Summary of previous parts In Part-I, and Part -II we saw Reflector can very accurately de-compile .NET apps and we also studies some basics of packing and obfuscation. We saw that Reflector cannot directly de-obfuscate obfuscated assemblies, so we need external tools to de-obfuscate .NET assemblies. {smartkill} is one such tool that can de-obfuscate assemblies obfuscated by {smartassembly}. Reflector can't also unpack/decrypt packed/crypted exes, so we need unpackers and decryptors. PEiD can identify huge array of packed and crypted exes, to make our lives easier. Introduction I think readers are starting to lose interest as I cover only theoretical aspects of cracking. So, in this part I will actually CRACK an app :wink: As I have written only about .NET cracking till now, so I would crack a .NET app. We will cover other languages gradually. Here is the link to the target of this article is : http://crackmes.de/users/w02057/crackme5_by_w02057/download Background The target is a Level-2 crackme, solved my me. This crackme was an easy one, designed by w02057. The author(w02057) gives the following information about his crackme: (*) Language : .NET (*) Platform : Windows The rule that the author stated for solving are: (*) No Patching : that would make it too easy :p (*) No Bruteforcing : that would make it time-consuming and boring ACTION (1) Load the crack me into Reflector: File --> Open --> Select crackme location (2) Expand the following: CrackMe5 --> frmMain --> btnValidate_Click (3) See that it works with the following: ..(*) txtKey : a textbox obviously ..(*) txtSerial : another textbox obviously ..(*) getserial : a function ..(*) check : a function (4) Click 'check' to goto the associated code. You will see this: Code: Private Function check(ByVal str As String) As Boolean Dim num As Integer = 0 Dim num2 As Integer = 0 Dim startIndex As Integer = 0 Dim flag2 As Boolean = False Try num2 = 1 Do Dim num4 As Integer = (str.Length - 1) startIndex = 0 Do While (startIndex <= num4) If (Conversions.ToDouble(str.Substring(startIndex, 1)) = num2) Then If flag2 Then Return False End If flag2 = True End If startIndex += 1 Loop flag2 = False num2 += 1 Loop While (num2 <= 9) If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(0, 1)), Conversion.Int(str.Substring(4, 1))), Conversion.Int(str.Substring(8, 1))), 15, False) Then num += 1 End If If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(2, 1)), Conversion.Int(str.Substring(4, 1))), Conversion.Int(str.Substring(6, 1))), 15, False) Then num += 1 End If If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(0, 1)), Conversion.Int(str.Substring(3, 1))), Conversion.Int(str.Substring(6, 1))), 15, False) Then num += 1 End If If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(1, 1)), Conversion.Int(str.Substring(4, 1))), Conversion.Int(str.Substring(7, 1))), 15, False) Then num += 1 End If If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(2, 1)), Conversion.Int(str.Substring(5, 1))), Conversion.Int(str.Substring(8, 1))), 15, False) Then num += 1 End If If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(0, 1)), Conversion.Int(str.Substring(1, 1))), Conversion.Int(str.Substring(2, 1))), 15, False) Then num += 1 End If If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(3, 1)), Conversion.Int(str.Substring(4, 1))), Conversion.Int(str.Substring(5, 1))), 15, False) Then num += 1 End If If Operators.ConditionalCompareObjectEqual(Operators.AddObject(Operators.AddObject(Conversion.Int(str.Substring(6, 1)), Conversion.Int(str.Substring(7, 1))), Conversion.Int(str.Substring(8, 1))), 15, False) Then num += 1 End If Catch exception1 As Exception ProjectData.SetProjectError(exception1) ProjectData.ClearProjectError Return False End Try Return (num = 8) End Function (5) Press Back to go back and click 'getserial' to goto associated code. You will observe this: Code: Private Function getserial(ByVal str As String) As Object Dim obj2 As Object Try Dim str3 As String = Conversions.ToString(Me.hash(str)) Dim str2 As String = "" Dim startIndex As Integer = 0 Dim num2 As Integer = (str.Length - 1) startIndex = 0 Do While (startIndex <= num2) str2 = (str2 & str3.Substring(CInt(Math.Round(CDbl((Conversions.ToDouble(str.Substring(startIndex, 1)) - 1)))), 1)) startIndex += 1 Loop str = Conversions.ToString(Me.hash(str2)) str2 = "" startIndex = 1 Do str2 = (str2 & str.Substring((startIndex * 4), 4)) If (startIndex <> 5) Then str2 = (str2 & "-") End If startIndex += 1 Loop While (startIndex <= 5) obj2 = Strings.UCase(str2) Catch exception1 As Exception ProjectData.SetProjectError(exception1) obj2 = Nothing ProjectData.ClearProjectError End Try Return obj2 End Function (6) You see another unknown funtion 'hash' used in the first line of 'Try' block. So click 'hash' in Reflector and you go here: Code: Private Function hash(ByVal str As String) As Object Dim provider As New MD5CryptoServiceProvider Dim bytes As Byte() = Encoding.ASCII.GetBytes(str) bytes = provider.ComputeHash(bytes) str = "" Dim num As Byte For Each num In bytes str = (str & num.ToString("x2")) Next Return str End Function NOTE FOR NEWBIES : Don't be afraid of seeing so much code. Your work is simple (as you will see when you read further). At max, you'll have to write about 10 lines of code. (7) So, now we have enough information, about what's going on inside :wink: : ..(*) The check function checks if the entered key is valid. ..