Cracking the Windows Screen Saver Password
This is an interesting hack and not many people know about it. This requires no
canned hacking tool, we will crack the password manually!!! First of all, why do we need to crack the Windows Screen Saver? How does it restrict us? If a Screen Saver is password protected, then whenever it is turned on, then in order to turn it off, you need to enter a password. It does not allow us to do anything on a system until and unless we enter the password. We will keep seeing
the screen saver until we authenticate ourselves by entering the password. No not even CTRL+ALT+DEL works in this case. An average user encounters around 20 different places where he needs to type in the password. Most people fin dit very difficult to remember even more than a single password, hence to make life easier for themselves, they use the same password in all the places. And also on some systems the Login password is same as the Screen Saver Password. Hence it is very useful to crack the Screen Saver Password.
Now let's move onto cracking the Screen Saver Password. For this example, protect your screen saver with the password, 'DOPE'. Windows stores the Screen Saver password in the user.dat file in the Windows directory. If you have multiple profiles on your system then it is stored in the user.dat file in the c:\windows\profiles\username directory.(On Win 3x systems it is stored in the control.ini file). The user.dat file constitutes the registry of the Windows system, thus we can say that the Windows Screen Saver Password is stored in the registry. First of all, you need to change the attributes of this file and make it editable by right clicking on it and unselecting the Read Only Option else you will not be able to edit it.
Once this is done, open this file in WordPad (Any text editor will do except MS WORD And Notepad.)Now look for the string: ScreenSave_Data
You will find an even number of characters after Data, this is the Screen Saver
Password encrypted and stored in the hex system. Each pair or hex values represent a single ASCII plain text character. This means that if there are 10 hex values then the password is of 5 characters, each pair of Hex values standing for a single plaintext ASCII character. So in order to get the Plaintext password you just need to decrypt these hex values into ASCII.
There are many screen Saver Password de crypters around which decode the password for you but I believe that it would be better if we could do it manually without using a third party canned hacking tool. And hey it is really simple once you get the hang of it. The only thing you need to know is the various number systems. This means that you need to know The Hex system, The Decimal System and also The Binary System.
For example ASCII character 'A' is 41h(ex), 65 Dec(imal) and 01000001 binary.
One could also get hold of a good ASCII chart which has all the number systems and their conversions. Make sure that the ASCII chart you get has Hex, Decimal, Binary and of course plaintext ASCII.
XOR
Before I go on let me introduce you to XOR. The following is the chart you need to refer to when you need to evaluate the XOR value.
input value A input value B Output
+--------------------------------------+
0 0 0
0 1 1
1 0 1
1 1 0
+--------------------------------------+
Example
Question: Answer:
00001100 00001100
00101001 00101001
-------- <--XOR -------- <--XOR
???????? 00100101
You may ask how did that happen? Well it's easy. Take the case of the first digits. The Input Value A is 0 and the Input Value B is also 0. Now refer to the XOR chart. You find that the Output when both the Input values are 0 is also 0. Similarly consider the third values. Input Value A is 0 and the Input value B is 1. If we refer to the XOR chart, we find that the Output is 1. However the conventional method is to start from the right, as we are taught in school.
**********************
Hacking Truth: The Screen Saver Password cannot be longer than 14 characters because if it is longer the system will not either prompt for the password or will hang and reboot.
**********************
It's an even string containing letters and numbers. This is your password. If you've read everything
you should have changed your password to 'DOPE' which is 4 characters
long, and your encrypted password is 8 characters long, (0CA12658)
Hmmm. so D O P E is the same as 0C A1 26 58.
So
D= 0C
O= A1
P= 26
E= 58
Am I right? Ok, and now listen carefully; the 0 represents 4 and C represents
4 too after decryption. Put those two number together and you get
44(h). This is the way you have to do that, with every decrypted couple.
Ok grab an ASCII table and look at 44 HEX. That's 'D' like in DOPE
know what I mean?
So now I'll show you how to get the encryption scheme:
0C --> 44h --> ASCII char 'D'
That means 0 --> 4
C --> 4
ok, now the binary
0 = 00000000
????????
