Answer:
31
Step-by-step explanation:
Solution:
First we count the powers of $2$. We can compute $2^9=512$ and $2^{10}=1024$, so the largest power of $2$ which is smaller than $1000$ is $2^9$. Remembering that $2^{-n}$ is defined as $\dfrac{1}{2^n}$, we can see that the smallest power of $2$ which is larger than $\dfrac{1}{1000}$ is $2^{-9}$. Thus, we have the following list of powers of $2$:
\[2^{-9}, 2^{-8}, 2^{-7}, 2^{-6}, 2^{-5}, 2^{-4}, 2^{-3}, 2^{-2}, 2^{-1}, 2^0, \]\[2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8, 2^9.\]
There are $19$ numbers in this list ($9$ with negative exponents, $9$ with positive exponents, and one more -- $2^0$).
We can count powers of $3$ in a similar way. The largest power of $3$ smaller than $1000$ is $3^6 = 729$. The smallest power of $3$ larger than $\dfrac{1}{1000}$ is therefore $3^{-6}$, or $\dfrac{1}{729}$. So we have these powers of $3$:$$3^{-6}, 3^{-5}, 3^{-4}, 3^{-3}, 3^{-2}, 3^{-1}, 3^0, 3^1, 3^2, 3^3, 3^4, 3^5, 3^6.$$We don't want to count $3^0$ again, though, because $3^0$ is $1$, which we already counted as $2^0$! There are $12$ other numbers in our list ($6$ with negative exponents and $6$ with positive exponents).
We don't have to worry about any other duplication between the lists, since $\allowbreak 2^1,2^2,2^3,\allowbreak \ldots,2^8,2^9$ are all even and $\allowbreak 3^1,3^2,3^3,\allowbreak 3^4,3^5,3^6$ are all odd. So, we have $19+12 = \boxed{31}$ different numbers in our two lists.
