Can you find the lighter and heavier marbles? You have $10$ bags with marbles. Each bag has $2000$ marbles. Standard marbles weight $10$g. However, one bag has marbles that all weight $9$g and one bag has marbles that all weight $11$g. If you can only use a scale once (one that displays the numerical weight), how can you determine which bag has the $9$g marbles and which bag has $11$g marbles?
I've been mulling this one over for a bit and discovered plenty of wrong ways to do this, but not the right way yet. Anyone have any ideas on how to tackle this problem?
 A: Mark the bags $1$ to $10$. Pick $2^1$ balls from the first bag, $2^2$ balls from the second bag and in general pick $2^k$ balls from the bag marked $k$. If all bags had balls of equal weight, the sum would be $20460$g. If bag $i$ has $9$g balls and bag $j$ has $11$g balls, the sum of weight of all balls would be
$$20460-2^i + 2^j  = x$$
Any integer of the form $2^j-2^i$ can be expressed only in a unique way, which would identify the bags for us.

EDIT:
To be a bit more clear, if $x>20460$, we would then end up with $2^j-2^i = x-20460$, i.e., $2^i(2^{j-i}-1)$. Note that $2^{j-i}-1$ is odd. Hence, $i$ is the highest power of $2$ in $x-20460$. Now this also fixes $j$. Similarly, for $x<20460$, $i$ and $j$ interchange roles.

FURTHER EDIT:
We could also take a very higher power, i.e., $f(k) = k^m$, though $m=2$ will not work since we have $\color{red}{8^2-7^2=4^2-1^2}$.
A: Line up the bags, and take a set number of marbles from each bag and weight those, using the difference from the weight of that many standard marbles to uniquely identify the two outlier bags.
To do this you need to find a set of ten numbers such that $x-y=\Delta w$ has only one solution for any $\Delta w$.
Can you find ten such numbers?
