Answer to Puzzle #1: 12 Coins, Old Fashioned Balance

1. You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weigh the same or that the dish that falls lower has heavier contents than the other.

The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more.

Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light.

Firstly let me stress that this puzzle is very difficult, and worse the answer is not particularly satisfying. It is not a good way to be introduced to my site. Questions 3 & 5 are quite easy. 9 is accessible and interesting and 13 is my favourite. You have been warned.

My answer is something of an amalgam between the Dr. Math web-site, myself, and various emails I received, including a correction from Ben Cronin. It is now rigorous but as easy to follow as is possible.

old fashioned balance puzzle Before reading the answer can I interest you in a clue?

Most people seem to think that the thing to do is weigh six marbles against six marbles, but if you think about it, this would yield you no information concerning the whereabouts of the only different marble or the nature of it's difference as we already know that one side will be heavier than the other. It's worth mentioning that if this is used as an interview question this observation, by itself, is likely to get you a good portion of the way there.

So that the plan can be followed, let us number the marbles from 1 to 12. For the first weighing let us put on the left pan marbles 1,2,3,4 and on the right pan marbles 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different marble is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different marble, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different marble is 9, 10, or 11, and that that marble is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy marble. If not, the heavier one is the heavy marble.
(3) 9,10,11 is light. Proceed as in the step above, but the marble you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these marbles could be the different marble. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different marble is either 3 or 4. Weigh 4 against 9, a known good marble. If they balance then the different marble is 3, otherwise it is 4. The direction of the tilts can tell us whether the offending marble is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy marble, or 1 is a different, light marble.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different marble. If they balance, then 1 is the different marble. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy marble or 2 is a light different marble. Weigh 5 against 6. The heavier one is the different marble. If they balance, then 2 is a different light marble.

Number Theory

Steve Mencinsky emailed me with a number theory approach, subsequently Nelson Gimenez da Motta has also been in contact, he took Steve's work and ran with it. It is not my intention that anyone using this site would have to have really any formal advanced maths training. I aspire to be able to explain everything through common sense and logic, even if what we are secretly doing is using a mathematical technique. This number theory approach does introduce an important technique to which I would like to return, the ability to quickly identify if we have enough information to solve a problem.

The first approach, my approach, uses three dependent weighings, that is what we weigh on the second weighing is dependent on the results of the first and so on. He demonstrates that with each individual weighing having 3 possible outcomes, (left, right & balanced,) and there being 3 weighings there are 33=27 possible results in total. Reflections are not useful to us so there is enough information from 3 independent weighings to determine the results for 13.5 marbles. This is an important first step.

The implications of a half marble I'll leave to Steve and Nelson to describe. Below are the emails they sent me.

Neither of them got this completely right. Bard came the closest.

If you're curious what Bard made of this puzzle...

If you're curious what ChatGPT made of this puzzle...

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