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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?

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    $\begingroup$ Not entirely sure why you can't measure something that is 9 + 3 + 1 = 13 pounds? Also not sure why you didn't mention your measuring methods in your lists. (e.g. balancing 3 + object against 9 to get a 6 pound object.) This would seem to be the only "trick" to the riddle? $\endgroup$
    – Graylocke
    Mar 11 at 12:08
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    $\begingroup$ This is a very confusingly worded puzzle - it sounds like it's asking you to use a single weighing operation to determine the weights of 10 objects at the same time. $\endgroup$ Mar 12 at 0:40
10
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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.

enter image description here
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.

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  • $\begingroup$ Don't we also need to prove that among these 13 states, there are no 2 states which will be measuring the exact same weight ? I feel that this also needs to be proven. $\endgroup$ Jul 28 at 22:31
  • $\begingroup$ @JaapScherphuis bah. My error. Sorry about that. $\endgroup$
    – Ben Barden
    Jul 30 at 13:27
  • $\begingroup$ In base 3, the weights are 1, 10, 100. Consider the state where all of them are on the left. Now how much can you increase the difference between right and left? Removing a weight increases the difference by the amount of that weight. Moving a weight to the right increases the difference by twice the amount of that weight. So the possible differences are 000, 001, 002, 010, 011, 012, ..., 222 (or, returning to base 10, 0 through 26). And the initial difference was -13, so the full set of differences is -13 through +13. $\endgroup$
    – Ed Murphy
    Jul 31 at 0:38
  • $\begingroup$ Jaap, I feel that for completeness sake, it also needs to be proven that all of these 13 states will measure objects of different weights. And no two states will measure an object of the same weight. For example, if the weights were 1, 5 and 11 pounds, then the state 5 pounds on one pan and the object to be weighed on the other pan and the state, 1 and 5 pounds weights along with the object to be weighed on one pan and the 11 pounds weight on the other pan, both measure an object of 5 pounds. Thus, in this case, there will be less than 13 possibilities. $\endgroup$ Aug 6 at 1:51
  • $\begingroup$ My question regarding the 1,3 and 9 pounds weights is, do we need to list all the possibilities and check that each of the states measure an object of a different weight or can it be proven that all the states measure a different weight without manually checking all the possibilities ? $\endgroup$ Aug 6 at 1:57
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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.

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