Think "Twice"

Let $a < b < c$ , as they are distinct. It can be seen that $\{s\} = 2^a + 2^b + 2^c$ are binary numbers. For example, the smallest $s = 2^0 + 2^1 + 2^2 = 111_2 = 7_{10}$ . When $c = 2$ , the number $n$ of elements is: $n = \left( \begin{matrix} 2 \\ 2 \end{matrix} \right) = 1$ ; when $n = 3$ , $n = 1 + \left( \begin{matrix} 3 \\ 2 \end{matrix} \right) = 4$ and so on. This implies that:

when $c = 8 \quad \Rightarrow n = 1 + 3 + 6 + 10 + 15 + 21 + 28 = 84$

The $84^{th}$ element $s_{84} = 2^8+2^7+2^6 = 448$ ,

and $85^{th}$ element $s_{85} = 2^9+2^1+2^0 = 515$

Similarly, when $c = 9$ and $b = 5$

$\Rightarrow n = 84 + \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) + \left( \begin{matrix} 2 \\ 1 \end{matrix} \right) + \left( \begin{matrix} 3 \\ 1 \end{matrix} \right) + \left( \begin{matrix} 4 \\ 1 \end{matrix} \right) + \left( \begin{matrix} 5 \\ 1 \end{matrix} \right) = 84 + 1 +2 +3 + 4 + 5 = 99$

The $99^{th}$ element $s_{99} = 2^9+2^5+2^4 = 560$ ,

and $100^{th}$ element $s_{100} = 2^9+2^6+2^0 = \boxed{577}$

Kindly mention in the question: set of all "distinct" integers.

C(9, 3) = 84 and C(10, 3) = 120

Therefore, one of the powers is going to be 9.

Again, C(6,2) = 15. Thus, (84+16)th will have remaining powers as 0 & 6.

Therefore, $2^{0} + 2^{6} + 2^{9} = 1 + 64 + 512 = 577.$