Numbers in Equilibrium: The Art of Reframing Problems

Introduction

Mathematics often reveals unexpected connections between seemingly unrelated ideas. A system devised for counting can illuminate the design of an ancient scale; a weighing puzzle can mirror the structure of a numeral system. What begins as a practical question—How few standard weights do we need to weigh anything from one to one hundred pounds?—turns out to be a lesson in the efficiency of number representation itself.

To see why, imagine early merchants measuring grain or metal with a simple balance: a lever resting on a fulcrum with two pans, one for the object and one for the standard weights. Each transaction required physical reasoning—adding or moving small masses until equilibrium was achieved. Yet this mechanical process hides an abstract mathematical idea: how to represent every quantity within a range by combining a fixed, minimal set of basic units.

This is the same problem that ancient mathematicians faced when developing numerical systems. The Roman numerals, for instance, required new symbols for each order of magnitude, while our modern positional system compresses those possibilities into a handful of digits through place value. Likewise, a clever choice of weights can represent every integer from $1$ to $100$ just as efficiently as the digits $0–9$ can represent any number in base $10$.

The following problem embodies this elegant interplay between number systems and physical balance. It invites you to think not only in terms of pounds and scales but also in terms of bases and positional symmetry.

A Weighing Puzzle

Illustrated below is a simple scale for weighing objects. The scale consists of a with weighing pans at each end of the lever, equidistant from the fulcrum. Suppose the objects to be weighed may range in weight from $1$ pound to $100$ pounds at one-pound intervals: $1, 2, 3, ..., 98, 99, 100$. After placing one such object on either of the two weighing pans, one or more precalibrated weights are then placed in either or both pans until balance is achieved, thus determining the weight of the object. If the relative positions of the lever, fulcrum, and pans may not be changed, and if one may not add to the initial set of precalibrated weights, what is the minimum number of such precalibrated weights that would be sufficient to bring into balance any of these one hundred objects?$^{1}$

Answer

On a balance scale, where you may put each precalibrated weight (simply referred to as "weights" hereafter) on the same pan as the object (to subtract) or on the opposite pan (to add), each weight has three "states": "not used", "placed with the object", or "placed against the object". Assume there are $n$ such weights, where $n\in \mathbb{Z^{+}}$. Since each weight can be in three possible states, by the , there are $3^{n}$ possible combinations of weight placements. Accordingly, we can reframe this problem as one of by assigning values from the set $\{ -1,0,1 \}$ to each weight state:

$-1$ if the weight is placed with the object (say, on the left pan)
$0$ if the weight is not used (on neither pan)
$1$ if the weight is placed against the object (say, on the right pan)

Moreover, the weight of any object to be weighed may be written as a unique sum of powers of $3$ (see a proof here). Thus, for weights $1,3,9,..., 3^{n-1}$, every combination yields a total value:

$$\sum_{i=0}^{n-1}d_{i}\cdot3^{i} \;\text{where} \; d_{i}\in \{ -1,0,1 \}$$

Each such signed sum gives a unique integer from

$$-\frac{3^{n}-1}{2} \; \text{up to} \;\frac{3^{n}-1}{2}$$

That's because the maximum value occurs when all digits are $1$, and the minimum when all digits are $-1$. Both of those give of the form:

$$\sum_{i=0}^{n-1}3^{i}= \frac{3^{n}-1}{2}$$

Hence, the range of possible integer weight coverage goes from $-\frac{3^{n}-1}{2}$ to $\frac{3^{n}-1}{2}$, inclusive. That is

$$\frac{3^{n}-1}{2}-\left( -\frac{3^{n}-1}{2} \right)+1=3^{n}$$

distinct integer values—exactly as many as there are combinations! On the other hand, to cover any integer weight from $1\;\text{lb}$ up to some positive bound $W\;\text{lb}$, we need

$$\frac{3^{n}-1}{2}\ge W$$

Note that this is completely equivalent to putting all of the weights on one pan, giving you the maximum amount of measurable weight since it is the maximum amount one could "add" against the object (or the maximum one could "subtract" if one moved them all to the object's side, with the opposite sign of course, in any case, it is the maximum absolute value). Setting $W=100 \;\text{lb}$ gives,

$$\frac{3^{n}-1}{2}\ge 100 \Longrightarrow 3^{n}\ge 201 \Longrightarrow n\ge 5 $$

Hence, $n=5$ is both necessary and sufficient since

$$3^{0}=1,3^{1}=3,3^{2}=9, 3^{3}=27, 3^{4}=81, 3^{5}=243$$

Therefore, only $5$ precalibrated weights are necessary and sufficient to weigh any integer weight between $1$ and $100$. Take the weights of $1,3,9,27,81$ pounds, which are the powers of three that can represent every integer number from $1$ through $100$, with a balanced trinary number of no more than five digits.

How it works in practice

Suppose you want to weigh a $100 \;\text{lb}$ object. Place the object on the left pan. Then convert the decimal number $100$ into a ternary number using an algorithm (you can look one up online). Successive divisions by $3$ gives the remainders; $1,0,2,0,1$ so $100_{10}=10201_{3}$. Then, we use a from ternary to balanced ternary, which gives $1,1,-1,0,1$. Observe what this means, you need to place the $81 \;\text{lb}$, $27\;\text{lb}$, and $1\;\text{lb}$ weights on the right pan (against the object), and you need to place the $9\;\text{lb}$ weight on the left pan (with the object). Then the pans balance, confirming the object weighs $100\;\text{lb}$. This scheme works for any of the given objects you wish to weigh.

References:

1. Adapted from Hoeflin, R. K. (1985, April). Problem 38 in The Mega Test. , 7(4), p. 132.

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