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From: swinfo on 23 Oct 2009 19:00

Thanks all for suggestion, i've solved myself.

Really, it was'nt so difficult... ;)

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From: Ant on 23 Oct 2009 20:27
"swinfo" wrote:

> crypted (HEX)
> 70 70 7E 6C 7E 76 65 55 59 60 30 42 40 3B 3C 3D 3F
> p p ~ l ~ v e U Y ` 0 B @ ; < = ?
>
> decrypted (HEX)
> 63 61 6E 61 72 69 56 45 4E 54 ? ? ? ? ? ? ?
> c a n a r i V E N T 300000

It's a 5-byte SUB mask (0D 0F 10 0B 0C) applied to each line. Subtract
from the first 5 bytes, then the next 5 until the end of line is
reached. The numbers have no special treatment.

63 61 6E 61 72 69 56 45 4E 54 23 33 30 30 30 30 30
c a n a r i V E N T # 3 0 0 0 0 0


From: swinfo on 29 Oct 2009 12:30
Ant ha scritto:

> "swinfo" wrote:

> > crypted (HEX)
> > 70 70 7E 6C 7E 76 65 55 59 60 30 42 40 3B 3C 3D 3F
> > p p ~ l ~ v e U Y ` 0 B @ ; < = ?
> >
> > decrypted (HEX)
> > 63 61 6E 61 72 69 56 45 4E 54 ? ? ? ? ? ? ?
> > c a n a r i V E N T 300000

> It's a 5-byte SUB mask (0D 0F 10 0B 0C) applied to each line. Subtract
> from the first 5 bytes, then the next 5 until the end of line is
> reached. The numbers have no special treatment.

> 63 61 6E 61 72 69 56 45 4E 54 23 33 30 30 30 30 30
> c a n a r i V E N T # 3 0 0 0 0 0

Thanks Ant, i've found the solution same day i post the question, when i
realize that the numbers havo no special treatment.

Thanks anyway for the answer.

Best wishes.

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