Inspired by a puzzle posted not that long ago.

A man was killed in an office. We have five suspects:

Isaac - takes care of trash in the company

Caleb - he once stole two dollars from a colleague's cabinet

Nathan - was once caught torrenting at work

David - takes care of IT stuff (he's a weirdo, by the way)

Gabriel - well, picked just to have five suspects

As you can clearly see, their roles doesn't matter. What matters is the note that was found on the crime scene:

"IISAC 14202410212202211102311"

So, who killed him?

I'm giving a hint:

Nothing separates the letters.

IISAC thing hint:

If I took the letters, and rearranged them a bit...

  • 5
    $\begingroup$ Welcome to Puzzling! Don't mind the downvotes. you do you. (+1) $\endgroup$
    – lois6b
    Aug 12, 2016 at 12:34
  • 2
    $\begingroup$ As you can see from style of my comment, I don't really care :P And that's my first puzzle here. $\endgroup$
    – Xirate
    Aug 12, 2016 at 12:35
  • 1
    $\begingroup$ Thanks for the edit, looks much better now (y) $\endgroup$
    – Xirate
    Aug 12, 2016 at 12:55
  • 1
    $\begingroup$ Doesn't seem to be any of the usual ciphers $\endgroup$ Aug 12, 2016 at 12:57
  • 3
    $\begingroup$ Another question where the similar logos of The Workplace and Puzzling make things interesting in the HNQ. $\endgroup$
    – Fiksdal
    Aug 12, 2016 at 17:15

4 Answers 4


I think, it's


The numbers

separated by 0 (nothing) are
142, 241, 2122, 22111, 2311

should be decoded as

alternating 1s and 0s, their numbers given by the strings above
1000011 (1 times 1 followed by 4 times 0 followed by 2 times 1, hence 142), 1100001, 1101100, 1100101, 1100010

which is

ASCII (anagram of IISAC) codes for Caleb

  • $\begingroup$ You sure about this? :P $\endgroup$
    – Xirate
    Aug 12, 2016 at 13:17
  • 1
    $\begingroup$ You sir deserve a cookie :P And I like how your nickname suits my character naming convention :D $\endgroup$
    – Xirate
    Aug 12, 2016 at 13:20
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    $\begingroup$ @Xirate sounds like you found suspect #6 $\endgroup$
    – CQM
    Aug 12, 2016 at 20:41

Partial answer in case it's helpful to other people. Still working on it :)


I got nothing on this yet.


Splitting this on the zeroes gives 5 substrings:
142 - 241 - 2122 - 22111 - 2311
All 5 of the subgroups have digits which add up to 7.

This suggests the answer is more likely to be a 5-letter name, which would be Isaac, Caleb or David, but that's just a guess at this point.

Based on the hint, which I've just seen, "nothing" separates the letters. This confirms that the subgroups are letters, separated by the zeroes, so we're talking a 5-letter name.

  • $\begingroup$ Really close :) $\endgroup$
    – Xirate
    Aug 12, 2016 at 13:11
  • $\begingroup$ "nothing" thing confirmed :) $\endgroup$
    – Xirate
    Aug 12, 2016 at 13:15

I think it might be


Because the clue

spells out Caleb in ASCII, as elias mentioned.


I'm kinda suspicious that either a) the guy killed had enough time to encode and transcribe the clue before being killed, and/or b) that the killer left or allowed the clue to be left when it leads back to him. Doesn't seem likely. Even if the killer didn't know what it said, it was typed after the killed guy knoew he was going to be murdered - the killer should have known it was dangerous to leave, obviously!


David, the weirdo, becomes my suspect, because he does "IT" stuff - which might include ASCII, coding, and so is likely to be creating the clue. None of the other suspects has as good a connection with the code (though Nathan, who torrented, came close - but I don't think torrenting requires ASCII). Caleb, who stole two dollars, is being framed (possibly the two dollars belonged to David). David left the clue to point the finger at someone else!

  • $\begingroup$ Dude, made my day. $\endgroup$
    – Xirate
    Aug 14, 2016 at 20:46

It is :


Numbers show all 7 , and he is the only one with 7 letters in his name.

  • $\begingroup$ Nope. Sorryyy :) $\endgroup$
    – Xirate
    Aug 12, 2016 at 13:17
  • $\begingroup$ A good approach! :) $\endgroup$ Aug 13, 2016 at 16:58

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