The Scales Problem


A market vendor sells dried cooking herbs in whole-number amounts from 1 to 40 grams. The vendor has an old-fashioned two pan weigh scale, and has exactly four weights of different amounts that allows them to weigh out any of these amounts of herbs -- without using the herbs or any other object as an auxiliary weight.

1.     What are the values of the four weights?
-       Since the lightest weight is 1 gram, one of the four weights must be 1 gram.
-       Now, we have the following weights to measure herbs that weigh from 1 to 40 grams: 1, a, b, and c, where a < b < c.
-       I couldn’t see the pattern right away, but I was sure that there must be a pattern for this problem.
-       I did lots of trial-and-error methods and figured out that the four weights have to be 1, 3, 9, and 27.
-       We place herbs on the left and the weights on the right. Assume that the sign becomes negative when we place some weights on the left to balance. The following table is the results that I came up with:
Herb Weights (g)
Weight Combination (g)
1
1
2
3-1; 1 on the left & 3 on the right
3
3
4
1+3
5
9-3-1
6
9-3
7
9+1-3
8
9-1
9
9
10
9+1
11
9+3-1
12
9+3
13
9+3+1
14
27-9-3-1
15
27-9-3
16
27-9-3+1
17
27-9-1
18
27-9
19
27-9+1
20
27-9+3-1
21
27-9+3
22
27-9+3+1
23
27-3-1
24
27-3
25
27-3+1
26
27-1
27
27
28
27+1
29
27+3-1
30
27+3
31
27+3+1
32
27+9-3-1
33
27+9-3
34
27+9-3+1
35
27+9-1
36
27+9
37
27+9+1
38
27+9+3-1
39
27+9+3
40
27+9+3+1


2.     Are there several correct solutions?
-       I don’t think there is any other correct solution.

3.     How could you extend this puzzle to help your students understand the mathematics more deeply?
-       I would ask students if they can make their own problem that is different from the original problem, but follows the same weight combination rules.

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