-------- <--XOR
4 = 00000100
Can you still follow me? It might sound a bit weird, but trust me, it is quite simple. Read it again to make it clear.
0 = 00000000
00000100
-------- <--XOR
4 = 00000100
Ok now you know that for the first part 00000100 is used
to decrypt the password, right? But with the second one it
goes different. Then the second part of the hex number, ok C must become 4 too,
so that's easy ;
C = 00001100
????????
--------
4 = 00000100
After performing XOR you will get
C = 00001100
00001000 <-- we found our encryption scheme for the second char and
-------- of the first encrypted character
4 = 00000100
Ok, so far so good, we now know how 0C gets decrypted to
'D' and that the second part uses 00001000
So we must check if it really works. Yeah. So we'll check it,
change your password to 'ERIKA' and the string in the user.dat
will be 0DBC3F5626. Ok, 0D = E
so check it out,
0 = 00000000
00000100 <-- Found decryption scheme
-------- <-- XOR
00000100 <-- 4!
D = 00001101
00001000 <-- Found decryption scheme
-------- <-- XOR
00000101 <-- 5!
So combine the 2 answers and you'll get 45! 45 HEX is ASCII 'E'!! Just like
in 'ERIKA'! So we now know how to decrypt the 1 letter/number of a
password! BUT, as you see and as you know I'll repeat this all shortly
The first password was DOPE with a first character 'D'
the 'D' was encrypted as '0C'. We knew that those two characters
represented the Hex code of the ASCII code 'D', 44! So that means
that 0C has to become 44, we did that with XOR and to make 0
a 4 you had to use 00000100, and to make C a 4 you needed to use
00001000. So that means That if you don't know the decrypted password,
but you found '0D'as first two characters of the password you need
to use the same two binary numbers, 00000100 and 00001000. So you
did that and 0 came out as 4, which is logical, and D came out as
5, using 00001000.
Encrypted password:
09 AC 35 59 22 2F E6 53 33 C6 0C B4 19 DB
Decrypting...
+-----------+
[09] AC 35 59 22 2F E6 53 33 C6 0C B4 19 DB
0=
00000000
00000100 <--- We found that one earlier
--------XOR
00000100 = 4
9=
00001001
00001000 <--- This one too
--------XOR
00000001 = 1
09 = 41 = A
Password until now: A
~
09 [AC] 35 59 22 2F E6 53 33 C6 0C B4 19 DB
A=
00001010
00001110 <--- You didn't knew this one yet, did you? hehehe
--------XOR
00000100 = 4
C=
00001100
00001110
--------XOR
00000010 = 2
AC = 42 = B
Password until now: AB
~
09 AC [35] 59 22 2F E6 53 33 C6 0C B4 19 DB
3=
00000011
00000111
--------XOR
00000100 = 4 (yes, it is a coincidence. Don't expect 4 to come out always)
5=
00000101
00000110
--------XOR
00000011 = 3
35 = 43 = C
Password until now: ABC
~
09 AC 35 [59] 22 2F E6 53 33 C6 0C B4 19 DB
5=
00000101
00000001
--------XOR
00000100 = 4 (*sighs*)
9=
00001001
00001101
--------XOR
00000100 = 4
59 = 44 = D
Password until now: ABCD
~
09 AC 35 59 [22] 2F E6 53 33 C6 0C B4 19 DB
2=
00000010
00000110
--------
00000100 = 4
2=
00000010
00000111
--------
00000101 = 5
22 = 45 = E
Password until now = ABCDE
~
09 AC 35 59 22 [2F] E6 53 33 C6 0C B4 19 DB
2=
00000010
00000110
--------XOR
00000100 = 4
F=
00001111
00001001
--------XOR
00000110 = 6
2F = 46 = F
Password until now: ABCDEF
~
09 AC 35 59 22 2F [E6] 53 33 C6 0C B4 19 DB
E=
00001110
00001010
--------XOR
00000100 = 4
6=
00000110
00000001
--------XOR
00000111 = 7
E6 = 47 = G
Password until now: ABCDEFG
~
09 AC 35 59 22 2F E6 [53] 33 C6 0C B4 19 DB
5=
00000101
00000001
--------XOR
00000100 = 4
3=
00000011
00001011
--------XOR
00001000 = 8
53 = 48 = H
