As I mentioned in the question, this is an age old problem, but I was still intrigued by the twist provided to this puzzle by a colleague’s daughter. Many people gave a correct answer to the first two parts of the puzzle, only Pratik Poddar came back with the right answer for the 3rd part.

Here are the answers (and I will add explanations later):

Part 1: Weights need to be 1, 2, 4, 8, ….essentially all the powers of 2

Part 2: Weights need to be 1, 3, 9, 27, 81….essentially powers of 3

Part 3: Weights need to be twice that of powers of 3, i.e. 2, 6, 18, 54, ….

The basic principle to be adopted is that you choose weights in a manner, so that no weight can be represented in two different ways. For example, if you had two weights of 5 kg each, then you can make 10 kg with it (in Part 1). However, there is a bit of “wastage” here as well, as 5 kg itself can be formed by two different weights.

In Part 1, if you begin with a weight of 1 kg, you can only weigh 1 kg with it. For 2 kg, you need to add a weight of 2 kg. With the two of them, you can weigh 3 kg also, but not 4 kg, so you add a weight of 4 kg. Now with these three weights, you can go up to 7 kg, but not 8 kg…you can continue this way and realise that if you have weights that are a power of 2, you can create any weight with these combinations, by effectively using the binary notation.

In a balance, you now have a special property, that you can put weights on both sides and therefore not only add weights, but also subtract weights. So lets begin again:

– you start with a weight of 1 kg

– Add a weight of 2 kg. Now you can measure 2 kg with it. However, you can now create 1 kg in two different ways – 1 kg by itself and also by putting 2 kg on one side and 1 kg on the other side (difference). That is “wastage”

– So instead of 2 kg, we try 3 kg. With 3 kg, you can measure 2 kg through a difference of 3 kg and 1 kg. You can also measure 3 kg, and also 4 kg. But you cannot measure 5 kg.

– Through trial and error, you will realise that there is wastage by trying 5 kg, 6 kg, 7 kg and 8 kg. Once you try 9 kg, you can measure 5-8 kg by subtracting 1-4 kg from 9 kg.

– A better way to think about this is – since you can measure all the weights up to 4 kg, the next weight you need to chose is 4 x 2 + 1 (9 kg), because all the intermediate weights can be measured by subtracting from 9. With these weights, you can measure all the way up to 13, and hence the next weight needs to be 2 x 13 + 1 = 27 kg…and so on.

Part 3: The trick here is that once you know that whatever you are trying to measure is heavier than lets say 15 kg and lighter than 17 kg, and you know that it can only be a whole number, then you know that it is 16 kg, without actually having a combination of weights that gives you 16. This is what Arushi Gupta pointed out.

When you think about this further, you can conclude that if you can measure all alternate numbers, then the ones inbetween can be inferred. The easiest way to do this is from Part 2, you know that you can measure all #s, so by multiplying the weights by 2, you can measure 2, 4, 6, 8, 10, ….and infer the odd ones!

Hope you enjoyed the puzzle!