Prime powers method

Send the guest in room ${\displaystyle i}$ to room ${\displaystyle 2^{i}}$, then put the first coach's load in rooms ${\displaystyle 3^{n}}$, the second coach's load in rooms ${\displaystyle 5^{n}}$; in general for coach number ${\displaystyle c}$ we use the rooms ${\displaystyle p_{c}^{n}}$ where ${\displaystyle p_{c}}$ is the ${\displaystyle c}$th odd prime number. This solution leaves certain rooms empty (which may or may not be useful to the hotel); specifically, all numbers that are not prime powers, such as 15 or 847, will no longer be occupied. (So, strictly speaking, this shows that the number of arrivals is less than or equal to the number of vacancies created. It is easier to show, by an independent means, that the number of arrivals is also greater than or equal to the number of vacancies, and thus that they are equal, than to modify the algorithm to an exact fit.) (The algorithm works equally well if one interchanges ${\displaystyle n}$ and ${\displaystyle c}$, but whichever choice is made, it must be applied uniformly throughout.)

Prime factorization method

Each person of a certain seat ${\displaystyle s}$ and coach ${\displaystyle c}$ can be put into room ${\displaystyle 2^{s}3^{c}}$ (presuming c=0 for the people already in the hotel, 1 for the first coach, etc.). Because every number has a unique prime factorization, it is easy to see all people will have a room, while no two people will end up in the same room. For example, the person in room 2592 (${\displaystyle 2^{5}3^{4}}$) was sitting in on the 4th coach, on the 5th seat. Like the prime powers method, this solution leaves certain rooms empty.

This method can also easily be expanded for infinite nights, infinite entrances, etc. ( ${\displaystyle 2^{s}3^{c}5^{n}7^{e}...}$ )

Interleaving method

For each passenger, compare the lengths of ${\displaystyle n}$ and ${\displaystyle c}$ as written in any positional numeral system, such as decimal. (Treat each hotel resident as being in coach #0.) If either number is shorter, add leading zeroes to it until both values have the same number of digits. Interleave the digits to produce a room number: its digits will be [first digit of coach number]-[first digit of seat number]-[second digit of coach number]-[second digit of seat number]-etc. The hotel (coach #0) guest in room number 1729 moves to room 01070209 (i.e., room 1,070,209). The passenger on seat 1234 of coach 789 goes to room 01728394 (i.e., room 1,728,394).

Unlike the prime powers solution, this one fills the hotel completely, and we can reconstruct a guest's original coach and seat by reversing the interleaving process. First add a leading zero if the room has an odd number of digits. Then de-interleave the number into two numbers: the coach number consists of the odd-numbered digits and the seat number is the even-numbered ones. Of course, the original encoding is arbitrary, and the roles of the two numbers can be reversed (seat-odd and coach-even), so long as it is applied consistently.

Triangular number method

Those already in the hotel will be moved to room ${\displaystyle (n^{2}+n)/2}$, or the ${\displaystyle n}$th triangular number. Those in a coach will be in room ${\displaystyle ((c+n-1)^{2}+c+n-1)/2+n}$, or the ${\displaystyle (c+n-1)}$ triangular number plus ${\displaystyle n}$. In this way all the rooms will be filled by one, and only one, guest.

This pairing function can be demonstrated visually by structuring the hotel as a one-room-deep, infinitely tall pyramid. The pyramid's topmost row is a single room: room 1; its second row is rooms 2 and 3; and so on. The column formed by the set of rightmost rooms will correspond to the triangular numbers. Once they are filled (by the hotel's redistributed occupants), the remaining empty rooms form the shape of a pyramid exactly identical to the original shape. Thus, the process can be repeated for each infinite set. Doing this one at a time for each coach would require an infinite number of steps, but by using the prior formulas, a guest can determine what their room "will be" once their coach has been reached in the process, and can simply go there immediately.

Arbitrary enumeration method

Let ${\displaystyle S:=\{(a,b)\mid a,b\in \mathbb {N} \}}$. ${\displaystyle S}$ is countable since ${\displaystyle \mathbb {N} }$ is countable, hence we may enumerate its elements ${\displaystyle s_{1},s_{2},\dots }$. Now if ${\displaystyle s_{n}=(a,b)}$, assign the ${\displaystyle b}$th guest of the ${\displaystyle a}$th coach to the ${\displaystyle n}$th room (consider the guests already in the hotel as guests of the ${\displaystyle 0}$th coach). Thus we have a function assigning each person to a room; furthermore, this assignment does not skip over any rooms.

Infinite layers of nesting

Although a room can be found for any finite number of nested infinities of people, the same is not always true for an infinite number of layers, even if a finite number of elements exists at each layer.