, we get a position that's the same pattern away from F0 as a knight would be able to move.

Bots used to be fun, now they just sound like disturbed people.

What is with these new-ish accounts conversing with themselves on this thread. They sound like crypto markov chain chatbots.

check it))

how you made it like that?

maybe Private Key is somewhere in here:

Never mind, the addition to the pool was from stolen BTC. https://twitter.com/coin_artist/status/933755410061787136

Funny I see the prize is now a cool even 5 BTC as more was just added on Nov. 17, the reawakening of the thread may have had something to do with that or was it the other way around?

https://blockchain.info/address/1FLAMEN6rq2BqMnkUmsJBqCGWdwgVKcegd

When you delete repeating pattern (green and red numbers) you get:
101 5 100 4 001 1 101 5 011 3 111 7 101 5 111 7 000 0 010 2 001 1 111 7 111 7 111 7 010 2 101 5 101 5 100 4 001 1 111 7 000 0 011 3 001 1 010 2 101 5 0

That is:
5415375702177725541703125(0) in octal with optional zero at the end. Converted to decimal and hex:

- (with zero)
208779943814743312839336 (decimal)
2C35FBC23FF56C3C32A8 (hex)

- (without zero)
26097492976842914104917 (decimal)
586BF7847FEAD878655 (hex)

Those numbers in black look like a bunch of octal numbers to me. Edit: each one between two red zeros.

In case anyone wants to take crack at this, here is the raw data as Java arrays.  I've checked these over many times but there's always a chance for error.  No warranty provided.


// 0 = short
// 1 = long
String lengths[] = {
  "001101100111001101", // outer top, left to right
  "001111000101", // outer right, top to bottom
  "0011111011010001", // outer bottom, right to left
  "01101101", // outer left, bottom to top
  "0110110110100010110110110011", // inner top, left to right
  "110111110110110111110010", // inner right, top to bottom
  "100011100010110111110111110111110", // inner bottom, right to left
  "0111001101101", // inner left, bottom to top
};

// 0 = yellow
// 1 = red
String outer_colors[] = {
  "010011101110001101", // outer top, left to right
  "011101000011", // outer right, top to bottom
  "0001111000110011", // outer bottom, right to left
  "01100101", // outer left, bottom to top
  "0001010100101101110010110000", // inner top, left to right
  "000010111011100011110000", // inner right, top to bottom
  "101011100001110001011111110101110", // inner bottom, right to left
  "0111101001100", // inner left, bottom to top
};

// 0 = green
// 1 = blue
String inner_colors[] = {
  "000101010010110110", // outer top, left to right
  "100101111110", // outer right, top to bottom
  "0001101111111110", // outer bottom, right to left
  "10001001", // outer left, bottom to top
  "1000110100101011110000111011", // inner top, left to right
  "011111011001000010110100",  // inner right, top to bottom
  "000011001101110001100010101001111", // inner bottom, right to left
  "0010000001110", // inner left, bottom to top
};

As stated in an earlier post, the length bits when appended have a pattern in them: That pattern is 0?1?1? when the order is taken this way: inner top, inner right, inner bottom, inner left, outer left, outer bottom, outer right, outer top

01101101101000101101101100111101111101101101111100101000111000101101111101111101111100111001101101101101101000101101111100101000111100101100111001101100

 the flames could indicate which hex character each tile is supposed to represent and the meanders could represent the order the hex characters are supposed to be read off the board.