⇥ Challenge #3: simpler, but not too much

It turns out that more people were trying to crack Challenge #2 than I thought, and some were looking for solutions that were far more esoteric than anything I thought of.

For Challenge #3, I have devised a new riddle that is simpler to solve than the previous one—because I’m giving you a key:

UXULCPZEFRZD = SUPERCILIOUS

Given this key, send me e-mail with the plain-text version of this encrypted text:

IYZ BIXBBIBGWX PMRB AXCIRM

As with the previous challenges, everything you need to solve the puzzle is right here (including, this time, an explicit decryption key), although it helps if you have followed my Twitter timeline (assuming Twitter is working for a change). The solution is really simple this time—so I wouldn’t be surprised if I wake up tomorrow and find out that one of the five people who read my blog has already e-mailed me a solution.

Here are the rules:

• The winner is the first person from whom I receive an e-mail with the correct answer. I am not responsible for e-mail messages that get lost in any way, and the timestamp on my end is the final judge of who gets here “first”. If you win, you agree to allow me to post your name (and URL if you like) here—just so that everyone else knows I’m not making the winners’ names up.
  
  
• The riddle does have a solution—but it is, of course, possible that I have made a mistake in calculating it. In that case, I will void the contest and post a correct one.
  
  
• The winner will receive a $50 gift certificate that can be spent at the php|architect online store on any product, without any limitation, so long as the entire $50 is spent on a single order (sorry, that’s the way our store works). The coupon will expire after one year from the date of issue.
  
  
• The contest closes on June 30, 2008, at which point I will post the correct solution if no-one has sent it to me yet.