The Scales Problem
My thought process to get to the solution
- Initially, I thought I found the answers when I tried using only weights 1 and 2 to weigh all the herbs after thinking:
- Use 1 to measure 1, 2 to measure 2, 1 and 2 to measure 3, then use measured herbs to weight all weights.
- But, after rereading the question I remembered that I was not allowed to use the herbs as auxillary weights.
- Next, I tried using 1, 2, 4, 8 but that did not get all the herbs.
- Then, I realized this was a base 2 pattern and thought that I could just measure even weights since the scale would tip down or up after putting in a 2. That is, I tried using 2, 4, 8, 16. But, this only got me up to 30.
- Then, I tried using 1, 3, 9, 27 since I was thinking in different bases. This looked really promising since I could find 40 but I was having a difficult time measuring the 5 herb weight.
Getting the solution
Dwelling on 1, 3, 9, and 27 I finally had the epiphany that I could use the 1, 3, and the 5 herb weight and compare that to the 9 using the scale. This breakthrough is what got me to the solution.
Here is the grunt work I did to determine the solution where the bolded numbers are the herbs being weighed.
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Left Scale
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Right Scale
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1
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1
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1
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2
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1 2
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3
|
3
|
3
|
3
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4
|
4
|
1 3
|
5
|
1 3 5
|
9
|
6
|
3 6
|
9
|
7
|
3 7
|
1 9
|
8
|
1 8
|
9
|
9
|
9
|
9
|
10
|
10
|
1 9
|
11
|
1 11
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3 9
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12
|
12
|
3 9
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13
|
13
|
1 3 9
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14
|
1 3 14 9
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27
|
15
|
3 15 9
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27
|
16
|
3 16 9
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1 27
|
17
|
1 17 9
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27
|
18
|
18 9
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27
|
19
|
9 19
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1 27
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20
|
1 9 20
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3 27
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21
|
9 21
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3 27
|
22
|
9 22
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1 3 27
|
23
|
1 3 23
|
27
|
24
|
3 24
|
27
|
25
|
3 25
|
1 27
|
26
|
1 27
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27
|
27
|
27
|
27
|
28
|
28
|
1 27
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29
|
1 29
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3 27
|
30
|
30
|
3 27
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31
|
31
|
1 3 27
|
32
|
1 3 32
|
9 27
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33
|
3 33
|
9 27
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34
|
3 34
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1 9 27
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35
|
1 35
|
9 27
|
36
|
36
|
9 27
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37
|
37
|
1 9 27
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38
|
1 38
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3 9 27
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39
|
39
|
3 9 27
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40
|
40
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1 3 9 27
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