Fermats little theorem on permutations

  • #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
 

Answers and Replies

  • #2
tiny-tim
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… the permutation (1, 2, 3, ..., n, n+1, n+2, n+3, ...2n) to (2, 4, 6, ..., 2n 1, 3, 5, ...2n-1) …

Hi Nicole! :smile:

Hint: what is k/2 (mod 53)? :wink:
 

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