A math puzzle aimed at 9 to 12 year olds

[Jackamus]

I'm 82 and was browsing through a old comic annual that was published back in the 1950's (Eagle Annual No 3). It had a puzzles page and one of the puzzles was quote:
'I have a pair of scales and 3 weights and with these 3 weights I can weigh any number of pounds from 1 up to 13.(I may, of course, find it necessary to put one or more weights in both scales.) What are my 3 weights?

I have to say that I was stumped trying to calculate it. The best I could do was to assume that one weight (a) would have to be 1lb and that all 3 weights (a + b+ c) would add up to 13. Therefore b + c = 12. Beyond that I was stuck!

This shows the level thought, back in those days, that was considered OK for children of this age.

Can this be calculated or only solved empirically?

 

[Office_Shredder]

How much can we measure with a 1 pound weight?

Obviously 1 pound. We can also identify something is 2 pounds by observing that it weighs more than 1. We can't distinguish 2 vs 3 so we need a weight for this. We want the heaviest weight possible (this is me making an intuitive assertion)

Obviously 2 pound weight works. 3 pound weight lets us identify that x <3 and x>1 hence x=2.

4 pound weight also works, and takes advantage of the double sided scales. If it's bigger than 1 and smaller than 4, you can compare x+1 against 4 and see if it's equal or smaller. This also lets you check if an object is 5 or 6 pounds (but can't distinguish 6 vs 7). I'll leave it at an exercise to you to figure out what is the heaviest new weight we can add that gives us more continuous coverage.

 

[Jonathan Scott]

I think there's an algorithmic solution, in that each new weight has to be one more than double the sum of all previous weights (and that scheme would continue for four or more weights). That means that the next weight above the sum of the previous weights can be weighed as the difference between the new weight and the sum of the previous weights. If the previous weights can represent all values up to that sum, then adding the new weight allows all higher values to be represented up to the new sum, firstly as differences from the new weight then as additions to it.

 

[Jonathan Scott]

(And looking at the solution, I note that there's a clear connection with powers of 3 and base 3 arithmetic).

 

[Jackamus]

OK so how do you calculate the 3 weights?

 

[256bits]

1, 3, 9, ...

 

[Jackamus]

Correct but how did you do it? I'm not a mathematician so I would like to steps of your process.

 

[Jonathan Scott]

Assume that the first weight is 1.

That can weigh up to 1 pound.

The second weight must be such that the difference from the first is 2 pounds, so that the difference is the next higher amount needed. So the second weight must be 1+2=3 pounds.

Those two can now weigh up to 1+3=4 pounds.

The third weight must be such that the difference from the first plus second is 5 pounds, so it must be 1+3+5=9 pounds.

Those three can now weigh up to 1+3+9=13 pounds.

At each step, the new weight must be one pound more than twice the sum of the existing ones.

This leads to the series 1, 3, 9, 27, 81, ..., i.e. powers of 3.

 

[Ibix]

If I've got some weights that can do everything up to $w$ then the next weight I need is $2w+1$. Then I can put my new weight on one pan and do everything from $w+1$ to $2w$ by putting combinations of my existing weights on the other pan (effectively subtracting them from $2w+1$) and everything up to $3w+1$ by adding them to the same pan. With my new weight I can weigh anything up to $3w+1$, so that's my updated $w$ value to find my next new weight.

I need a 1 unit weight to start, so my first $w$ is 1. My next weight is 3, which makes my new $w$ 1+3=4. My next weight is therefore 9, which makes my new $w$ 1+3+9=13.

 

[Jonathan Scott]

To clarify, if some combination of weights (on both pans) gives a certain amount, then switching them to the opposite pans is equivalent to changing the sign of that amount. They are not necessarily all in the same pan.

 

[Jackamus]

Assume that the first weight is 1.

That can weigh up to 1 pound.

The second weight must be such that the difference from the first is 2 pounds, so that the difference is the next higher amount needed. So the second weight must be 1+2=3 pounds.

Those two can now weigh up to 1+3=4 pounds.

The third weight must be such that the difference from the first plus second is 5 pounds, so it must be 1+3+5=9 pounds.

Those three can now weigh up to 1+3+9=13 pounds.

At each step, the new weight must be one pound more than twice the sum of the existing ones.

This leads to the series 1, 3, 9, 27, 81, ..., i.e. powers of 3.

Thank you Johnathon for that simple explanation. I still think it is expecting a lot from 9 - 12 year olds but that is what like in those days.

 

[PeroK]

Thank you Johnathon for that simple explanation. I still think it is expecting a lot from 9 - 12 year olds but that is what like in those days.

It's certainly asking a lot of a nine year old, but not quite so much of a twelve year old.