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Thi#s me$ssag@e* cont@ai%ns$ a sec@ret m%@es#*sa@$g$e, hid@d@en wi@th%in% it. Yo*%u ma#y enj@oy t#ryi@ng t#o so@lve$ this% p$uzz#l$e, may@be not.
Once you have solved it, you should find that decrypting the title of this article is easy.
Solution:There are five symbols that stand out: #$@*%. These occur between letters; they do not replace letters. The number of times that each symbol occurs are: @=12, $=7, %=6, #=6, *=3. The total is 34 symbols. We can guess that the hidden message contains 34 letters. Here are some possibilities concerning these five symbols:
Other, more complicated, schemes are possible. The above ideas give us many many possibilities to wade through. The encryption scheme may have been very complicated indeed. For example, maybe the hidden message was encrypted using some cipher, and then the symbols were applied using a different cipher. We probably want to examine relatively simple schemes first.
If a symbol modifies a letter, it would change it into another letter. G might become H or F or some more distant letter. How would the message look if the symbols modified letters? The first word in the message is THI#S. If # modifies a letter, it would seem that T and H were rejected; why? Probably none of the symbols could modify T to become the first letter of the hidden message, as there are only five symbols. Maybe the symbols add or subtract a fixed amount to the modified letter, for example: @=+1, $=+2, %=+3, #=+4, *=+5. Of course, other numbers are possible. Why was H rejected, while I was accepted. Maybe the first letter of the hidden message is I+5=N, as none of the symbols could modify H to get N. Near the end of the message, we have T#O SO@LVE$. If @ modifies the second O, why didn't it modify the first O instead? So maybe we are on the wrong track.
Let's assume that a symbol modifies the following letter. A ways into the message is the word HID@D@EN. If @ modifies the second D, why didn't it modify the first D instead? So maybe we are again on the wrong track.
So, maybe @ modifies the preceding letter, while # modifies the following letter. Then we still have the problem of T#O SO@LVE$. So, either the encryption scheme is more complicated than that, or we are again on the wrong track.
How about the physical position of the symbols? The number of letters between symbols are 3, 3, 4, 1, 4, 2, 2, 4, 4, 0, 2, 0, 2, 0, 1, 4, 1, 4, 2, 2, 4, 0, 3, 3, 3, 3, 3, 3, 4, 1, 4. That is fairly consistent. Maybe some of these numbers are slightly different because spaces and/or punctuation were counted, along with the letters. We will consider that (unlikely) complication later. If we have 5 symbols, and 5 spacings between symbols (0-4), then we can encrypt 25 letters. Then here are the letters of our message: 3#, 3$, 4@, 1*, 4@, 2%, 2$, 4@, 4%, 0@, 2#, 0*, 2@, 0$, 1$, 4@, 1@, 4@, 2%, 2%, 4*, 0%, 3#, 4@, 3#, 3@, 3#, 3@, 3$, 4%, 1$, 3#, 0$, 4@. This is a simple substitution cipher (See Solving Cryptograms), in which 4@ may be E (as it occurs 7 times, much more often than other combinations). That looks promising, doesn't it. The message is rather short; so it may be difficult to solve. But remember that E is the fifth letter of the alphabet, and 4@ may be the fifth number/symbol pair (0@, 1@, 2@, 3@, 4@). Maybe the encryption scheme is as simple as that: 0@=A, 1@=B, 2@=C, 3@=D, 4@=E. We have to try that first. Here is the message, so far: 3#, 3$, E, 1*, E, 2%, 2$, E, 4%, A, 2#, 0*, C, 0$, 1$, E, B, E, 2%, 2%, 4*, 0%, 3#, E, 3#, D, 3#, D, 3$, 4%, 1$, 3#, 0$, E. Then F, K, P and U are probably 0x, where x stands for the other symbols. We can guess that Z is 5x for the 26th letter. We can now experiment with the other four symbols. But we see that @#$% are four consecutive keys on the keyboard. We have to try 0@=A, 0#=F, 0$=K, and 0%=P. That leaves 0*=U. And, by substituting letters for number/symbol pairs, we see that this solves the puzzle. The hidden message is "I never met a huckleberry pie I didn't like." We didn't have to consider the complication of spaces and punctuation complicating the cipher.
See if you can decrypt the title of this article. You should find that it is easy.