(*) If key is valid, it generates a serial and matches the entered serial with it. Simple, isn't it ? (8) So, what we do to crack this easily is.. we try to understand what the check function expects as a GOOD key. Lets's analyze : OBSERVATION : Observe the Do...Loop While(num2 <= 9) loop inside the 'Try' block. CONCLUSION : It ensures that, there are not repeated digits in the key. Thus, note that, the key consists only of DIGITS no alphabets. OBSERVATION : Next move to the sequence of 'If' Checks. [ There are 8 'If's ]. If an 'If' is satisfied, num is increased by 1 ( 'cuz num += 1). At the end the function checks if num = 8. CONCLUSION : So, ALL 'If's must be satisfied. (9) Study the 'If's now. BTW, hover mouse over 'Substring' and 'Conversions' to know about them. See that, the author extracts the i'th character in the string str (which is the key), by using the command 'str.Substring(i-1, 1)'. So, 'str.Substring(1, 1)' would give the 2nd character in the key. Note that, the author uses i = 9 at max because he uses 'str.Substring(8, 1)' at maximum. So, we got a hint -- the author checks ONLY the first 9 chars of key. (10) Now, what does he check ?? Note again that, 'Conversion.Int(c)' returns the digit contained in the character c. As the key consists only of digits, 'Conversion.Int(str.Substring(i-1, 1))' would return the i'th digit in the key. (11) So, finally we get to know that the function 'check' does this: Code: (*) Checks that first 9 digits in key are unequal. (*) Checks that the key satisfies the following: (1) 0 + 4 + 8 = 15 (2) 2 + 4 + 6 = 15 (3) 0 + 3 + 6 = 15 (4) 1 + 4 + 7 = 15 (5) 2 + 5 + 8 = 15 (6) 0 + 1 + 2 = 15 (7) 3 + 4 + 5 = 15 (8) 6 + 7 + 8 = 15 [ Here 0 represents 1st digit, 4 represents 5th digit and so on... ;) ] (12) So we need a unique permutation of the digits 123456789 as the first 9 digits which satisfies above. Simple observation leads to a better representation of the above conditions [ as a MAGIC SQUARE ] Code: 0 1 2 <-- 15 3 4 5 <-- 15 6 7 8 <-- 15 / ^ ^ ^ \ / | | | \ 15 15 15 15 15 Sum of each Row, Column and Diagonal is 15. So, lets fill up the magic square. (13) Representation of 15 as sum of 3 distinct natural numbers : Code: ..(*) Using 1 : 1 + 6 + 8 [ as 1 can be present in 2 different combinations, it must be, ] 1 + 5 + 9 [ at the mid-point of a side of the magic square ] ..(*) Using 2 : 2 + 6 + 7 [ as 2 can be present in 3 different combinations, it must be, ] 2 + 5 + 8 [ at a vertex of of the magic square ] 2 + 4 + 9 ..(*) Using 3 : 3 + 5 + 7 [ as 3 can be present in 2 different combinations, it must be, ] 3 + 4 + 8 [ at the mid-point of a side of the magic square ] ..(*) Using 4 : 4 + 2 + 9 [ as 4 can be present in 3 different combinations, it must be, ] 4 + 3 + 8 [ at a vertex of of the magic square ] 4 + 5 + 6 We have enough hints now. So, we now have the magic square of the form: 2 x x x x 1 4 3 x Now we can fill up logically as follows: Code: (*) 1 and 3 cannot be opposite to each other as they are not together in any combination. (*) 4 should not be above or below 1 as they are not together in any combination. (*) Mid-Left is obviously 9. (*) Bottom-Right is obviously 8. Now we have: 2 x x 9 x 1 4 3 8 Code: (*) Center is of course 5 (*) Top-Right is 6. (*) Mid-Top is 7. So finally we have the grid as: 2 7 6 9 5 1 4 3 8 So, the key corresponding to this is 276951438. NOTE: All other keys can be formed by simply rotating and transposing. (14) Now we know the valid keys. The getserial function can fetch us a valid serial for a key, so we don't need to be bothered about calculating a serial. We can just copy the getserial function (and the hash function, because getserial uses it) to VB.NET and can pass the keys as arguments to get the serials :wink: (15) So, we are done ! We have just finished a keygen for w02057's Crackme :happy: All the valid keys and serial combination are : Code: -------------------------------------------------- KEYS SERIALS -------------------------------------------------- 276951438 : 7DFF-7430-C0FE-FB5E-5675 294753618 : C528-6819-5244-62B2-9FA4 438951276 : CDB6-32A6-C696-62DA-E1BA 492357816 : 3AE9-0717-B6B6-7AFC-D688 618753294 : 8DC0-E6A2-D499-0F1E-29D5 672159834 : C520-15B6-6AFA-6B40-DFA1 816357492 : CE71-5143-D977-A8EA-C38E 834159672 : 60D7-0189-3A7E-5172-CE21 I hope you enjoyed making the keygen, as much as I enjoyed it. Greets to shabbir and all my friends here at G4EF and thanks to you for reading this long article ! Take care and good bye :smile:

Im new at this. Could you tell me how you plugged in the numbers for the magic cube? The 2, 3 and 4. I understood they equal 15 but not how they were placed where they were. Thanks

You can do that in two ways : (1) Use a logical approach as I mentioned in POINT (13) : You calculate "in how many different valid combinations, does a digit occur". From that knowledge, you can easily know its position. (2) Brute-force ! Write a program in any language of your choice to brute-force all the valid combinations. If you still have doubts, I can post a simple C++ Bruter here. Or I can explain POINT 13.

I want to know these things:- STEP(8) = OBSERVATION : Observe the Do...Loop While(num2 <= 9) loop inside the 'Try' block. CONCLUSION : It ensures that, there are not repeated digits in the key. Thus, note that, the key consists only of DIGITS no alphabets. how can i find this that there is only digits no alphabets and even digits are not repeated. STEP(14)= i am newbie so i don't know "how to " and "where is to" put getserial function. And by the way this is an amazing article on cracking