CrackMe® Practices for Newbies
Project 9: CrackMe 2 by Cronos

Re: Re: Joseph's Thread
Thursday, 01-Apr-99 13:29:54

    "7777h mod (2^32) = 7777h and if so then the statement 79dfh +(5bfh * X) = 7777h mod (2^32) is meaningless. The same is true for the other statement 5bfh mod (2^32). "

    Hmm. I think you still misunderstand me. Perhaps I have assumed too much here. OK. The equation is:

    (79df + 5bf * X = 7777h ) mod 2^16

    (Sorry, my fault for saying 2^32 not 2^16 in a previous post I think, since of course we are using 16-bit numbers and not 32 bit numbers.)

    5bf * X = 7777-79df mod 2^16

    giving

    5bf * X = fd98 mod 2 ^16

    now consider multiplying by k, to be found later, (and non-zero),

    (5bf * k) * X = fd98*k mod 2^16

    basically what i mean by looking for the inverse of 5bf mod 2^16 is to find a k such that

    5bf * k = 1 mod 2^16, which it is possible to do since 5bf and 2^16 are relatively prime.

    The inverse is clearly EA3F and if you multiply it by 5bf and take the remainder after division by 2^16 you will find it is 1.

    This is your value for k, and should now enable you to proceed.

    Cronos



    Cronos


Message thread:

Joseph's Thread (Question to Cronos) (30-Mar-99 23:45:59)

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