Hello EMC! A new familiar face

Discussion in 'Introduce Yourself' started by EyeCar, Dec 31, 2012.

  1. I was gonna guess ob1bob69 after I saw he was right next to the res and a donator.
  2. did i get the start right?
  3. Exactly what I think, VITIRI doesn't own the account although he won't give me a straight answer on whether bob owns it or not. I think ob1bob69 owns it, however it is likely jrlizard or chickeneer own the account maybe it's tinkerbell's account...
  4. I am not going to come outright and say I am not anybody; But when choosing my residence I first chose smp4 and than picked an already open residence. It would only make sense to choose well-known members of EMC (not be surrounded by a bunch of people that may not be around next week)
    Equinox_Boss likes this.
  5. No.. nothing about what you did was correct.
  6. But then how do you know all of these Aikar things... You must have been watching EMC for a while before joining...
  7. Who then? Don't say it on the thread PM it...
  8. I want to see people solve this code; I am a noob at this type of stuff. The only advantage I would have is knowing the answer before hand.

  9. Are those lines I's or lines?
  10. Just lines. Like these lines. | | | | | |
  11. Ok…not much help, but should help in solving the code.
  12. Your really trying to solve the code? It's crazy 1080p's it's uncrackable!
    607 likes this.
  13. So?
    Worth a shot.
    Equinox_Boss likes this.
  14. Give us a hint, too difficult.
  15. I don't care to figure out something super complicated, so most things are 1 to 1 substitutions... except for the last step, which is evil. Pure evil.

    Generally, letters = Numbers.
  16. Yeah I guess...
  17. Trust me, Try smp4 :)
  18. He claimed a residence on SMP4 :)
  19. There's surely a multiplication by 4 or 8 in there.. as it can't be a coincidence that all the numbers are multiples of 8.
    It would make sense to be if all the sets (eg {ABC}) stood for letters, but I'm not sure about what the sets with just a single member are doing (perhaps they stand for case, as post division by 8, it leaves the single-membered sets as binary (just 0 and 1)). The sum of the members of the first and second "large" sets are the same, and the 4th and last "large" sets are identical. One weird thing is how the 4th and last "large" sets only have 2 groups of 3 members.

    So that changes it to something of the form: {Aa} 1 {Ab} 1 {B} 1 {D} 0 {C} 0 {D}, where the (sum of Aa)=(sum of Ab).

    There's a rather badly worded list of my observations, and an attempt to make sense of them. I need to go and do some proper maths revision, so if my observations are of any use to anybody, there they are!

    ..and if I'm barking up a severely wrong tree, then there's 10 minutes of my life wasted :p
    607 likes this.
  20. WIN <3
    _Stads_ and Equinox_Boss like this.