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My ex-boyfriend sent me a message with these letters and said that 'he's finally brave enough to tell me', or some bullcrap like that. He said he wouldn't tell me the encryption type, but he'd tell me that the key is 'acdc23b'. I don't know about any of this stuff. Please help.

Message --> I nrxj epu Lhux O kuuw yfy oexht gxbvg hptahh vr ettgeuv

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    $\begingroup$ I suspect that the first four words decipher to "I love you Jess", but I can't reverse-engineer a cipher for that that matches the key. $\endgroup$
    – F1Krazy
    Commented Jun 28, 2018 at 12:50
  • $\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$
    – Rubio ♦
    Commented Jul 4, 2018 at 23:36

1 Answer 1

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Try this:

If you treat this as a Vigenere cipher (https://cryptii.com/vigenere-cipher, change to DECODE) and use the key 'acdcfgb' (Thanks @IanMacDonald, I had it backwards), you get:

i love you jess i just was never brave enough to confess

Thanks to @F1Krazy for nailing the starting words

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    $\begingroup$ Key could also be acdcfgb, which is closer to what the person originally suggested (acdc23b). $\endgroup$
    – Ian MacDonald
    Commented Jun 28, 2018 at 13:15
  • $\begingroup$ @IanMacDonald Oops, I actually had it on ENCODE when I got that key. Yes, that would actually be the key. I updated my post to reflect that. Thank you! $\endgroup$
    – hagfy
    Commented Jun 28, 2018 at 13:19
  • $\begingroup$ How did you get the 'fg' from 23, if I may ask? $\endgroup$
    – Jess
    Commented Jun 28, 2018 at 13:40
  • $\begingroup$ @Jess You can just try different letters at that point. The Vigenere cipher is pretty commonly used, so I started with that. The problem is that it must be only letters, no numbers. If you assume the key is still 7 letters long, and use acdcXYb, you can just start playing with the X and Y to make it work. In this case, you'd see "I lov...", so just change the X around using that tool to make it an 'E'. I'm not sure why exactly fg=23, except that I'd also assume ABCDEFGH=ABCD1234. That's a bunch of assumptions, but I think it's mostly logical? $\endgroup$
    – hagfy
    Commented Jun 28, 2018 at 14:38

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