TITLE: Hilbert's Hotel Paradox: Guests moving to new room every day?
QUESTION [7 upvotes]: Suppose there are infinitely many coaches with infinitely many members in each coach. They stay at the hotel for infinitely many days. I know that guests can be accommodated using various methods like the prime powers method, but there's a slight variation in the question which is that the guests have to change their room every day such that one guest can't occupy the same room again (i.e., they have to occupy unique rooms every day). How can we achieve that? 
I tried solving the problem using the following method:

I allotted rooms using the prime powers method.
The next day, guests move from their current room $x$ to the new room $x+c$.

I'm struggling after this step. Can someone please help me out?

REPLY [3 votes]: There's another option.  With the Prime Powers method, you're relying on:

No prime, to any power, is equal to any other prime to any power.

Which makes if you think about it from a prime factorization perspective.  Each number has exactly one prime factorization.  So if you generate a number by taking a prime number X to the power of Y - it's factorization is X, Y times.  That'll obviously be a different prime factorization than a prime number A to the Bth power.
But once you think about it in terms of prime factorization, you'll realize you've got a lot of open space.  After all... the 6th room will never be filled.
Why?  Because 6's prime factorization involves two different primes - so you can't get to it by taking any prime X to the power Y.
So - you've got an infinite number of coaches with an infinite number of people... and they're arriving every day to stay forever?
On the first day, you're set.  Each coach gets a different prime number, and each person from that coach gets a different power of that prime.
The second day?  Each coach gets a different prime and the next prime higher than it, multiplied together.  So the first coach gets 6 - and each of its infinite occupants gets a different power of 6.  The next coach gets 15 (3x5).  The third coach gets 35 (5x7).
The third day?  Each coach gets a different prime and the second prime after it.  (So the coaches that day would be: 10, 21, 55, ...) Fourth day?  Each coach gets a different prime and the third prime after it.  (So the coaches would be: 14, 33, 65, ...)
You've got an infinite number of days covered.
In fact... we covered those infinite days simply by using only two different primes.
(We don't even have to stop there.  Imagine we were asked to solve this problem... across an infinite number of universes.  Universe #1 uses two primes.  Universe #2 uses three primes.  Universe #4 uses four primes.  Etc.)