Let $a$ be any positive integer. Consider set $$S=\{ a^i:i\in \mathbb N \cup \{0\} \}.$$ Let $S_1$ and $S_2$ be any two different finite subsets of $S$. Then show that sum of elements in $S_1$ and $S_2$ are different.
Attempt
The case $a=1$, which is very trivial because in this case $S=\{1\}$ and there's nothing to prove.
What can be done for $a>1$?
On
I'm assuming $S_1, S_2$ are both finite. Let $n_1, n_2$ be the appropriate sums from $S_1, S_2$, respectively, and choose the smallest $i$ such that $a^i$ is in exactly one of $S_1$ and $S_2$. Such an $i$ must exist because $S_1 \neq S_2$.
Note that for all indices $k \gt i, a^k \equiv 0 \pmod{a^i+1}.$ Also, by the choice of $i, n_1- m \neq n_2 - m \pmod{a^{i+1}}$, where we define $m = \sum \{a^k \mid k \lt i, k \in S_1 \} = \sum \{a^k \mid k \lt i, k \in S_2 \}$. Thus $n_1 \neq n_2 \pmod{a^{i+1}}$.
Note, by the way, that you're writing distinct sequences of $0$s and $1$s and reading them base $a$.
I will assume that $a>1$ and that $S_1$ and $S_2$ are finite.
Suppose that the largest element $N$ of $S_1$ is greater than any element of $S_2$. Then$$\sum_{n\in S_2}n<\sum_{k=0}^{N-1}a^k=\frac{a^N-1}{a-1}\leqslant a^N-1<a^N<\sum_{n\in S_1}n.$$
So, if the largest element of one of the sets is larger than the largest element of the other set, we are done.
Suupose now that the largest element $N$ of $S_1$ is also the largest element of $S_2$. Then all you have to do is to prove that the sum of all elements of $S_1\setminus\{N\}$ is not the sum of all elements of $S_2\setminus\{N\}$. Now you can start all over again with the largest element of these two sets. Since they are finite, the process must stop at some point.