I've never been a fan of ciphers/encryption. Truth be told, despite my love of puzzles (and the amount of time I spend here on Puzzling.SE), I'm just no good at them and haven't properly learnt the strategies for trying to crack them.

So in an effort to expand on the types of ciphers and hopefully broaden the audience, I offer you the following challenge:

I have encrypted a five-word phrase in the form _ _ _ _/ _ _ _ _, / _ _ _/ _ _ _ _ _ _/ _ _ _ _ _ _ [word lengths are (4) (4), (3) (6) (6)]

However, instead of just encrypting it once, I have done so six times, using six different methods.

The output for each encryption is as follows:
Encryption Output

The final answer has two parts to it. First, the completed five-word phrase (which I suspect will be discovered first) and second, the six different encryption methods, all of which must be detailed in the answer.

The decrypted phrase will tell you to do something, so make sure you do it :) You wouldn't want to make the puzzle sad, would you?

Some excellent community effort so far! The phrase and methods 2, 4 and 5 have been cracked, so I'll throw some subtle hints for the remaining three methods in

Method 3

How often do you see a 3-letter word have a higher value than a 4-letter word?

Method 6

The order of the letters doesn't seem logical in isolation, but if it ever got changed, many of us would struggle to adapt!

  • $\begingroup$ Shouldn't the phrase have 4 words? $\endgroup$ Nov 12, 2018 at 23:09
  • $\begingroup$ I have an idea for Method 4, but only if word 1 is 35 instead of 45. thanks! $\endgroup$ Nov 13, 2018 at 10:12
  • $\begingroup$ I'll need to double check in the morning $\endgroup$
    – Dmihawk
    Nov 13, 2018 at 10:18
  • $\begingroup$ Sorry I was wrong $\endgroup$ Nov 13, 2018 at 23:34
  • $\begingroup$ "Hash" might be a better word than "cipher" or "encryption", since you can't easily go back from the value to the word. $\endgroup$
    – Harfatum
    Nov 14, 2018 at 3:41

4 Answers 4


This is a summary of all answers provided by everyone, I have credited them, if I left anyone out, please state it in the comments. Thanks!

Complete Answer:

The phrase is:

WELL DONE, NOW UPVOTE PUZZLE (which I did), thanks to @DrXorile (approved by OP in comment)

Method 1 is

Sum of Scrabble values of each letter. So much thanks to @Braegh! (approved by OP in comment)
My original guess:
Ceasar shifting, the number being the rotation number, as all numbers are not larger than 26. 26 means the word is not shifted at all. (I believe that this is wrong...)

Method 2 is:

Sum of values in A1Z26 scheme, Thanks to @ImongMama (approved by OP in comment)

Method 3 is:

The sum of values of each letter according to the frequency (descending): etaoi nshrd lcumw fgypb vkjxq z, where e=1, z=26,
THANKS TO @NudgeNudge!!

Method 4 is:

the sum of values of each letter on a telephone keypad, like this:
e.g. for "DONE" 3+(6+6+6)+(6+6)+(3+3)=39 (approved by OP in comment)

Method 5 is

the sum of ASCII values of each letter, since they add up to around 80-90 per letter. (approved by OP in comment)

And method 6:

is the sum of the values for each letter according to its position on a QWERTY keyboard, like this:
qwerty encryption


I really enjoyed this puzzle. Thanks so much, @Dmihawk!

  • $\begingroup$ Good spotting! :) $\endgroup$
    – Dmihawk
    Nov 12, 2018 at 23:41
  • $\begingroup$ Method 2 and 6 are correct (I didn't know method 2 had an actual name - so that was cool to learn). For method 1, perhaps consider why one 6-letter word is "worth" more than the other... $\endgroup$
    – Dmihawk
    Nov 13, 2018 at 9:42
  • $\begingroup$ Method 6? @Dmihawk Thanks! $\endgroup$ Nov 13, 2018 at 9:57
  • $\begingroup$ Sorry, meant 5! $\endgroup$
    – Dmihawk
    Nov 13, 2018 at 10:14
  • 2
    $\begingroup$ For method 3, I believe rot13(gur ahzrevpny inyhr bs gur yrggre vf rdhny gb vgf cbfvgvba ba gur yvfg bs yrggref beqrerq ol eryngvir serdhrapl va Ratyvfu (r svefg, gura g, n, b, v...)) $\endgroup$
    – NudgeNudge
    Nov 14, 2018 at 20:57

Method 2 is

Sum of A1Z26 values


I believe Method 1 is

the (English) Scrabble value for each word.

  • $\begingroup$ Well done! Good find :) $\endgroup$
    – Dmihawk
    Nov 14, 2018 at 0:35

Based on @Omega Krypton's method 5, the phrase is:


I figured this out by going through all the english words that would fit those sums. The last word stood out, and I got the remaining ones except for the fourth quite quickly. The fourth wasn't in my dictionary, but easy enough to guess and check...

  • $\begingroup$ Nicely done :) now you just need to deduce the other 5 encryption methods! $\endgroup$
    – Dmihawk
    Nov 13, 2018 at 0:37
  • 1
    $\begingroup$ I upvoted the puzzle lah :) $\endgroup$ Nov 13, 2018 at 2:57

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