Four weights for forty pounds
You want to weigh objects of every whole number of pounds from 1 to 40 on a two-pan balance. You choose a set of weights, and for each weighing you may put weights on either pan, including the pan with the object.
What is the smallest number of weights that makes every weighing from 1 to 40 possible?
Fractions, decimals with at least 3 significant digits, and expressions such as sqrt(2)/2, e^-1 or pi/4.
Each weight can be on the object's pan, on the other pan, or off the balance. How many different amounts can weights measure?
Try weights that are powers of 3.
h
s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.
Idea. Each weight has three states, so weights can measure at most about positive amounts. Three weights are too few for 40 amounts, and the weights 1, 3, 9 and 27 do the job.
Step 1: the bound. A weight on the other pan counts times its value, on the object's pan times, off the balance 0. So weights give at most signed sums. One of them is 0, and the others come in pairs , so at most positive amounts. With that is 13, fewer than 40. So at least 4 weights are needed.
Step 2: four weights suffice. Take 1, 3, 9 and 27. Every whole number from to has a unique expansion with : the balanced ternary expansion. For example, : put 27 and 3 on the other pan, and 9 and 1 with the object.
So the answer is 4, and is exactly the largest amount reachable.
If weights may go on one pan only, each weight is just in or out, and you need powers of 2: 1, 2, 4, 8, 16 and 32, six weights.