One arm of a weighing scale is twice as long as the other.
Suppose you want to create a set of weights so that any object weight from 0.5 and its multiples up to 40 pounds can be exactly balanced there by placing a certain combination of these weights onto that scale.
What is the fewest number of weights you need, and what are their weights?
This question is different from the existing one. This one has a twist: the length of the arms and it must be from 0.5 and all its multiples.
Answer
If the object is in the pan with the long arm, then we have $$object+(some\ weights) = (some\ other\ weights)/2$$ If the object is in the other pan then we have the similar $$object+(some\ weights) = (some\ other\ weights)*2$$ The solution I have chosen only uses the first, i.e. the object is only ever on the long-armed side.
I use the seven weights $1$, $1$, $5$, $5$, $25$, $25$, $125$. The pattern obviously extends, being the powers of $5$, each weight occurring twice.
As proof that these weights work, here are all possibilities:
0.5 = 1 /2
1 = 1+1 /2
1.5 +1 = 5 /2
2 +1 = 5+1 /2
2.5 = 5 /2
3 = 5+1 /2
3.5 = 5+1+1 /2
4 +1 = 5+5 /2
4.5 +1 = 5+5+1 /2
5 = 5+5 /2
5.5 = 5+5+1 /2
6 = 5+5+1+1 /2
6.5 +5+1 = 25 /2
7 +5+1 = 25+1 /2
7.5 +5 = 25 /2
8 +5 = 25+1 /2
8.5 +5 = 25+1+1 /2
9 +5+1 = 25+5 /2
9.5 +5+1 = 25+5+1 /2
10 +5 = 25+5 /2
10.5 +5 = 25+5+1 /2
11 +5 = 25+5+1+1 /2
11.5 +1 = 25 /2
12 +1 = 25+1 /2
12.5 = 25 /2
13 = 25+1 /2
13.5 = 25+1+1 /2
14 +1 = 25+5 /2
14.5 +1 = 25+5+1 /2
15 = 25+5 /2
15.5 = 25+5+1 /2
16 = 25+5+1+1 /2
16.5 +1 = 25+5+5 /2
17 +1 = 25+5+5+1 /2
17.5 = 25+5+5 /2
18 = 25+5+5+1 /2
18.5 = 25+5+5+1+1 /2
19 +5+1 = 25+25 /2
19.5 +5+1 = 25+25+1 /2
20 +5 = 25+25 /2
20.5 +5 = 25+25+1 /2
21 +5 = 25+25+1+1 /2
21.5 +5+1 = 25+25+5 /2
22 +5+1 = 25+25+5+1 /2
22.5 +5 = 25+25+5 /2
23 +5 = 25+25+5+1 /2
23.5 +5 = 25+25+5+1+1 /2
24 +1 = 25+25 /2
24.5 +1 = 25+25+1 /2
25 = 25+25 /2
25.5 = 25+25+1 /2
26 = 25+25+1+1 /2
26.5 +1 = 25+25+5 /2
27 +1 = 25+25+5+1 /2
27.5 = 25+25+5 /2
28 = 25+25+5+1 /2
28.5 = 25+25+5+1+1 /2
29 +1 = 25+25+5+5 /2
29.5 +1 = 25+25+5+5+1 /2
30 = 25+25+5+5 /2
30.5 = 25+25+5+5+1 /2
31 = 25+25+5+5+1+1 /2
31.5 +25+5+1 = 125 /2
32 +25+5+1 = 125+1 /2
32.5 +25+5 = 125 /2
33 +25+5 = 125+1 /2
33.5 +25+5 = 125+1+1 /2
34 +25+5+1 = 125+5 /2
34.5 +25+5+1 = 125+5+1 /2
35 +25+5 = 125+5 /2
35.5 +25+5 = 125+5+1 /2
36 +25+5 = 125+5+1+1 /2
36.5 +25 +1 = 125 /2
37 +25 +1 = 125+1 /2
37.5 +25 = 125 /2
38 +25 = 125+1 /2
38.5 +25 = 125+1+1 /2
39 +25 +1 = 125+5 /2
39.5 +25 +1 = 125+5+1 /2
40 +25 = 125+5 /2
40.5 +25 = 125+5+1 /2
41 +25 = 125+5+1+1 /2
...
93.5 = 125+25+25+5+5+1+1 /2
I'm not sure this is optimal, but it is at least a generic solution that can be extended to any weight range.
EDIT:
Electric_monk's answer shows that you can use the same number of weights but with a lower maximum weight for the range given in the problem.
Keeping all the weights on the short arm, opposite the object, you have the traditional solution with weights $1$, $2$, $4$, $8$, $16$, $32$, $64$. This allows one to weigh any object up to $127/2=63.5$ in weight with seven weights. So for the range given in the problem, there is no need to use a subtractive system of weights yet (unless there is some more optimal answer with 6 weights only). It seems therefore that it would have been better for a larger range to have been used, e.g. $0.5$ to $80$.
EDIT2:
Oray was the first to announce that
6 weights is enough for the given range of object weights.
But I found something similar.
The weights $1$, $1$, $5$, $5$, $25$, $50$ suffice. Almost the same configurations can be used as in my 7 weight solution, except that when 25+25 was used before, you now use the 50, and instead of -25 + 125/2 we use (25+50)/2. The heaviest object that this set can handle is 43.5.
You can also use $1$, $1$, $5$, $6$, $28$, $56$ to get objects up to 48.5 pounds, but the various arrangements are rather more haphazard.
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