x0rz Posted June 23, 2017 Share Posted June 23, 2017 Difficulty : 7-8Language : DelphiPlatform : Windows x32OS Version : All Windows Packer / Protector : None and No Crypto Description : Rules : no patch, only working keygen or a valid serial for your name... Good luck Screenshot : x0rz Keygen ME.rar 2 Link to comment

sama Posted June 27, 2017 Share Posted June 27, 2017 (edited) Just a quick overview thanks for the challange Name: John Key: F6FBF830-0D910F71-0A79EF4F-0D631D6E enjoy Edited June 27, 2017 by sama 3 Link to comment

s3rh47 Posted June 28, 2017 Share Posted June 28, 2017 15 hours ago, sama said: Just a quick overview thanks for the challange Name: John Key: F6FBF830-0D910F71-0A79EF4F-0D631D6E enjoy Thx for Solution. Can you please explain, how to you solved it? Link to comment

Solution sama Posted June 28, 2017 Solution Share Posted June 28, 2017 4 hours ago, s3rh47 said: Thx for Solution. Can you please explain, how to you solved it? sure there is no secret in it serial parts calling P1-P2-P3-P4 P1 is based on name with a bit xor, the tricky is P2 and P3 but simple to recognize anway P2 and P3 are xored with name based value then turned into decimal now for P2 take last 6 digits which must be a Carmichael pseudo prime (ODD, at least 3 factors) and must be > then 100.000 to pass the first check and to pass second check the whole P2 (after the xor) must be prime! Same is for P3 except that the Carmichael pseudo prime are now the first 6 digits! as for P4 is a simple calculation xor and add (if i remember right) and game is done. have a nice day 2 Link to comment

x0rz Posted June 28, 2017 Author Share Posted June 28, 2017 5 minutes ago, sama said: sure there is no secret in it serial parts calling P1-P2-P3-P4 P1 is based on name with a bit xor, the tricky is P2 and P3 but simple to recognize anway P2 and P3 are xored with name based value then turned into decimal now for P2 take last 6 digits which must be a Carmichael pseudo prime (ODD, at least 3 factors) and must be > then 100.000 to pass the first check and to pass second check the whole P2 (after the xor) must be prime! Same is for P3 except that the Carmichael pseudo prime are now the first 6 digits! as for P4 is a simple calculation xor and add (if i remember right) and game is done. have a nice day Thank you for the solution. Link to comment

s3rh47 Posted June 29, 2017 Share Posted June 29, 2017 On 28.6.2017 at 2:03 PM, sama said: sure there is no secret in it serial parts calling P1-P2-P3-P4 P1 is based on name with a bit xor, the tricky is P2 and P3 but simple to recognize anway P2 and P3 are xored with name based value then turned into decimal now for P2 take last 6 digits which must be a Carmichael pseudo prime (ODD, at least 3 factors) and must be > then 100.000 to pass the first check and to pass second check the whole P2 (after the xor) must be prime! Same is for P3 except that the Carmichael pseudo prime are now the first 6 digits! as for P4 is a simple calculation xor and add (if i remember right) and game is done. have a nice day hey Sama again thx for solution but i want to ask you something xorz has forgot a string in KeygenMe under Procedure of Carmichaeal Algorithm .. That was "CARMI". have you got a Tipp from this String or you have solved without this string ? Link to comment

sama Posted June 30, 2017 Share Posted June 30, 2017 (edited) indeed i saw that string but even without it here are the points get the squareroot then in next loop checking if CGD is 1 and then and this is according with what wikipedia tells us: (extract from wiki) "In number theory, a Carmichael number is a composite number n{\displaystyle n} which satisfies the modular arithmetic congruence relation: for all integers b which are relatively prime to n. ^{[1]} They are named for Robert Carmichael. The Carmichael numbers are the subset K_{1} of the Knödel numbers." further information look at https://en.wikipedia.org/wiki/Carmichael_number have a nice day Edited June 30, 2017 by sama 4 Link to comment

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