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RE: The Steemit Crypto Challenge #11 - CC2k17.2

in #steemitcryptochallenge7 years ago (edited)

Ok, I think I may have a grip on hint2.
The string is comprised of hex part, a few random characters (salt) and base64 part.
I thinkEnCt2 means Encrypted Cipertext 2, irrelevant.
Then we have a SHA-1, 34eec5006fba77efa79febc136a5d9ccc1ea9f66, which starts repeating itself (padding to sha256 size).
Then we have a random part, tL=iqshH, which is used by hashing function (pbkdf2_hmac) as salt.
Lastly we have the ciphertext, sgIxIOZm0lgb/kyoe4AN7tsp1O5EqUkKwPsCG2Mv+fudbW1FmVRXycQXoZeI3NgoJwCXyi8ztg==, which is Base64 encoded output of a 256bit AES in MODE_CTR. We want to crack this part.

I wrote a little python script to decrypt it, but now we need a password, which is either in hint1, or some other hint in the puzzle.

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try numbers maybe, thanks for doing that :) cryptography is fun after all :) who is ben swann btw :D
30 @digicrypt/36 @jaki01 is what people have counted below, the morse code says count the bonds @digicrypt,

so amount and type of bonds, x,y,z triple double single so one triple, two doubles, 15 so infinity dunno

Fuck it I'm out of hints and so flew my hopes :D good luck guys :) Anonocon :D 2017 :D

Ben Swann is a TV journalist who gained notoriety for covering "alternative" news and challening MSM narratives. That is the only Ben Swann I know off.

Best comment so far

My previous count of 36 bonds is wrong because at first I thought the red color was a hint that there was a mistake in the molecule, so that I made another molecule out of it.
No, if the molecules are sketched correctly, we have altogether 34 bonds:

  • X = N2 and has 3 bonds (0 single bonds, 0 double bonds and 1 triple bond).
  • Y = 1,3-Cyclohexadiene (C6H8) and has 16 bonds (12 single bonds, 2 double bonds and 0 triple bonds).
  • Z = benzene (C6H6) and has 15 bonds (9 single bonds, 3 double bonds and 0 triple bonds).
  • That are altogether 34 bonds (21 single bonds, 5 double bonds and 1 triple bond).
  • I come to a higher number of bonds than @digicrypt because everywhere in Y, where you see a H2, there are 2 bonds, because every H atom has exactly 1 bond.

well timsaid the password is in the counts furion has the "box" you have the key, any hints mate :) can either of you open the pass with either 1215 or 3415 dunno the exact idea behind the count, but he did say it was simple and 1215 sounds simple :D

On point! Sweet :) the password is given within hint 1

Will this be a strange random numeric key or password generated when our answer was? If not have we tried the type of bond(s)? Maybe something like nonpolarcovalenthydrogen or? I am clumsy with the AES 256 been trying to crack. Should be working the whole string or just the section mentioned by @furion?

Hint 3: X = number of tripple bonds, Y = number of double bonds Z = number of single bonds

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