Originally Posted by

**jac1999** The problem is when infinitely many coaches turn up with infinitely many people on them. The solution, as I'm led to believe, is this:

1) Move all the current guests from their room m to 2m, freeing up all the odd numbered rooms.

2) Label each coach q = 1, 2, 3, 4, ...

3) Label each seat on each coach n = 1, 2, 3, 4, ...

4) Place people from the 1st coach into rooms numbered 3^n, 2nd coach into 5^n, 3rd coach into 7^n and qth coach into the (p+1)^n where p = (q + 1)th prime number.

Thus everyone is accomodated. My problem is that the hotel is now not full. In fact, every odd numbered room which has 2 or more distinct factors is empty.