Forgotten two digits of an encrypted disk image's passphrase!

[J@ffa]

Stupidly, I set up an encrypted sparse disk image and used OS X's password assistant thingy to think up a password for it. I'd forgotten it, but I found the piece of paper where I scribbled a reminder. It basically goes like this;

longbunchofwordsgoeshere####anotherfewlongwordshere

I've forgotten the second two numbers of the ## in the middle. Is there anything I can use to, I guess, bruteforce just those two numbers without having to sit there typing all the possible combinations? Thanks!

 

[VoodooDaddy]

Stupidly, I set up an encrypted sparse disk image and used OS X's password assistant thingy to think up a password for it. I'd forgotten it, but I found the piece of paper where I scribbled a reminder. It basically goes like this;

longbunchofwordsgoeshere####anotherfewlongwordshere

I've forgotten the second two numbers of the ## in the middle. Is there anything I can use to, I guess, bruteforce just those two numbers without having to sit there typing all the possible combinations? Thanks!

Isnt there just 100 possible combinations, 00-99? How long would that take to do?

 

[notsofatjames]

Isnt there just 100 possible combinations, 00-99? How long would that take to do?

cant automater do it ? I've never found a use for it yet, but this sounds like the sort of thing its for... cracking passwords.

 

[CRAZYBUBBA]

cant automater do it ? I've never found a use for it yet, but this sounds like the sort of thing its for... cracking passwords.

lol, it sounds easy enough but automator actually can only do things that its been trained to do.. useless for me but I love the cute icon.

 

[jellomizer]

Well there is no easy way. But here is one way you can at lest get it.

Copy and past the code below. Tabbing is very important.
replace longbuchofwordsgoeshere and anotherfewlongwordshere with the two parts of your phrase.

#!/usr/bin/python

FirstPart="longbunchofwordsgoeshere"
SecondPart="anotherfewlongwordshere"
ValidRange = ["(", ")", "*", "+", ",", "-", ".", "/", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", ":", ";", "<", "=", ">", "?", "@", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "[", "\\", "]", "^", "_", "`", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "{", "|", "}", "~"]
File = open("FindPossiblecCmbination.txt","a")
cnt = 0
for i in ValidRange:
	for j in ValidRange:
		File.write(FirstPart+i+j+SecondPart+"\n")
	cnt=cnt+1
	print "%s%%"%(float(cnt)/float(len(ValidRange))*100.0) 

File.close()

save the file in raw text mode as in your home directory as FindKey.py
go into the terminal app Applications->Utilities->Terminal

Type

chmod 755 FindKey.py
./FindKey.py

Use text edit and open up the file called FindPossiblecCmbination.txt located in your home directory Open that up and Cut and Past each combination.

 

[119576]

Isnt there just 100 possible combinations, 00-99? How long would that take to do?

00-99 is a whole lot more than 100 combinations haha!

you have to consider 01, 02, 03, 56, 58, 72, and other numbers where both digits are different.

 

[mac-convert]

00-99 is a whole lot more than 100 combinations haha!

you have to consider 01, 02, 03, 56, 58, 72, and other numbers where both digits are different.

And, uh, how many combo's would you think there would be between zero and 99?