Number of subsets of $X = $ { $ 1,11,21,31,41,..., 551 $ } such that no two elements sum to $552.$ Question :

Find the number of subsets (say) $A$ of $X = $ { $1,11,21,31,41,..., 551  $ } such that no two elements in the subset sum to $552$. 

My attempt :
I divided the original set $X$ into two parts (or subsets, say $P,Q$) each containing $28$ elements. ( i.e containing first half of the element and other part containing other half of the set.)
Now,Each of the sets $P,Q$ would have $\displaystyle 2^{28}$ total subsets. Thus in all, $2^{29}$.
Now, for each element in $P$ there would be an element corresponding to the element in $Q$. 
Thus it is a bijective function $f:P \rightarrow Q$. 
Now we are left with $27$ elements in $P$ that do not correspond to a element in $Q.$ 
So there would be $\displaystyle 2 \times {28 \choose 1} \times \sum_{k=0}^{27} {27 \choose k}$ subsets.
And then considering that set $A$ can have one element at least we need to add $56$ to the total number of subsets to get the set $A.$

Total number of subets $A$ of $X$ will be $$\displaystyle \ 2^{29} + 2 \times {28 \choose 1} \times \sum_{k=0}^{27} {27 \choose k} + 56 = 8,053,063,736.$$

Please verify it.
 A: You are correct that $\sum\limits_{k=0}^{27}\binom{27}{k}=2^{27}$.  Now... as for your original question, I like your idea to partition your set into the "big" numbers and the "little" numbers, noting that for each little number there is one big number such that they add to $552$, for instance $1+551$ or $11+541$ or $21+531$ etc...
Perhaps easier to visualize for the final answer however is to partition it into the $28$ sets $\{1,551\},\{11,541\},\{21,531\},\dots$
Now... in constructing a valid subset, for each of the $28$ subsets in the partition choose whether you wish to include the smaller number, the larger number, or neither.  For each of these subsets in the partition we have three choices and for each unique sequence of choices we get a unique final resulting subset fitting our requirements.
We have then applying multiplication principle a.k.a. the rule of product, that there are a final total of:
$$3^{28} = 22,\!876,\!792,\!454,\!961$$
Note: this does include the empty set and the single-element sets in the overall count.  Depending on preference, these might be removed from the count if so desired.
A: In each of the $28$ pairs $(i, 552-i)$, $i = 1, 11, 21, \ldots, 271$, you can have either $0$ or $1$ member (which can be either member of the pair).  So there are $3^{28}$ possibilities.
