The Market Scales Puzzle

How does the scale work?

Before doing any math, I had to remind myself how a two pan scale works.
The herbs go on one pan. The weights can go on either pan.
So:

--> If a weight is on the opposite pan, it adds weight.

--> If it is on the same pan as the herbs, it subtracts weight.

--> Or I don’t use it at all.

That means each weight has three options: add, subtract, or ignore.

Part 1:

1 g
To weigh 1 gram, I need a 1 g weight. 
So my first weight is: 1 g

3 g

I figured out the second weight through brute force

--> if my second weight is 2 g then I get:

       1 g

       2 g

       1 g + 2 g = 3 g

--> if my second weight is 3 g then I get:

       1 g

       3 g - 1 g = 2 g

       3 g

       3 g + 1 g = 4 g

--> if my second weight is 4 g then I get:

       1 g

       4 g - 1 g = 3 g

       4 g + 1 g = 5 g

       No way to get 2 g :(

9 g

Instead of starting small I started big this time

--> if my third weight is 10 g

       1 g

       3 g - 1 g = 2 g

       3 g

       3 g + 1 g = 4 g

       10 g - 3 g - 1 g = 6 g

       10 g - 3 g = 7 g

       10 g - 3 g + 1 g = 8 g

       10 g - 1 g = 9 g

       10 g 

       10 g + 1 g = 11 g

       10 g + 3 g - 1 g = 12 g

       10 g + 3 g = 13 g

       10 g + 3 g + 1 g = 14 g

       No way to get 5 g :(

--> if my third weight is 9 g

       1 g

       3 g - 1 g = 2 g

       3 g

       3 g + 1 g = 4 g

       9 g - 3 g - 1 g = 5 g

       9 g - 3 g = 6 g

       9 g - 3 g + 1 g = 7 g

       9 g - 1 g = 8 g

       9 g

       9 g + 1 g = 10 g

       9 g + 3 g - 1 g = 11 g

       9 g + 3 g = 12 g

       9 g + 3 g + 1 g = 13 g

27 g

This is my last weight so I know the total of all four weights should add up to 40 g.

--> if my fourth weight is 40 g - 9 g - 3 g - 1 g = 27 g

       1 g

       3 g - 1 g = 2 g

       3 g

       3 g + 1 g = 4 g

       9 g - 3 g - 1 g = 5 g

       9 g - 3 g = 6 g

       9 g - 3 g + 1 g = 7 g

       9 g - 1 g = 8 g

       9 g

       9 g + 1 g = 10 g

       9 g + 3 g - 1 g = 11 g

       9 g + 3 g = 12 g

       9 g + 3 g + 1 g = 13 g

       27 g - 9 g - 3 g - 1 g = 14 g 

       27 g - 9 g - 3 g = 15 g

       27 g - 9 g - 3 g + 1 g = 16 g

       27 g - 9 g - 1 g = 17 g

       27 g - 9 g = 18 g

       27 g - 9 g + 1 g = 19 g

       27 g - 9 g + 3 g - 1 g = 20 g

       27 g - 9 g + 3 g = 21 g

       27 g - 9 g + 3 g + 1 g = 22 g

       27 g - 3 g - 1 g = 23 g

       27 g - 3 g = 24 g

       27 g - 3 g + 1 g = 25 g

       27 g - 1 g = 26 g

       27 g

       27 g + 1 g = 28 g

       27 g + 3 g - 1 g = 29 g

       27 g + 3 g = 30 g

       27 g + 3 g + 1 g = 31 g

       27 g + 9 g - 3 g - 1 g = 32 g

       27 g + 9 g - 3 g = 33 g

       27 g + 9 g - 3 g + 1 g = 34 g

       27 g + 9 g - 1 g = 35 g

       27 g + 9 g = 36 g

       27 g + 9 g + 1 g = 37 g

       27 g + 9 g + 3 g - 1 g = 38 g

       27 g + 9 g + 3 g = 39 g

       27 g + 9 g + 3 g + 1 g = 40 g

So the four weights are 1 g, 3 g, 9 g, 27 g

I notice that each weight is a power of 3. A simpler way to solve this problem without using brute force was to go back to the original idea that a weight has three options: add, subtract, or ignore. So I am really just working with digits of −1, 0, and 1. A base-3 system!!! 

All that work just for this simple solution, but at least it looks pretty. :)

Part 2:

Now the herbs go on one side and the weights only go on the other. No subtracting allowed.

So each weight can only be: used or not used.

This looks like base-2 system!

Instead of brute force I can just use powers of 2:

--> 1 + 2 + 4 + 8 + 16 = 31 grams

Any number from 1 to 31 can be made by adding some of these weights.
Example:
8 g + 4 g + 1 g = 13 g
16 g + 4 g + 2 g = 22 g

So the five weights are 1 g, 2 g, 4 g, 8 g, 16 g

How I would extend this puzzle for students?
- Ask students to find the fewest weights needed for a different range
- Let them test what happens if one weight is missing

- Ask students to design their own “weight system” with different constraints:

    - Only even weights
    - Only odd weights

Connections to number theory and bases?
- The one pan scale connects to base 2, where numbers are built using powers of 2
- The two pan scale connects to balanced base 3, where digits can be negative
- It also ties into place value and representation of numbers

Overall:

What I liked most is that this problem makes those ideas feel concrete. You are not just writing numbers, but you are physically balancing them. That moment where everything clicks is the part I would want my students to experience too. I remember I was just staring at my paper when I realized the weights weren’t random.