Let $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Find the number of subsets $A$ of $S$ such that $x \in A$ and $2x \in S \implies 2x \in A$. Let $S=\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Find the number of subsets $A\subseteq S$ such that $x\in A$ and $2x\in S$ $\implies 2x\in A$.
My Attempt
I broke the problem into cases. I made pairs $(1,2),(2,4),(3,6),(4,8),(5,10)$.
If considering these pairs only there are
$$\binom{5}{1}+\binom{5}{2}+\binom{5}{3}+\binom{5}{4}+\binom{5}{5}=2^5-1=31$$
Considering solely elements of $\{6,7,8,9,10 \}$ there are similarly $31$ subsets.
Now there may be cases where both occur.
$$2^4\binom{5}{1}+2^3\binom{5}{2}+2^2\binom{5}{3}+2^1\binom{5}{4}+2^0\binom{5}{5}=211$$
Clearly answer after summing all and including empty set is $31+31+211+1=274$.
It is not likely to be correct please give a systematic proceeding of solution.

Edit: On making an important comment by @Desparado
I would change tuples to $(\pmb1,2,4,8),(\pmb2,4,8),(\pmb3,6),(\pmb4,8),(\pmb5,10),(6),(7),(8),(9),(10)$.
Boldface indicates the entry which must be accompanied by the following in that particular tuple in the subset.
Correct answer is 180 given by @lulu and @Desperado in the comments
 A: A subset $A$ of $S$ satisfying your property is exactly the same as a set that satisfies the following properties :

*

*If $A$ contains 1, then $A$ contains $1$,$2$,$4$ and $8$ ;

*If $A$ contains 2, then $A$ contains $2$,$4$ and $8$ ;

*If $A$ contains 3, then $A$ contains $3$, $6$ ;

*If $A$ contains 4, then $A$ contains $4$ and $8$ ;

*If $A$ contains 5, then $A$ contains $5$ and $10$ ;

Five families of numbers appear : $(1,2,4,8)$, $(3,6)$, $(5,10)$, $(7)$ and $(9)$ (two numbers which can or cannot be in $A$ without any consequence on the property).
$A$ is therefore given by exactly 5 independant informations, which are the four smallest numbers that $A$ contains in these families (or if it contains none of the numbers of a family).
We have $5$ choices for the first family ($A$ contains no one / $A$ contains only $8$ / $A$ contains $4$ and $8$ / $A$ contains $2,4,8$ / $A$ contains $1,2,4,8$) ; $3$ choices for the 2nd family ; $3$ choices for the 3rd family and 2 choices for each of the last two families ($A$ contains 7 or $A$ does not contain 7). Which gives a total of
$$5 \times 3 \times 3 \times 2 \times 2 = 180 \text{ possibilities}.$$
