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Fermats little theorem on permutations

  1. Jan 27, 2009 #1
    I'm looking at a card shuffle. And in shuffle would be the permutation (1, 2, 3, ..., n, n+1, n+2, n+3, ...2n) to (2, 4, 6, ..., 2n 1, 3, 5, ...2n-1) I know that it would take 52 perfect shuffles to get the deck of cards back in the original order. I think that I'm supposed to show this using Fermats Little Theorem.
    Fermats little theorem says a p-1 ≡ 1(mod p) or a p ≡ a(mod p). Now I'm pretty sure that using the permutation from above plugging in for Fermat'l Little Theorem looks like: 52 ≡ 2(mod 2n+1)
    If that is right I'm not sure how to show that I came up with that 52.
    Can somebody please help?
    Thanks you so much!! Nicole
     
  2. jcsd
  3. Jan 30, 2009 #2

    tiny-tim

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    Hi Nicole! :smile:

    Hint: what is k/2 (mod 53)? :wink:
     
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