last question

$2^9=512$

$3^6=729$

Those are the largest positive powers of 2 and 3 under 1000.

$2^{-9}=0.001953.....$

$3^{-6}=0.0013717.....$

These are the smallest negative powers greater than  $\frac{1}{1000}$

Therefore, we have $9-(-9)+1=19$ different powers of 2 between $\frac{1}{1000}$ and 1000.

We also have $6-(-6)+1=13$ different powers of 3 between $\frac{1}{1000}$ and 1000.

This gives us $19+13=32$ total integers...

32 is the answer I got but someone is going to have to check over my work :b

The problem is symmetric in x vs. 1/x so we can find the number of integers between 0 and 1000 that are either a power of two or 3 and just double the answer.

Note that any non-zero power of 2 is even and any non-zero power of 3 is odd so these two sets are disjoint.

There are 10 powers of 2, 0-9, and 6 non-zero powers of 3 that are less than 100.

We use non-zero power for 3 since we only want to count 1 = 2⁰ = 3⁰ once

That gets us 16 from 0 to 1000.  Doubling this we get 32 but we don't want to count $1 = \dfrac{1}{1}$

twice so we subtract 1 from this getting 31 as you found.

That is an interesting question and good answers from both guest and Rom.

Thanks :)