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Math Help - solving for x in 78^x =x (mod 10^12)

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    solving for x in 78^x =x (mod 10^12)

    Hi Guys,

    Any idea how to solve the equation 78^x = x (mod 10^12) ? Assume that x is a 12 digits number
    I've looked over every congruence and can't seem to find anything useful.
    Last edited by ShadowKnight8702; March 13th 2013 at 04:58 PM.
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    Super Member ILikeSerena's Avatar
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    Re: solving for x in 78^x =x (mod 10^12)

    Quote Originally Posted by ShadowKnight8702 View Post
    Hi Guys,

    Any idea how to solve the equation 78^x = x (mod 10^12) ?
    I've looked over every congruence and can't seem to find anything useful.
    Hi ShadowKnight8702!

    Let's see... you're looking for a number x such that the last 12 digits of 78^x is the same as the last 12 digits of x.
    Perhaps we should start with just the last digit.
    Can you solve 78^x = x (mod 10)?
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    Re: solving for x in 78^x =x (mod 10^12)

    Im not quite sure how that is solved. But one piece of info i forgot to mention is that x is a 12 digits number.
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    Super Member ILikeSerena's Avatar
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    Re: solving for x in 78^x =x (mod 10^12)

    Quote Originally Posted by ShadowKnight8702 View Post
    Im not quite sure how that is solved. But one piece of info i forgot to mention is that x is a 12 digits number.
    The way to solve it, is to try each modulo class of x.
    That's quite doable since there are only 10.
    For instance for x \equiv 3 \pmod{10}, you get:

    78^3 \equiv 3 \pmod{10}

    (78 \pmod{10})^3 \equiv 3 \pmod{10}

    8^3 \equiv 3 \pmod{10}

    2 \equiv 3 \pmod{10}

    This is a contradiction, so any number x that ends in 3 will not satisfy the equation.
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