#1
|
|||
|
|||
Some equation for a hasp passwords
There is a next equation for the nibbles of a hasp passwords:
a4 = (a1^9 - a0^1 -((a1^9 + a2 + ((a3^9 + a2) >= 15)) <= 15 )) %8 (see forum.ru-board.com) for pass 1177:0a28 a0 = 1 , a1 = 1 , a2 = 7 , a3 = 7 a4 = 0 , a5 = a , a6 = 2 , a7 = 8 what it means? |
#2
|
|||
|
|||
we have 8 var., but have only one equation.
i can't imagine the solution without bruteforing. have any idea?? |
#3
|
|||
|
|||
It was found by analysis about 500 passwords
|
|
|
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Sth. about InnoSetup's passwords | cnbragon/iPB | General Discussion | 10 | 02-10-2006 08:17 |
Bypassing rar passwords? | Rhodium | General Discussion | 2 | 11-04-2003 21:34 |
Encpyted passwords | SLIM SLIM | General Discussion | 5 | 12-17-2002 23:28 |