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

Re: Re:Question
Wednesday, 31-Mar-99 14:21:43

    Andy is quite right, and the problem you are asking about is a problem in modular arithmetic.
    You asked about:
    79dfh + (5bfh * X) = Y, Y = 7777h
    and how to find X. In fact you really need to consider
    79dfh + (5bfh * X) = 7777h mod (2^32)
    which is a completely different problem. You no longer consider the division, but the inverse of 5bf mod 2^32

    Cronos


    Cronos


Message thread:

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

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