Let's determine how many shuffles it takes to return 52 cards to their original order if we shuffle them as described in the article about Faro Shuffle. Solve this problem in the case where we start the shuffle with the bottom card of the right-hand deck, meaning the card originally in the 26th position ends up at the bottom.
Solution. 1. case. We number the cards as described in the article - proceeding from top to bottom - with the numbers $0,1,2, \ldots, 51$. Initially, the first and last card always remain in place; otherwise, after one shuffle, the $k$-th card moves to the $y_{k}$-th position, where $y_{k} \equiv 2 k(\bmod 51)(k=1,2, \ldots, 50)$. If we shuffle $n$ times in succession, then the card originally in the $k$-th position moves to the $2^{n} k$-th position (mod 51). Each card will be back in its original position if for every $k=1,2, \ldots, 50, 2^{n} k \equiv k(\bmod 51)$ holds, that is, $2^{n} \equiv 1(\bmod 51)$. Writing down the powers of 2 modulo 51 in sequence:

$$
\begin{array}{ll}
2^{1} \equiv 2 \quad(\bmod 51), & 2^{5} \equiv 32 \quad(\bmod 51) \\
2^{2} \equiv 4 \quad(\bmod 51), & 2^{6} \equiv 64 \equiv 13 \quad(\bmod 51) \\
2^{3} \equiv 8 \quad(\bmod 51), & 2^{7} \equiv 2 \cdot 13=26 \quad(\bmod 51) \\
2^{4} \equiv 16 \quad(\bmod 51), & 2^{8} \equiv 2 \cdot 26=52 \equiv 1 \quad(\bmod 51)
\end{array}
$$

According to this, $n=8$ is the smallest number for which this is true, that is, the original order is restored for the first time after the eighth shuffle.

2. case. Now we can most easily track the movement of the cards if we number the cards from 1 to 52. Then the $k$-th card moves to the $2 k$-th position $(\bmod 53)$, and after the $n$-th shuffle, it moves to the $2^{n} k$-th position $(\bmod 53)$. This time, we are looking for the smallest $n$ for which $2^{n} \equiv 1(\bmod 53)$. Writing down the powers of 2 in sequence $(\bmod 53)$ would take a long time until the 1 first appears. We can speed up the calculation if we narrow down the set of $n$ numbers that could be considered. Since 53 is a prime number, by Fermat's little theorem, $2^{52} \equiv 1(\bmod 53)$, so the powers of 2 modulo 53 are periodic, and the period length is 52. Our task is to find the smallest period (let's denote it by $p$), because on the one hand, $2^{p} \equiv 2^{p-p}=1(\bmod 53)$, so $n \leq p$, and on the other hand, because of $2^{n} \equiv 1(\bmod 53)$, the powers of 2 are periodic modulo 53, so $p \leq n$; thus indeed $p=n$. The smallest period is a divisor of every period, including 52, so it can only be $1,2,4,13,26$ or 52. Consider the following powers of 2:

$$
\begin{aligned}
2^{4} & \equiv 16 \quad(\bmod 53) \\
2^{8} & \equiv 16^{2}=256 \equiv-9 \quad(\bmod 53) \\
2^{16} & \equiv(-9)^{2}=81 \equiv 28 \quad(\bmod 53)
\end{aligned}
$$

therefore

$$
2^{26}=2^{16} \cdot 2^{8} \cdot 2^{2} \equiv 28 \cdot(-9) \cdot 4 \equiv 6 \cdot(-9)=-54 \equiv-1 \not \equiv 1 \quad(\bmod 53)
$$

This means that 26 is not a period, so 13 is not either, and 1, 2, 4 are clearly not. Therefore, the shortest period is 52, that is, the original order is restored for the first time after the fifty-second shuffle.