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Exponentiation of mods

  1. Mar 2, 2015 #1
    Could somebody at least tell me about mods in exponents?
     
    Last edited: Mar 2, 2015
  2. jcsd
  3. Mar 2, 2015 #2

    Mentallic

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    Tetration already has a formalized expression.

    [tex]^n3=3^{3^{...^3}}[/tex] with n 3's in the stack.

    Or you can even use Knuth's up arrow notation. Exponentiation is denoted by 1 up arrow, and tetration by 2, so

    [tex]^n3 = 3\uparrow\uparrow n[/tex]

    So what's your question?
     
  4. Mar 2, 2015 #3
    Thanks for the quick reply, but I wasn't finished typing. I accidentally hit enter. Sorry.
     
  5. Mar 2, 2015 #4

    Mentallic

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    I could only guess what you're asking for, but if I were to guess, say you want to prove that [itex]^n3 \mod 10 = 7, n>1[/itex] we would do an inductive proof. I'll make it short and you can fill in the details, but essentially you start with the base case [itex]3^3 = 27[/itex] hence [itex]27 \mod 10 = 7[/itex] and we can then show by calculation that [itex]3^7 \mod 10 = 7[/itex] hence it must be true for all n.
     
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