Maximum number of masses you can weigh given 3 weights

Question

The puzzle is as follows:

Assume that you have a two pan scale and three weights, one of 1 pound, the other 3 pounds and the last one 9 pounds. How many objects of different weights can you weigh at maximum, using just these 3 weights ?

For instance,

1. We can measure 6 pounds by keeping the 3 pound weight and the 6 pound object on one pan and the 9 pound weight on the other pan.

2. We can measure 4 pounds by keeping the 3 pounds and 1 pound weights on one pan and the 4 pound object on the other pan, etc.

The choices given are:

1. 10 objects
2. 12 objects
3. 13 objects
4. 14 objects

This puzzle appears to be an adaptation from a reprinted copy of an intelligence psychometry JPA exam of the late 1990s which might be based on Weschler IQ tests from that time period, a similar problem has been also used in Thurstone's exams of the 1960s.

I assume that a strategy to solve this puzzle is to find the possible weights that we can have with all these weights combined.

Assuming that we use the 1 pound weight, we can only measure one object.

Using the 1 pound and/or 3 pound weight, we can measure:

2, 3 and 4 pounds. Thus there are three additional objects.

Using the 1 pound and/or 3 pound and/or 9 pound weight:

5, 6, 7, 8, 9, 10, 11, 12 pounds objects

This accounts for 8 additional objects.

If we add all these up:

1+3+8=12 objects

Therefore choice 2 would be correct

I'm not sure if this is the right answer, but to me is the answer with the most logical sense to me. But I could be wrong. Could someone else help me?

Answer 1

There are

$26=3^3-1$

ways to place the three weights because

each has three possibilities - left pan, right pan, or neither - and we exclude the case where no weights are placed in the pans at all since we do have an object we want to weigh.
The object to be weighed is added to whichever pan is the lightest. This double-counts since swapping the contents of the two pans does not make a difference.

This leaves exactly

$26/2=13$

possibilities. It is easy to verify these all give different weights for the object, but here is a picture that illustrates how that works.

For each new weight that is considered, the range of weighable values triples. Each new weight is exactly large enough for the ranges to join end to end without gaps.

Answer 2

13 objects.

• 1 pound
• 3-1 = 2 pound
• 3 pound
• 3+1 = 4 pound
• 9-3-1 = 5 pound
• 9-3 = 6 pound
• 9+1-3 = 7 pound
• 9-1 = 8 pound
• 9 pound
• 9 +1 = 10 pound
• 9+3-1 = 11 pound
• 9+3 = 12 pound
• 9+3+1 = 13 pound

That's all.