Market Scales Puzzle

Let's say we use 1, 2, 3, 4g of weights to measure

Left: Herb 1 Right: 1

Left: Herb 2 Right: 2

Left: Herb 3 Right: 3 or 1 + 2

Left: Herb 4 Right: 4 or 1 + 3

Left: Herb 5 Right: 2 + 3 or 1 + 4

Left: Herb 6 Right: 4 + 2

Left: Herb 7 Right: 4 + 3

Left: Herb 8 Right: 1 + 3 + 4

Left: Herb 9 Right: 2 + 3 + 4

Left: Herb 10 Right: 1 + 2 + 3 + 4

It seems we have used up all the weights and we're stuck. Let's try a different approach.




Let's say we use 1, 3, 6, 10g for weights

Left: Herb 1 Right: 1

Left: Herb 2 + 1g Right: 3

Left: Herb 3 Right: 3

Left: Herb 4 Right: 3 + 1

Left: Herb 5 + 1 Right: 6

Left: Herb 6 Right: 6

Left: Herb 7 + 3g Right: 10

Left: Herb 8 + 1g Right: 6 + 3

Left: Herb 9 + 1g Right: 10

...

Left: Herb 20 Right: 10 + 6 + 3 + 1

Left: Herb 21 Right: X

We have used up all 4 numbers for weights, even if we add all four numbers together, the heaviest weight we will get is 20g, so we're stuck again.


Let's do the other way around this time, starting with the total weight that gives us 40. Since we need 4 numbers that would give us the combination of 40g for the heaviest weight, we can safely eliminate the weight of 35 to 40g.

Let's start with weights of 1, 2, 3, 34g.

Left: 34 + 1 + 2 + 3 Right: 40

Left: 34 + 2 + 3 Right: 39

Left: 34 + 1 + 2 Right: 37

Left: 34 + 2 Right: 36

Left: 34 + 1 Right: 35

Left: 34 Right: 34

Left: 33 + 1 Right: 34

Left: 32 + 2 Right: 34

...

Left: 28 + 3 + 2 + 1 Right: 34

Left: X Right: 34

We can quickly see that we ran out of options for giving us weights lower than 28g.

What about 33g? The combination we can get from 40g is 1, 2, 4, 33g. However, we ran out of options when we get to weights lower than 26.

Left: 26 + 1 + 2 + 4 Right: 33

Left: X Right: 33


I find that when we take away the heaviest weight out of the four, the herb plus the remaining three smaller weights added together must be larger or equal to the biggest weight. So we must figure out a way to form the combination of the remaining smaller weights plus the herb added together to be equal to or larger than then heaviest weight, or the remaining 3 smaller weights can form different combinations to find the weight of the remaining herbs.

Let's take a leap and jump straight to a lower number, say 28. Let's use the combination of 1, 2, 9, 28.

However, we're stuck when we get to herb equal to 15 grams because:

Left: Herb 15 + Any weight combination Right: X

We also know that 9 + 2 + 1 = 12, so we can see that we can have herbs up to 12 grams.

The other combinations including 28grams couldn't work either.

But it seems we are getting close!


Let's try a lower number 27. Say we use the combination of 27 with 3 numbers that gives us 40 is 1, 3 and 9.


Left: Herb 1 Right: 1

Left: Herb 2 + 1g Right: 3

Left: Herb 3 Right: 3



Left: Herb 4 Right: 3 + 1

Left: Herb 5 + 1 + 3 Right: 9

Left: Herb 6 + 3 Right: 9

Left: Herb 7 + 3 Right: 9 + 1

...

Left: Herb 13 Right: 9 + 3 + 1

Left: Herb 14 + 9 + 3 + 1 Right: 27

Left: Herb 15 + 9 + 3 Right: 27

Left: Herb 16 + 9 + 3 Right: 27 + 1

...

Left: Herb 39 Right: 3 + 9 + 27

Left: Herb 40 Right: 1 + 3 + 9 + 27




So it seems like we have our answer in 1g, 3g, 9g and 27g.

This confirms my theory, that in order for the scale to work without gaps, the sum of the 3 smaller weights with the herb has to be equal to or larger than the heaviest weight. The smaller weights can also form different combinations with the herbs to find their respective weights. Essentially, the smaller weights have to fill the gaps between lighter herbs and the heaviest weight.


There were other combinations such as 1, 2, 10, 27, but the numbers don't add up to the weight of some herbs, or we encounter a dead end when calculating.




What if it was a one-pan scale?

Let's say this time, we only have a one-pan scale. We need to find 5 total weights that give us 31 grams together.

Let's say we use 1, 2, 3, 4, 21. We can quickly see that this doesn't work because we can't make a combination between 11 and 20.

Since we know that we 1 + 2 = 3, we can safely disregard 3 as one of our numbers and use the next number 4 as one of the 5 weights. Let's try 1, 2, 4, 8, 16. The reason we do this is because we can make every number weights from 1 to 7 through numbers 1, 2 and 4, therefore we can jump straight to the next number after 7, which is 8 in this case.


Herb 1 = 1

Herb 2 = 2

Herb 3 = 1 + 2

Herb 4 = 4

Herb 5 = 1 + 4

Herb 6 = 2 + 4

Herb 7 = 2 + 4 + 1

Herb 8 = 8

Herb 9 = 1 + 8

Herb 10 = 2 + 8

...

Herb 15 = 1 + 2 + 4 + 8

Herb 16 = 16

...

Herb 31 = 1 + 2 + 4 + 8 + 16


We know this works because we can use number 1, 2, 4 and 8 to add up to any numbers up between 1 to 15, which also means as long as the numbers we're trying to make is 15 or less it will work, which is why 1, 2, 4, 8 and 16 works, because numbers between 16 and 31 is 15 or less.

One interesting fact I noticed is that 1, 2, 4, 8, 16 are all results from powers of 2:

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

Similarly, for the two pan scale questions, numbers 1, 3, 9, 27 are also results from powers of 3:

3^0 = 1

3^1 = 3

3^2 = 9

3^3 = 27




By working with this puzzle, students will develop logical thinking by using a set of given numbers, starting with 1, and finding the combinations that would eventually lead us to the correct weight. Students can be asked "What's the next number that fits the combination after 1?" This encourages them to dive deep and explore the different combinations they can create. The puzzle also allows students to understand how pan balance scale works, while they develop their thinking and problem solving skills.




This puzzle connects in a way where I'm already familiar with both how the two-pan scales and the one-pan scales work. We have to try different combinations of weight in order to find the correct total that involves working with logical thinking. On the other hand, the patterns that I discovered from the puzzle are also related to exponents. This also connects to the idea of different number bases where numbers are represented with powers of 2 and 3, which is quite fascinating!