The Clues

All the information you might need is here in the post



(All the information you need can be seen in the above section. You do not need to look anywhere else for clues.)

What is this?

This puzzle is a sort of "puzzle-ception," inspired by Cicada 3301. Once you solve the puzzle in the clue section, you will find another puzzle. Upon solving that, you will find another puzzle, and so on.

If no one is able to progress in a day, I will post a hint in the hint section below. Once someone solves a layer, I will also add it to the hint section.

Collaboration is Key!

This puzzle is fairly difficult, and will probably take several days minimum for the community to solve. I highly recommend that you collaborate, and if you make a breakthrough, post as an answer!


2 hints for the first phase.

Hint #1: AES-128bit. The key is 32 hex characters long (16 ASCII characters).
Hint #2: The key is in "The Clues" section. It is almost literally told to you.

0 hints for the second phase.

0 hints for the third phase.

The question has been solved! I am working on a much more elaborate puzzle in the same style of this, which I will try to release in a couple of days.


It turns out that I was using a faulty algorithm for the AES-128 encryption. No one has posted the correct key though, so I fixed it. (You can use http://aesencryption.net/ to encrypt/decrypt with the algorithm I meant to use). Very sorry about this slip, and I hope that it didn't affect your analyses.

  • $\begingroup$ In what way do you mean "layers"? $\endgroup$ Mar 29, 2016 at 0:34
  • $\begingroup$ When you solve one puzzle, it leads to another. $\endgroup$
    – AMACB
    Mar 29, 2016 at 0:35
  • $\begingroup$ So the first line is one puzzle, and the second line is another? Or are they part of the same puzzle, and the indentation matters? $\endgroup$ Mar 29, 2016 at 0:36
  • 1
    $\begingroup$ There are subtle details in implementing AES, you can't exactly just throw it out there. What mode? Now, I think we can guess this one since the size of the bytestream is divisible by a block size and no initialisation vector is required. But what about padding? If the original key is not 16 bytes, does it have to go through a KDF? Is it intentional that we might not need these details, or be able to figure them out without you having to say anything? $\endgroup$
    – Reti43
    Mar 29, 2016 at 22:17
  • 1
    $\begingroup$ Also, most recent edit says the all information can be found in above section. But one important clue was found outside of the above section! $\endgroup$ Mar 29, 2016 at 22:21

4 Answers 4


There's an image (base64 encoded) embedded in the file ffao linked to

enter image description here

The image contains these links:


The "number" in the first link is actually

a hex encoded version of the WAV file, with a little bit appended on at the end: 546865206d6573736167652069733a20636f6e67726174756c6174696f6e7321

This bit reveals our coveted hidden message using the standard Hex-to-ASCII Conversion Ritual™

The message is: congratulations!

  • $\begingroup$ Hold up a second, where did you find the embedded file? $\endgroup$ Mar 30, 2016 at 2:39
  • $\begingroup$ @DevilApple227 I opened the PDF in Notepad++ - it's the object indexed 19 on lines 136-138. $\endgroup$
    – Will
    Mar 30, 2016 at 2:40

The first string is, Base64 decoded:

Remember that at first you are off by one.

The second also bears the characteristics of a Base64 string, but decodes to binary garbage.

The question's source contains an unused link to the URL:


  • $\begingroup$ Good start! I can confirm that both of the things you found are intentional. $\endgroup$
    – AMACB
    Mar 29, 2016 at 0:48
  • 6
    $\begingroup$ @AMACB I would be surprised if they weren't; odds of a random string of that length correctly decoding as Base64 to English text are around a billion billion billion billion billion billion billion billion to one. And the odds of the second phrase occurring at random I would put at a flat zero. $\endgroup$ Mar 29, 2016 at 0:54
  • $\begingroup$ I don't have time to throw together a program to do it right now, but maybe subtract (or add) 1 from each digit of the second string and see if that causes it to decode to something nice? $\endgroup$
    – Tim C
    Mar 29, 2016 at 3:14
  • $\begingroup$ Given the way base64 works, it's unlikely that a small transformation like that would make complete binary garbage turn into something readable. I tried it anyway just to make sure. $\endgroup$
    – ffao
    Mar 29, 2016 at 7:10
  • 2
    $\begingroup$ @DevilApple227 you can view the edit log and source text by clicking edit or clicking the "edited 23 mins ago" (subject to change) part of the question. $\endgroup$
    – Z. Dailey
    Mar 29, 2016 at 14:36

Decrypting the second string we obtain


Key is "6865726520696e2074686520706f7374" (hex for here in the post)
The output comes out ROT-1'd, so that's what the first hint was for.

  • 1
    $\begingroup$ I thought that was the key, but it didn't work earlier. >.> $\endgroup$
    – f''
    Mar 30, 2016 at 2:08
  • $\begingroup$ @f'' that was my mistake. Read my edit. (Sorry about that) $\endgroup$
    – AMACB
    Mar 30, 2016 at 2:12

What we know as of now:

The first string mentioned in the puzzle, decoded from Base64, is

Remember that at first you are off by one.

The second clue we know:

At the bottom of the source, we find an unused link that says


Now, as of just a little ago, we know that the second string is encoded using AES 128-bit encryption, with a 32 byte hex key (16 ASCII characters).

Suggested course of action: start by writing a script to look through the source and find any 16 character strings that are surrounded by spaces.

  • $\begingroup$ You can take in my new hint with this. $\endgroup$
    – AMACB
    Mar 29, 2016 at 23:51
  • $\begingroup$ I managed to hack together a program to follow your suggestion, but no dice. The only key taken from anywhere in the question that correctly decodes is , you will find (it's not in the "The Clues" section, but neither was the cipher), and the output is another blob of binary garbage. Either the OP is using CBC mode and not telling us the IV, or the key is not literally in the question. At least we know the key is ASCII; assuming he means printable ASCII we only have around forty thousand billion billion billion keys to check. $\endgroup$ Mar 30, 2016 at 0:53
  • $\begingroup$ @2012rcampion Read my edit, and try the same keys you did before. (Really sorry if that messed you up!) $\endgroup$
    – AMACB
    Mar 30, 2016 at 1:47

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