The Scale Problem

 

Thinking process:

We want the scale to balance on both sides.

- To measure 1g of herb, we have 1g weight to balance the scale. We know the one of the four weights is 1g.

- To measure 2g of herbs, given that we already have 1g weight, we need can have another 1g of weight to balance both sides. However, the question states that the four weights are at different amounts, so the second weight cannot be 1g. Now if we put the 2g of herb to the 1g weight on the left side, we would need 3 g of weight on the right side. So we can say the second weight is 3g. 

- To measure 3g of herbs, we can use the 3g weight to balance the scale

- To measure 4g of herbs, we can use the 1g + 3g weight. 

- To measure 5g of herbs, we have 1g + 3g weight, we need another 1g to make the scale balance. Once again, we already have 1g, so it doesn’t work. Analogic to measuring 2g of herbs, we can move 5g of herbs to one side with 1g+3g weight, in total we have 9g. So the other side might be weigh 9g. We can say the third weight is 9g. Now, we have the weights to be 1g, 3g, and 9g. 

- To measure 6g, we have 1g,3g,and 9g. We can balance the scale by having 9g on one side, and 6g herbs + 3 g weight on the other side

- To measure 7g, we can have 7g herbs + 3g weight on one side, and 1g+9g on the other side

Similarly thinking process can be applied for the rest of the numbers until we get to 14g.

- To measure 14g, we will end up with the same situation in measuring 2g and 5g of herbs. From there, we can find our last weight to be 27g. 

I have provided a trial-and-error chart to support my thinking. Using the four weights at 1g, 3g, 9g, and 27g. We can balance the scale in weighing out whole-number amount of herbs from 1 to 40g. 

I believe there is only one correct solution to this problem. 

To extend on this puzzle, I would ask the students what they notice about the numbers, and if there is a quicker way of doing it more mathematically. In which number will they need to add a fifth weight? Sixth weight? What will be the values of the fifth and sixth weight