\[x\]

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

\[n =52\]

.