China Mathematical Competition

We will first prove that $|A\cup B|\le 66$, or equivalently $|A|\le 33$. For this purpose, we only need to prove that, if $A$ is a subset of $\{1,2,\dots ,49\}$ with 34 elements, then there must exist $n\in A$ such that $2n+2\in A$. The proof is as follows.

Divide $\{1,2,\dots ,49\}$ into 33 subsets:

$\{1,4\}$, $\{3,8\}$, $\{5,12\}$, $\dots$, $\{23,48\}$, 12 subsets; $\{2,6\}$, $\{10,22\}$, $\{14,30\}$, $\{18,38\}$, 4 subsets; $\{25\}$, $\{27\}$, $\{29\}$, $\dots$, $\{49\}$, 13 subsets; $\{26\}$, $\{34\}$, $\{42\}$, $\{46\}$, 4 subsets.

By the Pigeonhole Principle we know that there exists at least one subset with 2 elements among them which is also a subset of $A$. That means there exists $n\in A$ such that $2n+2\in A$.

On the other hand, let

$A=\{1,3,5,\dots ,23,2,10,14,18,25,27,29,\dots ,49,26,34,42,46\}$

$B=\{2n+2\mid n\in A\}$.

We find that $A$ and $B$ satisfy the condition and $|A\cup B|=66$.

Answer: B.