Password until now: ABCDEFGH
~
09 AC 35 59 22 2F E6 53 [33] C6 0C B4 19 DB
3=
00000011
00000111
--------XOR
00000100 = 4
3=
00000011
00001010
--------XOR
00001001 = 9
33 = 49 = I
Password until now: ABCDEFGHI
~
09 AC 35 59 22 2F E6 53 33 [C6] 0C B4 19 DB
C=
00001100
00001000
--------XOR
00000100 = 4
6=
00000110
00001100
--------XOR
00001010 = A
C6 = 4A = J
Password until now: ABCDEFGHIJ
~
09 AC 35 59 22 2F E6 53 33 C6 [0C] B4 19 DB
0=
00000000
00000100
--------XOR
00000100 = 4
C=
00001100
00000111
--------XOR
00001011 = B
0C = 4B = K
Password until now: ABCDEFGHIJK
~
09 AC 35 59 22 2F E6 53 33 C6 0C [B4] 19 DB
B=
00001011
00001111
--------XOR
00000100 = 4
4=
00000100
00001000
--------XOR
00001100 = C
B4 = 4C = L
Password until now: ABCDEFGHIJKL
~
09 AC 35 59 22 2F E6 53 33 C6 0C B4 [19] DB
1=
00000001
00000101
--------XOR
00000100 = 4
9=
00001001
00000100
--------XOR
00001101 = D
19 = 4D = M
Password until now: ABCDEFGHIJKLM
~
09 AC 35 59 22 2F E6 53 33 C6 0C B4 19 [DB]
D=
00001101
00001001
--------XOR
00000100 = 4
B=
00001011
00000101
--------XOR
00001110 = E
DB = 4E = N
COMPLETE PASSWORD: ABCDEFGHIJKLMN
I did this so you could see 14 encrypted characters, being decrypted.
Also you could see the decryption scheme that I used, which is always
the same as I used. But, for beginners who didn't paid attention or are
too lazy to look it up above here is the entire decryption scheme:
Number. in string 1st char of encrypted password : 2nd
+---------------------------------------------------------------------------+
1 00000100 00001000
2 00001110 00001110
3 00000111 00000110
4 00000001 00001101
5 00000110 00000111
6 00000110 00001001
7 00001010 00000001
8 00000001 00001011
9 00000111 00001010
10 00001000 00001100
11 00000100 00000111
12 00001111 00001000
13 00000101 00000100
14 00001001 00000101
+---------------------------------------------------------------------------+
So...I'll give another example, here I show how to use the scheme printed
above and how to decrypt an unknown password. If you already get it, just
skip this part and read the next part. Here we go;
Encrypted password;
18A1394D
As you can see it's 8 chars long.
Well, let's go!
1= 00000001
00000100 <-- look it up in the scheme above, pos 1,1
--------XOR
00000101 --> 5
8= 00001000
00001000 <-- Scheme positions 1,2
--------XOR
00000000 --> 0
Combine those two solutions and you'll get 50h(ex); ASCII char 'P'
Ok, second couple;
A= 00001010
00001110 <-- Scheme pos. 2,1
--------
00000100 --> 4
1= 00000001
00001110 <-- Scheme pos. 2,2
--------XOR
00001111 --> F
Combine those two solutions and you'll get 4Fh; ASCII char 'O'
Ok, third couple;
3= 00000011
00000111 <-- scheme..etc
--------XOR
00000100 --> 4
9= 00001001
00000110
--------XOR
00001111 --> F
Same as the previous one...4Fh = ASCII char 'O'
Next couple; Fourth one
4= 00000100
00000001
--------XOR
00000101 --> 5
D= 00001101
00001101
--------XOR
00000000 --> 0
And you'll get 50h = 'P' so the password was POOP. Got it?
The above process is quite not necessary and there is a simpler way to crack.
warning :--------------------------------------------------------------------------------
All the Tricks which had shown in this site is just for Teaching and Learning purpose.use all tricks on your personal risk.
--------------------------------------------------------------------------------
Mail me at masteragarwal@yahoo.com to receive latest updates and source cods and be in touch.
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