\[x\]
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
\[n =52\]
.