Puzzle Corner

Weighing in one go means you put whatever it is you want to weigh on the scale, then you add or subtract weights until it balances. 

Show contentSpoiler:

I wouldn't consider your description "weighing in one go", since it involves adding and subtracting product and keeping track of which product is part of the order and which is there to balance the scale. There's a simpler answer.

Ahh, thanks Linda. I have to be honest, I realise the question is/was vaguely worded and leaves it too open to interpretation/further clarification. I'm just going to tell myself that my original answer was so correct that it broke your question 

A little further response to your comment here though:

Show contentwhat's going on:

I'm sure my answers are just as vague and ambiguous as the question , but there would be no need to "subtract" product using the method I mentioned, so you wouldn't need to "keep track" of it. Any product placed on the scale would be part of the final total. You would only be adding, or moving, further weights, or moving the product from one scale to the other, but never subtracting it.

Regardless, it's clear I don't understand your question - or I understood it better than you   - but I am interested, are you saying for example weighing 20g weight against 20g product, then removing the weight and replacing it with 20g product, hence giving a total of 40g, is not a methodology you would allow?

I shall here leave the discussion, but look forward to reading the actual answer when it is revealed!

Thanks

Ahh, thanks Linda. I have to be honest, I realise the question is/was vaguely worded and leaves it too open to interpretation/further clarification. I'm just going to tell myself that my original answer was so correct that it broke your question 

A little further response to your comment here though:

Show contentwhat's going on:

I'm sure my answers are just as vague and ambiguous as the question , but there would be no need to "subtract" product using the method I mentioned, so you wouldn't need to "keep track" of it. Any product placed on the scale would be part of the final total. You would only be adding, or moving, further weights, or moving the product from one scale to the other, but never subtracting it.

Show contentwhat's going on:

Ah, I see. I was picturing it a bit differently. How would you weigh out 9 g?  

Regardless, it's clear I don't understand your question - or I understood it better than you   - but I am interested, are you saying for example weighing 20g weight against 20g product, then removing the weight and replacing it with 20g product, hence giving a total of 40g, is not a methodology you would allow?

I shall here leave the discussion, but look forward to reading the actual answer when it is revealed!

Thanks

This is a problem with any puzzle - because it is an artificial circumstance, it is often the case that one can offer a solution which ignores the point of the puzzle or assumes conditions which weren't available. You can always suggest that the product be taken to the local post office and the scale at the self-service kiosk be used to save the cost of any weights. It shouldn't count against me if I fail to stipulate that this isn't an option. 

Show contentSpoiler:

For example, your solution assumes that 40 g or lb of the product can be divided in 10 g or lb increments. But when it can't, your solution doesn't work. I'm looking for a solution which works for any situation, otherwise I would have stipulated that the product could be divided into 1 g or lb increments. In which case the answer would be obvious, trivial and inelegant. Buy a 1 g or lb weight and weigh out each g or lb one at a time. No thought or creativity needed.

Hint 2:

Show contentSpoiler:

Weights can be added to the pan with the product on it, as well as the other pan, so that the difference between two (or more) weights also effectively serves as a weight.

The merchant thinks on this a bit and says, "I can do this with 4 weights." What size weights do they order? 

Their spouse points out that with a little creativity, they can probably do it with one. 

Linda

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The merchant orders 1, 3, 9 and 27 pound weights. Each increment from 1 to 40 can be achieved by adding the product to one side of the scale, and by placing weights on both sides of the scale (when not weighing 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39 or 40 pounds). For example, to weigh 15 pounds, the 27 pound weight is placed on the other side of the scale, while the 3 and 9 pound weight are placed with the product.

Their spouse points out that ordering just one 1 pound weight as a reference, items in their store could probably be found to serve as substitutes for the 3, 9, and 27 pound weights (or any other convenient weights, for that matter).

[Laird]

will think on it overnight...

It turned out to be two or three nights, and then my internet went down, so my response has been delayed, but this, finally, is my solution:

Show contentMy solution:

1, 3, 9, 27

[Laird]

this, finally, is my solution

I now see that Linda revealed the solution in the spoiler in the post above mine, so there's no way for me to prove that I arrived at it independently, but... I did - though with the help of a brute-force algorithm. I now see the logic behind how it could have been arrived at without resorting to brute force, but lots of things like this seem much simpler - by which I mean "much more obvious" - in hindsight.

[Laird]

Oh, and the Spouse Solution is a good one.

[Laird]

Show contentSpoiler:

Laird, I don't think you need the 1 pound weight; just the 20, 10, 5, and 2 2's. You can create the following weights (maybe more, but they aren't needed) from the five weights:

2,4,5,7,9,10,12,14,15,17,19,20,22,24,25,27,29,30,32,34,35,37,39

Thus, there is no product weight between 1 and 40 pounds (presuming all weights are in integers) that we can't isolate from those 5 weights.  Right?

Show contentMy reply to Silence's "spoilered" question:

Nope.

There are two cases to consider: the first, being that in which weights may only be placed on one side of the scale, as I initially assumed, and the second, given Linda's later prompting that a scale has two sides, being that in which weights may be placed on both sides of the scale. In the first case, the obvious product weight which cannot be weighed is... one (ETA: others - non-exclusively - are as for the second case: 36, 38, and 40) - as the very list you provided demonstrates. In the second case, the (non-obvious, apart perhaps from the last) product weights which cannot be weighed are 36, 38, and 40 (all the others can be).

Show contentMy reply to Silence's \"spoilered" question:

Nope.

There are two cases to consider: the first, being that in which weights may only be placed on one side of the scale, as I initially assumed, and the second, given Linda's later prompting that a scale has two sides, being that in which weights may be placed on both sides of the scale. In the first case, the obvious product weight which cannot be weighed is... one (ETA: others - non-exclusively - are as for the second case: 36, 38, and 40) - as the very list you provided demonstrates. In the second case, the (non-obvious, apart perhaps from the last) product weights which cannot be weighed are 36, 38, and 40 (all the others can be).

Show contentSpoiler:

Linda gave us this: "The expectation is that either the weight of the product only comes in 1 pound increments, or that the weight of the product is altered to adhere to 1 pound increments."

Thus, if I place a 2 pound weight on the left side and a single product on the right side, if the right side goes up..... its a 1 pound product.  It can't weight "0" and if it weighed 2 pounds the scales would balance.  And if weights are fractional then the preferred answer wouldn't be able to isolate, say, a 1.x pound product either. Right?

Doesn't really matter as the 5 weights was bettered by the 4 weight option.

I now see that Linda revealed the solution in the spoiler in the post above mine, so there's no way for me to prove that I arrived at it independently, but... I did - though with the help of a brute-force algorithm. I now see the logic behind how it could have been arrived at without resorting to brute force, but lots of things like this seem much simpler - by which I mean "much more obvious" - in hindsight.

I assume people are honest about the source of their solutions. After all, there's not much incentive to lie - the pleasure comes from working out a solution, which is lost if you get the answer elsewhere.

Linda