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Okay, so below are a handful of ciphers. They're pretty easy to identify, ranging from simple substitution to complicated binary. Some may have the keyword given others you have to solve. Now you're probably wondering, "What's the catch?" Well, the catch is that you have to solve it BY HAND. No internet help, no looking up stuff online. Actually, no technology help PERIOD. This is purely to test your skills, to see how good you are, to test your knowledge. You may use the internet to help you out (like say you don't know how a polynomial cipher works) but SHOW ALL WORK. Take a picture as proof. Please don't cheat, okay? And since I wish not to come across as a hypocrite, I will also upload proof that I enciphered EVERYTHING by hand after all the ciphers have been completed. So, are you up to the challenge?


  1. "Why so serious, Batsy? Hee hee hee..."
    ZX OD TU FB IE QB KU KY

Your task: Detect the cipher type, figure out the keyword, MAKE ME A POLYBIUS SQUARE(s), hand solve the message. The deciphered message then is a semi-word play you have to solve.


  1. T EFJLD JE OBQKR

    AJVIQBHN TOJVI UR

    CURS T EFJLD JE OBQKR

BR SATS WATS YJU SABHD JE FJVI

Your task: This one's very straight forward. Good old keyed substitution cipher. Tackle this by doing the frequent letter analysis technique, then after deciphering, tell me the keyword. This is also the first four lines to a song. Tell me the name and the band/artist that wrote it.


  1. 101116107113116101116104120101104115105105121125114123

Your task: Straightforward octadecimal with no word divisions. You'll get an "equation" after transliterating this to English. This should be the easiest to solve, since I pretty much gave you all the hints you needed.


  1. 101001000101110011010010011110111001000010001101011010010010011001110111110100100011011010110101010111111001010111

A=1, B=2, C=3, etc.

Your task: This should be the most challenging one y'all. Binary code except with no divisions. And to make your lives more miserable, letters are in numbers (a=1, b=2, etc.), so you don't have your comfortable eight digit range. Here are the clues to this cipher:

There are thirty decimal equivalent numbers in here. There are seven words in this cipher, in addition to 3 letters standing alone, with one of them being a "w". The "w" is located at the end.

I will accept partial answers.

Bounty shall be awarded to the person that has answered the most amount of question. If there is a tie, I'll award it to the user with the lowest reputation.

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    $\begingroup$ i think a no-computers tag would be suitable. $\endgroup$
    – Quintec
    Feb 19, 2018 at 0:25
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    $\begingroup$ Thank you! I didn't know such tag existed. $\endgroup$ Feb 19, 2018 at 1:09
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    $\begingroup$ Sigh I'm still doing this puzzle after like 30 minutes. I bet Rand'alThor or Deusovi could finish these in like < 10 minutes. $\endgroup$
    – NL628
    Feb 19, 2018 at 5:07
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    $\begingroup$ @NL628. Hey, its okay. You're not Desouvi or Rand'alThor, but you've still got quite a bit of experience. With practice you could become better than them someday. $\endgroup$ Feb 19, 2018 at 15:14
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    $\begingroup$ I believe there is an error in the second puzzle, YTU should be YJU. Pbyqcynl naq ybeq bs gur evatf ner njrfbzr! The others are a bit above my level, I'm afraid. $\endgroup$
    – Rick
    Feb 20, 2018 at 13:00

3 Answers 3

3
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For the 3rd, not sure if theres a typo in there, but I see the answer:

Luke. Per comments, a better answer would perhaps be Luke and Leia.

Based on

Group to sets of 3 digits, and each number is an ASCII value of a letter. Why 3? Since 64 is 100 in octal (and 65='a'), and the original text can be broken into all characters following the 1XX pattern, this it seems to make sense to group the ciphertext this way.
The string revealed (which appears to have a couple typos?) is ANGKNANDPADMEEQULS which I believe was meant to say ANIKINANDPADMEEQUALS (Anikin and Padme equals), appears to be a reference to the pair from Star Wars.

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    $\begingroup$ They had twins though.... $\endgroup$ Feb 21, 2018 at 15:15
  • $\begingroup$ "They had a twin though" Jack Pettinger $\endgroup$ Feb 21, 2018 at 15:55
  • $\begingroup$ quite true, i had thought about that after posting. but is there a typo, or am I wrong on that? $\endgroup$ Feb 21, 2018 at 17:38
  • $\begingroup$ Let me double check most likely yeah there probably is a typo. I just seem to make typos EVERYWHERE $\endgroup$ Feb 21, 2018 at 18:09
  • $\begingroup$ Bounty has been awarded to @thugsinuggs since he had the lower repuation of the two. Lets see if you can solve the rest! $\endgroup$ Mar 4, 2018 at 17:17
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Since no one has posted this, I might as well, for #2...

A FLOCK OF BIRDS
HOVERING ABOVE US
JUST A FLOCK OF BIRDS
IS THAT WHAT YOU THINK OF LOVE

Which is

A Coldplay song named "O".

Method

Basically note that T = I or A, JE and BR are two letter words, try cases, recognize the song.

Keyword:

TOLKIEN

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  • $\begingroup$ Great job! Do you kmow the keyword though? $\endgroup$ Feb 27, 2018 at 4:56
  • $\begingroup$ Also do you have the work (where you solved it on, refer to directions) $\endgroup$ Feb 27, 2018 at 5:20
  • $\begingroup$ @North Oops, I'll find the keyword but I don't have work because I solved this a long time ago(when I posted the comment correction) $\endgroup$
    – Quintec
    Feb 27, 2018 at 14:14
1
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The fourth message is: (- to separate letters, -- to separate words)

10100-1000-101--1100-1-10100-1001-1110--1-1100-10000-1000-1-10-101-10100--100-1001-100--1110-1111-10100--1000-1-10110-101--1010--10101--1111-10010--10111

Which translates to: THE LATIN ALPHABET DID NOT HAVE J U OR W

Method of finding:

Started by checking for spots that could only contain a specific letter and found the P in alphabet. Also highlighted areas with '10100' because T is much more common than BD or other letters possibilities is those areas. First word I found was the 'THE' at the beginning. Then I just kept playing around with the letters around the P until I discovered alphabet. Then went through and solved the rest based on what letters formed words in areas between letters already found.

P.S. This does actually have 31 decimal equivalent numbers but it seems likely that the OP made a counting error since this is an accurate statement and has the required 7 words and 3 letters.

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    $\begingroup$ Good job! Yeah my bad... this was one of my earlier puzzkes, there's a lot of errors $\endgroup$ Nov 13, 2018 at 3:27

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