PDA

View Full Version : Mind Challenge...(or not?!)


xfze
November 29th, 2003, 21:30
I would apreciate all answers, speculation, etc

Well...

Data:

>>we have an 8 digit key
>>we have to get an 8 digit serial that belongs to the key

>>example: the key: RXYW3R55 has the serial (one of them): 4MS2JB9A

>>the generation of the serials is somehow done by the string:
NAX8B9TY3ZCRQF2GSWVH4J7EDK5LM6PU

Clues:

>>as i said a key may have more than one serial, but i think that don't mind right now(does it?)

>>as we can see in the example the last 2 digits of the key are the same (5), but in the serial theyīre not (9 and A). so a single alfabet comparison won't work, maybe a multiple table like i did once with 8 strings (one for each digit) taking the example as a start point(something like: 1st digit R equals 4; 2nd digit X equals M;...and so on)

>>the generation string gave above contains the characters for the key, the key can only have 8 among those characters. and as you can see there arenīt any I;O;1;0; so neither the alfabet or the ten decimal digits are complete (wich makes dificult the making of the tables i mentioned)


So...does anyone can figure out who the serials are made?
if yes say how.

you can try to guess the serial for the key: AGYQTULM
report it to me and then i can tell if your solution is right or wrong

good luck

dELTA
November 30th, 2003, 08:59
Damn, people have become quite creative trying to find loop-holes in the FAQ rules and concealing their crack requests lately...

How about tracing the damned app?

ZaiRoN
November 30th, 2003, 10:54
Quote:
Damn, people have become quite creative trying to find loop-holes in the FAQ rules and concealing their crack requests lately...
LoL

xfze
November 30th, 2003, 14:55
LoL...
yeah...sorry about that
I'll try to trace it, but there are people who like beating challenges anyway

mike
December 2nd, 2003, 18:58
Quote:
[Originally Posted by xfze]LoL...
yeah...sorry about that
I'll try to trace it, but there are people who like beating challenges anyway


But attacking an unknown cryptosystem with a single ciphertext is pretty futile...

xfze
December 2nd, 2003, 20:33
Quote:
[Originally Posted by mike]But attacking an unknown cryptosystem with a single ciphertext is pretty futile...


no...i know now two sets of keys...so itīs more easy (although i didnīt get to it yet )
with the generation key (NAX8B9TY3ZCRQF2GSWVH4J7EDK5LM6PU)
and these two matches:
(RXYW3R55 <-> 4MS2JB9A)
(UWW4Y6ZG <-> NGTR5V74)

I think it isn't impossible to check it out just with pencil and paper (thus a little tracing would speed up the process, no doubt about it.)( i'm almost sure this cryptosystem is very basic so time is all we need )

yousky
December 3rd, 2003, 23:21
The generation key (NAX8B9TY3ZCRQF2GSWVH4J7EDK5LM6PU) seems to be a MD5 string generated with the other informations you've post because all MD5 code have a length of 32 and it is for your example. Perhaps it's a way to search an answer.

bye bye
yousky

mike
December 5th, 2003, 00:13
Quote:
[Originally Posted by yousky]The generation key (NAX8B9TY3ZCRQF2GSWVH4J7EDK5LM6PU) seems to be a MD5 string generated with the other informations you've post because all MD5 code have a length of 32 and it is for your example. Perhaps it's a way to search an answer.

bye bye
yousky
While the string is 32 bytes long, it's almost certainly NOT an MD5. The upper two bits are all zeros. There's a 1/2^64 = 1 in 10 quintillion chance of that happening by accident.