Help: Sets

Given that $B$ is a subset of $A$ { $1, 2, 3, \ldots , 2021$ }, what is the maximum number of elements of $B$ such that no element is twice the other?

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

By just taking all the odd numbers and all the even numbers which are not multiples of four, I got 1516 which looks quite good, do you know the answer because I can’t prove mine

Jason Gomez - 2 months, 3 weeks ago

@Jason Gomez If we take all even numbers which are not multiples of four, { $2, 6, 10, ..., 2018$ }, together with all the odd numbers, then the even numbers which are not multiples of four would be double of the odd numbers { $1, 3, 5, ..., 1009$ }.

Here's my thought about this:

Taking all odd numbers { $1, 3, 5, ..., 2021$ } together with odd numbered multiples of $2^{2}, 2^{4}, 2^{6}, ..., 2^{2n}$ .

$B=$ { $1, 3, 5, ..., 2021$ } $∪$ { $2^{2}\cdot1, 2^{2}\cdot3, 2^{2}\cdot5, ..., 2^{2}\cdot505$ } $∪$ { $2^{4}\cdot1, 2^{4}\cdot3, 2^{4}\cdot5, ..., 2^{4}\cdot125$ } $∪$ { $2^{6}\cdot1, 2^{6}\cdot3, 2^{6}\cdot5, ..., 2^{6}\cdot31$ } $∪$ { $2^{8}\cdot1, 2^{8}\cdot3, 2^{8}\cdot5, 2^{8}\cdot7$ } $∪$ { $2^{10}\cdot1$ }

$∣B∣=1011+253+63+16+4+1=1348$

Ryan Merino - 2 months, 3 weeks ago

That’s seems right, it might be the answer, if all odd numbers are supposed to be included(need to just somehow show this now)

Jason Gomez - 2 months, 3 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$