The Encrypted text is:

7q7f 93d4 6s03 8aua 3pn9 73wu 9l2l a1wx 8u52 9tld 6fhc axfx 8kn4 628q 4abk 314d 255s 5o5b 9cc1 5904 4sja aa1r 7fai apsw 80ph ai09 ejbs

I am currently (at the time of posting the question) not giving any hints, it may depend upon later (except this tiny one).

You may call this a hint according to your wish:

The encoding is surely a multi-step process. Anyways check the tags.

Edit: The previous cryptogram was (You can solve this too anyways):

8kn4 255s 3pn9 73wu kg2b 7q7f apsw kg2b 314d 8aua ai09 kg2b aa1r 6fhc 9l2l 628q kg2b 5904 6fhc a1wx 4sja kg2b 4abk 8aua axfx 4sja 80ph kg2b 7fai 6fhc 8u52 9tld 8aua 93d4 kg2b 6s03 9tld 5o5b 9cc1

  • $\begingroup$ Repeats: "kg2b" 7 times, "8aua" 2 times, "6fhc" 3 times, "4sja" 2 times, and "9tld" 2 times. Perhaps space and four of the five vowels? $\endgroup$ – Glen O May 3 '15 at 14:37
  • $\begingroup$ Also, if we interpret each one as a 4 digit number in base 36, then "8kn4" is 400000, "255s" is 100000, "apsw" is 500000, "6fhc" is 300000, and "4abk" is 200000. $\endgroup$ – Glen O May 3 '15 at 14:50
  • $\begingroup$ @ADG I think that GlenO was only trying to share his progress with the community, attempting to cooperate in order to solve your cryptogram. $\endgroup$ – leoll2 May 3 '15 at 16:36
  • 1
    $\begingroup$ ADG - You can't give people a cipher text with no context and no real hints of any sort in order to make it "harder", and then be upset that people start sharing minor observations. I had no idea whether my observations were relevant. @randal'thor - I wouldn't have called it a partial answer, or "some progress", at all - they were just a couple of minor observations that may or may not help. $\endgroup$ – Glen O May 3 '15 at 21:38
  • 1
    $\begingroup$ @GlenO sorry, I am new (mabe this is irrelevant to the discussion), but anyways I'll try to be cooperative, thanks. $\endgroup$ – RE60K May 4 '15 at 4:23

For the first cipher, the decoded message is:

MR JOCK TV QUIZ PHD BAGS FEW LYNX. which is a notable pangram.

The code is recognizable as base-36 because all letters and numbers appear to be used. After converting to base-10, I tried a few functions and noticed that squaring the numbers produced 2 significant figures that could represent the letters of the alphabet.

So the decoding steps are:
(i) convert each 4 digit group to base-10
(ii) square each base-10 number and divide by $10^{10}$
(iii) select characters of the alphabet (except for 46 which is an ASCII period)

Using the same method, the second cipher decodes to:

PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS which is also a notable pangram.

| improve this answer | |
  • $\begingroup$ But what does that mean? $\endgroup$ – Ian MacDonald May 3 '15 at 19:00
  • $\begingroup$ why don't you try the second one with the same technique? $\endgroup$ – RE60K May 4 '15 at 3:50
  • $\begingroup$ and sorry but can you add some thought how you solved/approached the problem? $\endgroup$ – RE60K May 4 '15 at 4:22

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