0+D4R2NBZ4JDCFF5OFBHH
SHR43ADRZK0BSEG0CEIS
0J50XKV00PKDDSL5DKFL
CHFCIF3JZDKECJSTF5J3
HR2NKPKDFSFK-1

Hint: 0 = 0. but 1 = 27, the line breaks don't matter, and the answer has four letters.

With the speed we are going towards the solution, this may very well become the most commented on problem.

Stick leading zeroes in front  and 3/4 of them have a 1 already and won't need two leading zeroes, and approximately 1/4 will need two leading zeroes

Ok, I should take the PID's and put zero's in front, that's clear.  Those with a 1 (in front???) don't need two zero's, they need one??? zero???? zero's.  The rest (1/4) needs two, this 1/4 would point to at least two numbers, so those are 51 and 67.  e can change the above sentence to: Those with a 1 in front don't need two zero's,...

Here are the two possibilities:

1160, 0123, 0051, 0149, 0186, 1021, 0101,0067, 1184, 1157
1160, 123, 0051, 149, 186, 1021, 101, 0067, 1184, 1157

Now what?  How do I binarise that and end up with four letters or a question?

 

 

 

 

 

Posted by Hugo on 2006-02-06 11:59:08