First of all, let me say to all my readers how fantastic it felt to post a puzzle, after not posting anything for two months, and get eighteen thoughtful, insightful comments in just three days; it’s every blogger’s dream. You all are fantastic and make this a lot of fun — thanks for reading!
I thought I’d take a post just to summarize some of the responses and solutions to the broken weight problem. As many commenters realized, the solution is that the weights are of sizes 1, 3, 9, and 27. Hmm, powers of three… coincidence? Of course not!
As several commenters noted, something involving base three readily presents itself if we realize that there are three possibilities for each weight: it can be on the left of the balance scale, on the right, or not on the scale at all. Since one side of the scale corresponds to adding the weight and the other side to subtracting, we are essentially writing numbers in base three, but using the digits -1, 0, and 1 instead of the usual 0, 1, 2. For example, 25 can be written as 10(-1)1, that is, (if we were really going to use this system we’d want to come up with a better symbol for -1). In fact, this is known as balanced ternary, and it is a fact (as proved by a few commenters) that n digits of balanced ternary allow us to uniquely represent every integer between . With four digits (as in the problem) we can uniquely represent every integer between -40 and 40. There was a bit of confusion in the comments about being able to represent some integers in more than one way, but I think if you try it out you will find that this is not the case.
From this problem (generalizing it to arbitrary numbers of weights), we can see that
JM noted that this generalizes to
and wondered whether this has anything to do with solving the problem, or with problems like it. Indeed it does; here’s one for you: suppose you are tasked with designing a set of weights. The weights should make it possible to weigh as many different integer weights as possible, without leaving any out, just like the weights 1, 3, 9, 27 make it possible to weigh every integer weight up to 40 without leaving any out. The one difference is that you want to have two identical copies of each weight. What is the best you can do?
I’ve left the problem slightly vague on purpose, but I hope you will have fun solving it and figuring out what it has to do with JM’s observation! Can you come up with other interesting generalizations of the problem?
Finally, Sam Shah posted a link to a description of his experience using the problem in a real-life problem-solving session.