3. How many of the integers from $2^{10}$ to $2^{18}$ inclusive are divisible by $2^{9}$ ?
Solution
511

We observe that $2^{10}=2 \times 2^{9}$, and $2^{18}=512 \times 2^{9}$. Therefore, on multiplying $2^{9}$ by any integer between 2 and 512 inclusive, we obtain an integer between $2^{10}$ and $2^{18}$ inclusive. There are $512-1=511$ such integers.