Fact#1: You are the nations top ranked secret agent. Your skills at surveillence and interceptions is like that of no other. It is you job to intercept the enemies most notorious secret stealing criminal on the planet.
Fact#2: The manhunt is on! The espionage agent that you were hired to watch is on the move and headed to the airport to board the earliest flight out of your country.
Fact#3: The espionage agent has been warned by a mole that s/he is being watch and tailed by yourself, so s/he does the only thing that can save their scrawny ass, and that is to ditch the message before it can be delivered.
Fact#4: Fortunately for you, we have the janitors on the take and they were able to remove the coded message just seconds before it made it to the septic take. Unfortunately for you, beside the rank odor, we do not have a cipher for the message and the message was written on a cylindrical piece of paper without punctuation.
Fact#5: What we do know is that each letter of the alphabet is represented by a two (2) digit number.
Clue#1 We know that this message has exactly twenty-nine (29) letters.
So, the first thing I do as the nation’s top ranked secret agent is look at the piece of paper to see just what it means by ‘cylindrical piece of paper’.
(That is- are you saying that the message has no easily-identifiable beginning, or that if we wrapped the piece of paper around a pencil, say, we’d get a different message than just reading the numbers in order?)
Well, I can give you what I’ve started with, which isn’t much.
We know there are 29 letters, each letter has 2 digits, and there are 58 digits. Straightforward. A safe guess (famous last words) are that adjacent digits form a letter.
As we don’t know where the message starts (that is, if the beginning ‘65’ is a number, or if the beginning ‘6’ is the second half of a number), we have
65 45 10 36 25 65 85 75 27 70 95 65 15 10 27 05 40 50 30 30 05 35 18 30 45 70 36 27 57
and
76 54 51 03 62 56 58 57 52 77 09 56 51 51 02 70 54 05 03 03 00 53 51 83 04 57 03 62 75
as potential lists.
Looking at those for repetitions, we see that the first set of numbers is more promising. It’s got 3 digits that appear 3 times, 5 digits that appear 2 times. The second set of numbers has 2 digits that appear 4 times, and 3 that appear 2 times. Looking for an average appearance ratio, we get 1.61 for the first set and 1.45 for the second set. (This means that the first set has more repeats, on average, which we’d expect. E is a very common letter.)
The first set has one doubled digit - that is, you see ‘30 30’ in there. This also appears in the second set as ‘03 03’, and the second set contains ‘51 51’ as well. In the second set, 03 and 51 appear 4 times. It’s another safe guess (more famous last words) that E is the most common letter in this string, and seeing it back-to-back is interesting. The 0303 and 5151 in the second set prevent us from discarding it out of hand.
Basically, at this point, it becomes a simple cryptogram. Assign arbitrary letters to the digits and solve. Given that I’ve never been very skilled at solving cryptograms, I leave this as an exercise for the reader.
After considerable forensic studies a break (literally) was finally discovered in the note left by the espionage agent.
It has been discovered that there are hidden markings that could only be seen by ultraviolet light produced by its mercury discharge, aka The Black Light. When held up against this magical mysterious secret light, The Black Light, the message was transformed.
LNO, you and I are on the same track. The only discoveries I’ve made are that 30 probably equals a letter that is doubled in words frequently; (36)(27)(57) probably does not mean “T-H-E”, respectively, because it occurs as the last word of the phrase and that is not common in the English language. Wait, this is in english, isn’t it? If not then I give up. (65), (27), and (30) each occur 3 times each, but that isn’t conclusive as to which one could possibly be “E”. The numbers that each occur 2 times are (45), (10), (36), (70) and (05). Are we assuming that this message is gramatically correct? i.e., no “The” at the end of a message?
Scratch the last part of my previous post, We don’t know where the beginning or the end is, do we. Okay, back to the drawing board…or in my case the drawing pad…
The frequency of letters in a given statement are as follows:
E T A O I N S H R D L U C M F G Y P W B V K X J Q Z
There are only 18 different lettters in the message, and by frequency alone we could eliminate W, B, V, K, X, J, Q and Z. But because the english language does not work on frequency alone, to eliminate those would not be logical.
(65) does in fact start the word, but not the message. This means that the word “(65)(45)(10)(36)” is not the beginning of the message.
So if we assume that 362757 is THE, (and spacing out the number pairs to make them more readable) we get:
65 45 10 t / 25 65 85 75 h 70 / 95 65 15 10 / h 05 40 50 30 30 05 / 35 18 30 45 70 / the
Now, it’s possible that’s right, but in that case, the only E in the message would be at the end of “the” which makes the message unusual, because of the number of times E shows up in words. So, lets make 362757= FOR
65 45 10 f / 25 65 85 75 o 70 / 95 65 15 10 / o 05 40 50 30 30 05 / 35 18 30 45 70 / for
Now, if you’ll notice, “30” is doubled in the 4th word…which means it’s most likely E, L, or S, because those are the letters, IIRC, most likely to double. Also, if 36=f, then either 65, 45, or 10 has to be a vowel. Lets say that 30=E. It is the number most common.
65 45 10 f / 25 65 85 75 o 70 / 95 65 15 10 / o 05 40 50 ee 05/ 35 18 e 45 70/ for
There’s my contribution…I’m tempted to say that 65451036 is half, but I’m not sure.
To make it easier (assuming that this is indeed a simple cipher) I substituted a letter for each number, giving:
MKBI EMPOFN QMCB FAJLGGA HDGKN IFR
Which at least looks like a cryptogram.
One thing I tried was guessing three letter words (for the last word). (36)(27)(57). Per Captain Amazing’s request, let’s make it “FOR”. (36)=F. So we count down and up at the same time. Numbers go down, and we go up the alphabet. (35)=G, and so on, until we get to #27. How about that, it turns out to be O…Going back to 36=F, and counting down the alphabet, we get to R=50, which doesn’t work for the code. Last word ends up being FOK. Doubtful. However, going back to (27)=O, counting up the alphabet and up with the numbers, we get to R=(56). Close, but no cigar. I’m thinking that this is the pattern. One letter=something, then numbers go up and alphabet goes down, then alternating with some kind of pattern. It was worth a shot.
