For all positive integers f, f◎ equals the distinct pairs of

Great question - and my strategy for something like this is to first look at what they're asking:

The greatest number...

Which means that I want to maximize the number of factors, meaning that I also want to minimize the value of each factor (so that I can use it more often):

For example, if I use 99 (9 * 11), that 11 takes up too much space...I can't use it very often. Whereas if I use smaller factors (2 and 3), I can use them more often. 2 can become 4 and then 8 before it takes up as much space as does 11.

So if my goal is to find a number less than 100 that has as many small factors as possible, I'm looking at 96, because:

99 = 3*3*11 (and 11 is too big a prime factor)
98 = 2*7*7 (and 7 is too big a prime factor)
97 = not divisible by 2 or 3, so you're not using your smallest available factors
96 = 2*2*2*2*2*3 (or 32*3) ---> this is perfect because you're getting maximum value out of your factors. The only other that would work is 64 (drop the 3 and add another 2)

So for 96, break out the factors into pairs:

1, 96
2, 48
4, 24
8, 12 (these are easy to do - just double one number and halve the other for a fresh pair)
16, 6
32, 3

And that's as far as you can go. My fear is that we may not have enough unique factors (there's a lot of repetitiveness with the 2s), so I'll check that by trying the next smallest prime factor, 5. The biggest I can go with that is 90, or 2*3*3*5, and that gets us:

1, 90
2, 45
3, 30
5, 18
6, 15
9, 10

We're still at 6 factors, and it's going to get harder and harder to incorporate larger prime factors and have room for multiple factors in between (we've seen that 7 and 11 limit us a lot), so the answer must be 6.