There are 5 bags with 100 coins in each bag. A coin can weigh 9, 10 or 11 grams. Each bag contains coins of equal weight, but we do not know what type of coins a bag contains. You have a digital scale (that tells the exact weight). How many times do you needd to use the scale to determine which type of coin each bag contains ?
Since there are 3 possible weights, upon centering we can work with $-1,0$ and $1$. It is possible to proceed in only one weighing by selecting $\lambda_1, \ldots, \lambda_5\in \{1,\ldots, 100\}$ such that the map
$\{-1,0,1\}\to \{-\sum_{i=1}^5\lambda_i,\ldots, \sum_{i=1}^5\lambda_i\}$
$(w_1,\ldots, w_5) \mapsto \sum_{i=1}^5 \lambda_i w_i$
is injective.
This can be accomplished by letting $\lambda_i = 3^{i-1}$. I don't understand the rationale behind this choice. Can someone provide some intuition ? Is there a simple mathematical reason as to why choosing powers of $3$ makes the map injective ?
The reason why choosing powers of 3 makes the map bijective is that you have 3 distinct possible weights.
If you had 10 possible distinct digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, then you'd use powers of 10 to have a bijection between lists of digits and numbers. If you had 2 possible distinct bits 0 and 1, then you'd use powers of 2 to write numbers in binary.
Here you have three possible weights; originally 9, 10 and 11. You already saw that it was easy to "recentre" them to -1, 0 and 1; you might as easily recentre them to 0, 1 and 2 if it makes the analogy with a number-writing system more obvious.