Homework: Scale Problem

Solution:

At first, I tried to use numbers without including 1. I thought if we had some x, we could take the next number, x + 1, and get 1 by subtracting the two but I realized that this created issues when trying to create numbers later on that may have required subtracting 1, but x and x + 1 were also needed in the construction of the number, so this wouldn't work (e.g. When trying 10, 11, 13, and 17, I couldn't achieve 22 because 22 could be taken by adding 10 + 11, but in order to get the extra 1, I would need 11-10. Of course there are more options, like 11 + 11, but each required doubling up on a weight, which is not possible in this situation):

Then, I tried using 1 and working up from there. I chose 3 and 9 because they resulted in consecutive numbers from 1-13. Then, I subtracted 1, 3, and 9 from 40 to get 27. I checked to see if this worked and it did:

Thus, the weights are 1 g, 3 g, 9 g, and 27 g.

Are there several correct answers?

I tried to do a bit of an analysis based off of my findings. I realized if the numbers were a <  b < c < d, then a + b + c + d >= 40 to reach 40, and d <= 2(a + b + c) + 1 to bridge the gap between the highest number a + b + c  could achieve before d would have to be used. However, I was unable to determine if there were any other correct answers. 

Extension:

I think it would be useful to have students try to make all combinations of weights with a lower number by using a smaller total number and real weights and scraps of recycled paper to avoid wastage of herbs. For example, if we took 1 + 3 + 9 = 13 to be the total grams achievable, this would cut the table down to:

p = 1g -> 1 = p

p = 2 g -> 3 = p + 1

p = 3 g -> 3 = p

p = 4 g -> 3 + 1 = p

p = 5 g -> 9 = 3 + 1 + p

p = 6 g -> 9 = 3 + p

p = 7 g -> 9 + 1 = 3 + p

p = 8 g -> 9 = 1 + p

p = 9 g -> 9 = p

p = 10 g -> 9 + 1 = p

p = 11 g -> 9 + 3 = p + 1

p = 12 g -> 9 + 3 = p

p = 13 g -> 9 + 3 + 1 = p

where p is the amount of paper, in grams and the numbers represent the 1, 3, and 9 gram weights.

If weights from 1 - 13 g were available, students could spend time trying to see if other combinations worked and tangibly grasp why or why not. I myself could probably benefit from an activity like this because I had trouble disproving other combinations to the four weight case.

I think it might be a bit excessive for students to do a five weight case. However, if the pattern of 3^0 = 1, 3^1 =3, 3^2=9, 3^3=27 continued, the next weight would most likely be 3^4 = 81. It might be worth having students guess what they think the next weight would be and justify their answer to show that they recognized a pattern.