Given a set $A \subseteq \{10,11,12,13...98,99\}$ such that $|A|=10$. Prove using Pigeonhole Principle there are 2 disjoint subsets with the same Sum. Given a set $A \subseteq \{10,11,12,...98,99\}$ such that $|A|=10$. Prove using Pigeonhole Principle there are 2 disjoint non-empty subsets of $A$ with the same Sum.
Direction or Hint would be appreciated.

The range of sums of a subset of $|A| \le 10$ is $\le 21 (10+11)$ and $945\ge (90+91+...+99)$, therefore $925$ different sums you can get from a subset of $\{10,11,...,98,99\}$ if $|A|\le10.$ 
In a set $A, |A| = 10$, there are $2^{|A|} - 1$ non-empty subsets = $1023$.
$1023 \gt 945$, therefore considering $945$ "pigeonholes", atleast $2$ disjoint sets will share the same sum. We can force our sets $A,B\, s.t \,\Sigma A = \Sigma B$ to be disjoint easily by - 
$$ A = A \backslash \{A\cap B\} $$
$$ B = B \backslash \{B\cap A\} $$
As required.
