2 Pans Puzzle

At first, I had a bit of trouble figuring out how to approach the problem. I didn't really know where to start. Whenever that happens, I write down what I know. So I began with \( a+b+c+d=40, a<b<c<d \). And I thought about it some more and then I realized 2 things: the first being that the smallest weight must be 1 (\(a=1\)) and that you can "add" negatives by putting the weights on the other pan. So then each weight has 3 "states". It can be on the left pan, off, or on the right pan. So then I wrote out some basic expressions: \( b+c+d=39 \Rightarrow b+c+d-a = 38 \). Then the rest of the weights followed from there: \(b=3, c = 9, d = 27\).

 After solving the weights, I had a conversation with Ross about encoding ternary numbers (\(1-40\)) as some combination of the weights. After messing around with "2's complement" (a way to represent negative ternary numbers inspired by 1's complement in binary) and some quick googling, I stumbled onto balanced ternary (https://en.wikipedia.org/wiki/Balanced_ternary) which is ternary but instead of \(2\), there's \(-1\) or sometimes denoted \(T\). After this discovery, Ross said the representation was possible by adding the ternary number to \(1111_3\) then the following encoding emerges \[ 0: \text{on the "negative" pan}\\ 1: \text{not used} \\ 2: \text{on the "positive" pan}\] or this can be converted to balanced ternary by subtracting 1 from each digit then we have:\[ -1(T): \text{on the "negative" pan}\\ 0: \text{not used} \\ 1: \text{on the "positive" pan}\]

This encoding provides a more intuitive interaction to the ternary expressions since, while there are 3 states the weights can take, ternary numbers with 2s can only be represented by adding the next weight up and subtracting the next power down (i.e. \(6= 9-3\)). With this in mind, when we add it to \(1111_3\) it'll carry over a 1 to the next place value leaving a 0/-1/T which represents subtracting the closest power down and adding the next power up. 

After Jimena's comment about finding an algorithm to solve for the locker problem, I would eventually like to find an algorithm which, given a decimal number, can output the combination of weights to properly weigh it. I will leave that for another day.

 The one-pan, 5 weight problem was much easier for me. After going through a couple of the same steps as above, I quickly noticed the relationship between 31(\(2^5-1\)), 5 weights(bits), and only having one pan means no "negative" numbers. As such, the weights are 1,2,4,8, and 16.

 What struck me in working through these problems was how connected they are to fundamental ideas in mathematics: number bases, representation systems, and algorithms. In the future, I’d like to come back to this and write an algorithm that, given any number, can output the combination of weights to balance it.

For teaching, I think this puzzle has a lot of potential. Students could start by physically experimenting with weights on a scale to build intuition, then move into noticing patterns and connections with binary or ternary systems. It could also be extended into coding activities for computer science classes (i.e. writing a program that translates a decimal number into a set of weight placements.) One of the most powerful lessons this puzzle offers is a concrete way to understand different number bases. In the one-pan version, the weights 1, 2, 4, 8, and 16 mirror powers of 2, which naturally connects to binary. In the two-pan version, the weights 1, 3, 9, and 27 link to powers of 3, which connects to ternary and even balanced ternary. What’s valuable here is that students don’t just hear about bases in the abstract—they see and feel how numbers can be built up systematically from powers, whether through addition alone (binary) or through addition and subtraction (balanced ternary). This makes the idea of number bases less about memorizing place-value systems and more about recognizing structure in problem-solving.

Below are some scratch notes from my process.