## How Many Perfect Faro Shuffles Are Required to Return a Pack of Cards to Original Order

In the perfect faro shuffle a pack of 42 playing cards is cut into equal halves of 26 cards each and the halves flicked into each other so that the half consisting of the first 26 cards now occupy positions 2, 4,..., 52 and the half consisting of the last 26 cards now occupies positions 1, 3, 5,..., 51. We can say that the card starting at position
$x$
moves to position
$2x \; (mod \; 53)$
. After
$n$
such shuffles the card will be in position
$2^nx \; (mod \; 53)$
.
After how many perfect shuffles will the cards return to their original arrangement?
We require the solution to
$2^n \equiv 1 \; (mod \; 53)$
.
From Fermat's Little Theorem,
$n=52$
is one possible solution, but is it the smallest? Any smaller solution must be a divisor of 52, so only 2, 4, 13 or 26 are possible.
$2^2 =4$

$2^4 =16$

$2^{13}=8192 \equiv 30 \; (mod \; 53)$

$2^{26} =(2^{13})^2 \equiv 30^2 \; (mod \; 53) =900 \equiv 52 \; (mod \; 53)$

Hence
$n =52$
. 