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Three high school friends who took different paths after high school are now all enjoying retirement. Each one settled for a different job but all enjoyed cryptography throughout their lives.
Now having reunited they send each other encrypted messages to keep their minds sharp and prevent their knowledge from dwindling.

  • One spent his life in the military.
  • Another in the police force.
  • The last a telecommunications company.

Currently they communicate using a homemade cipher they can all relate to.

!3B7BA2B3C 1CF 7A9A3B2ABD1F A7A1B 7A9A3B2ABD1F 7A2B7C2BD B7A1B.
!C4A 1C2E E1A2CB1F 3B7BA2B3C 2E7B 2AB 6A1AEE3C,
C7B9A2D 1F2B1A2D3A6A 4A7B2D 6A1AEEC7A2B1FF 0A1CDD 7A2B7C2BD B7A1B.

Can you crack their latest message???

To complete the problem you must:

 - Decrypt the message and find the author.
 - Explain clearly how the cipher works.
 - Explain what the cipher is based on.

Hints
After a few days I will post some more hints if this remains unanswered.

They all share practical knowledge, they have been known to mumble.
The last thing they want to do is meet up, they are in their element.
They also like to abbreviate, time is of the essence.
Read between the lines systematically, these hints are not what they seem.
What is the solution to verbal communication errors?
Shall I spell it out for you? Perhaps.
Nate! Oh My Gosh, Get It Over With!

PLEASE NOTE: I have updated the cipher text due to minor errors.

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The message says:

Money is numbers and numbers never end.
If it takes money to be happy,
your search for happiness will never end.

(which is a quote from Bob Marley)


The encryption process works as follows.

First, produce an alphabet based on the NATO phonetic alphabet, but by

grouping the alphabet by the last letter of the corresponding radio code (alpha, delta, india, ... bravo, echo, ...). Assign each group a number and each word a letter based on the order they appear (credit to Zandar for working out this step).
So A = alpha = 1A (first group, first word), B = beta = 2A (second group, first word), etc.
Resulting in the following dictionary (code determined by column, then row):
- 1 2 3 4 5 6 7 8 9 0
A: A B C F G H N Q U W
B: D E M J O X
C: I K Y V
D: L R
E: P T
F: S
(not sure what they'd do if they needed a Z, since that would require an 11th group...)

From here, you can encrypt your message easily, however:

Once you have converted each letter into it's corresponding number/letter pair, if consecutive pairs use the same leading number, you drop the number in subsequent pairs (so 2A2B becomes 2AB).

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  • $\begingroup$ OK, I followed this up until the last spoilerblock - how did you arrive at that? $\endgroup$ – question_asker Feb 18 '16 at 15:11
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    $\begingroup$ I used the same strategy, but didn't think of repeating the previous number; I assumed that each letter had a 'default' number. Regardless of the justification behind that particular mapping, you broke the cipher, and IMHO that's what matters. Nice work! $\endgroup$ – 2012rcampion Feb 18 '16 at 15:13
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    $\begingroup$ @question_asker - see my comment(s) on the question itself... I still don't know how it actually works, I just reverse engineered it. $\endgroup$ – Alconja Feb 18 '16 at 15:17
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    $\begingroup$ It looks like the letters are grouped by the last letter of their radio alphabet word, although I don't know whether there's a reason why they're assigned to specific numbers. $\endgroup$ – Zandar Feb 18 '16 at 15:44
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    $\begingroup$ @Daedric - fair enough, you've convinced me (frequency analysis, etc are most definitely valid techniques). And at the end of the day, the most difficult thing is getting the difficulty right. :) $\endgroup$ – Alconja Feb 19 '16 at 8:43

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