The set $\{1,2,3, \ldots, 3000\}$ contains a subset of 2000 elements in which no element is double the other?
Let's construct a subset of $\{1,2,3, \ldots, 3000\}$ in which no element is double another. We start by including all the odd numbers, 1, 3, 5, ..., 2999. Thus, we already have 1500 numbers, and none of them is double any other. Now we can add the numbers that are quadruple some odd number, that is, add

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-077.jpg?height=109&width=660&top_left_y=1596&top_left_x=732)

This list has 749 numbers, and none of them is double another; moreover, none of them is double an odd number.

Thus, our set already has $1500+749=2249$ elements. It suffices to take any subset with 2000 elements to obtain a subset of $\{1,2,3, \ldots, 3000\}$ in which no element is double another.