Let $A \subset \{1,2,...,99\}$, prove or disprove the following:
a. For $|A| = 27$
b. For $|A| = 26$
There are $2$ different numbers in $A$ that their sum or their difference can be divided with $50$.
I am sure the question points me to use Pigeonhole Principle, but I'm not quite sure which are the pigeons and which are the holes because the question deals with or so proving that a sum (or a difference) can be divided with $50$ closes the proof without handling the other case.
I've started with saying that there are $(99+98)-(1+2)+1=195$ different sums,
and $(99-1)-(1-99)+1=197$ differences (including negatives) in $\{1,2,...,99\}$.
Also, when dividing with $50$ you can get maximum 50 remainders (let $r \in \{0,1,...,49\}$ mark the reminders)
Can anyone hint me with how to proceed from here?