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Math Help - Quick question...

  1. #1
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    Quick question...

    Lets say you have a number like  2^{1000} . Then the last two digits of this number is the remainder of  \frac{2^{1000}}{100} in  \mathbb{Z}/100 \mathbb{Z} . And  \mathbb{Z}/100 \mathbb{Z} contains  100 residue classes.

    Is it okay to generalize, and say that the last  n digits of a number  k^{p} where  k > 0 and  n \leq p is the remainder of  \frac{k^{p}}{n} in  \mathbb{Z}/n \mathbb{Z} ? And the last  n digits are in one of the  n residue classes?
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  2. #2
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    Hello,
    Quote Originally Posted by particlejohn View Post
    Lets say you have a number like  2^{1000} . Then the last two digits of this number is the remainder of  \frac{2^{1000}}{100} in  \mathbb{Z}/100 \mathbb{Z} . And  \mathbb{Z}/100 \mathbb{Z} contains  100 residue classes.
    Yes

    Is it okay to generalize, and say that the last  n digits of a number  k^{p} where  k > 0 and  n \leq p is the remainder of  \frac{k^{p}}{n} in  \mathbb{Z}/n \mathbb{Z} ? And the last  n digits are in one of the  n residue classes?
    Huh ?
    The last n digits are the remainder in the Euclidian division of k^p by {10^n}

    What I don't understand is why you bother yourself talking about residue classes... ?
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  3. #3
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    thats what I meant divide by  10^{n} . Because the remainder is a residue class.
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