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116313333134261542156511321455121126113213151461123615112516114142

OK, so recently I made up a way of sending coded messages and I was just wondering how difficult it would be for someone to crack. So let's say, for example, that you are in the army and you caught a suspected spy and you found a slip of paper on him with those numbers on it, exactly as they are above.

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  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio Aug 5 at 7:44
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The first thing that's apparent is that

the numbers are all in the range 1..6.

Since

the number of letters in the alphabet is comparable to 6 squared,

it's natural to consider

pairs of digits,

especially as

the total number of digits is even. It's also a multiple of 3, but it doesn't look as if there's much repetition in the digit-triples, so that isn't a very promising line of attack.

An obvious guess is that

letters are mapped into pairs of digits in some consistent way (there's already a pretty obvious guess, but let's be a little methodical here).

We then notice that

there are quite a lot of 11 and 15, which would work nicely if those were A (first letter) and E (fifth letter).

So the obvious guess is

that at least 11...16 are a..f.

Looking at what that gives us

and considering whether then 21..26 are g..l, etc.,

we notice

ba.a.ced which could be "balanced" if 26 is L

and now

we just do the obvious thing, interpreting a digit pair ab as letter 6(a-1)+b,

and get

a_complete_and_balanced_breakfast where each _ is a (different, as it happens) "out-of-range" digit pair.

Seems like we've cracked it.

(Is there an egg joke somewhere around here?)

In answer to the question of how difficult it is to crack,

I think it took somewhere between three and five minutes to go through the process above, though I confess my very first thought on looking at the string of digits was pretty much the actual answer.

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