There are 17-element subsets of that form a geometric sequence. My solution follows.

Since we're dealing with such huge numbers, it may be better to think about things in a more generalized manner. That is, let's see how to find the number of -element subsets of the set that form a geometric sequence. Any such subset will have the form where and are natural numbers and . Notice that we will have a different geometric sequence for different values of . So for any given we will have a -element subset for each natural number given that the largest term does not exceed . Then for any , . That is,

This also means that for any given we have -element subsets of that form a geometric sequence with that particular ratio . Now we just sum those up over all possible . But notice that when exceeds (that is, we no longer are dealing with subsets of ), then the value of will equal zero. Thus we can simply sum each for to infinity.

In other words, let be a function which describes the number of -element subsets of the set that form a geometric sequence. Then

The problem with this is that the question asks for a final answer. Obviously we can't tell a computer to sum to infinity. So there are two ways to do it. We can either find a max value for or we can check each time to see if . When it does we can stop summing.

To find a max for , we want to know when exceeds . Then we have

That is,

Then,

Now we can use a program which can handle huge numbers to quickly find the solution. I used the wonderful, free PARI/GP. (Follow link to download or if using Linux check your distro's package manager for it.) Other programs/languages which handle huge numbers are Maple and Mathematica. Here is the code I used to obtain the solution in PARI/GP:

? f(n,k)=sum(r=2,floor(n^(1/(k-1))),floor(n/r^(k-1))); ? f(33^33,17) %2 = 1973093124882606298882104993483387674675038805