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.
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.
Here is an attempt at a mathematical analysis of the problem. I will consider two versions of the problem, the one as given and the version where the maximum weight $n=80$ half-pounds is allowed to grow to infinity, and we ask how to minimize the asymptotic growth of the minimum number of weights needed.
From the setup, we have a sequence $(w_i)_{i=1}^k$ of weights (that we can choose to be any positive real numbers), and the goal is to minimize $k$ such that for each $2c\in\{1,\dots,n\}$, there is some sequence $(\epsilon_i)_{i=1}^k$ where each $\epsilon_i\in\{-1,0,2\}$ such that $\sum_{i=1}^k\epsilon_iw_i$ is either equal to $c$ or $-2c$.
The idea is that $\epsilon_i$ represents whether the $i$th weight is on the long arm ($2$), the short arm ($-1$), or not on the scale ($0$), and the sum total of these weights must add to either $c$ (if the unknown weight is on the short arm) or $-2c$ (if the unknown weight is on the long arm). Alternatively we can write this equation as $\sum_{i=1}^k\epsilon_iw_i+\epsilon_cc=0$, where $\epsilon_c$ can only take the values $\{-1,2\}$ because the unknown weight must appear on the scale (otherwise we can't measure it).
Now there are $3^k$ choices for the values of the $\epsilon_i$'s, and $2$ choices for $\epsilon_c$, so at first blush any set of weights can weigh at most $2\cdot3^k$ distinct values, i.e. $2\cdot 3^k\ge n$. But this isn't quite right because it is impossible for $\sum_{i=1}^k\epsilon_iw_i$ to equal both $c$ and $-2c$. That is, no matter how the weights are arranged, without the unknown weight they are either unbalanced to the left or to the right, and so there is only one place you can put the unknown weight to have a chance at balancing it. Thus the choice of where to put the unknown weight is not really a choice at all.
So we find that $3^k\ge n$, just as in the case with balanced arms. Unfortunately, this is not a very good bound - for the present case $n=80$ this gives $k\ge 4$, while the best solutions elsewhere on this page find $k=6$ as the best case.
Since each new weight triples the number of possible values in the best case, we should also expect $n=O(3^k)$ (or $k=\log_3n+O(1)$) in the limit. But unlike the classical version of the problem, $\{-1,0,2\}$ does not form a complete set of residues mod 3, so a formula like balanced ternary will not suffice.
The currently accepted answer shows that it is possible to have a solution where $n=5^{k/2}\approx2.23^k$, and we can search among higher integer bases for solutions in the following manner: Choose a base $t$ and number of weights in this base $m$. The efficiency of this coding system will be $n=t^{k/m}$, so we want $t^{1/m}$ to be as large as possible. We pick numbers $x_1,\dots,x_m$ for the weights, and then our system of weights will be $x_1,\dots,x_m,x_1t,\dots,x_mt,\dots,x_1t^i,\dots,x_mt^i,\dots$, stopping when we have reached our target $n$. For such a system to work we must have that every number in the base is reachable using the $x_i$'s, that is $\sum_{i=1}^mx_i\epsilon_i\equiv0,\dots,t-1\pmod t$.
In the accepted answer, we have $t=5$, $m=2$, $x_1=x_2=1$, and we can verify that $0+0\equiv 0$, $2-1\equiv 1$, $2+0\equiv 2$, $-2+0\equiv 3$, and $-2-2\equiv 4$, so this is a valid coding system. Can we do better?
I conjecture that there are an infinite sequence of valid coding systems where $t^{1/m}\to3$, but it gets harder to describe them as $t$ and $m$ grow. The best solution I have found is at $t=7,m=2$, giving asymptotic growth $2.64^k$. There is nothing better at $m=3$, and $m=4$ is already quite hard to search but does not seem to have anything better.
The solutions for $t=7$ and $m=2$ essentially involve $x_1\equiv1,2,4$ and $x_2\equiv 3,5,6$. But this only deals with the "large scale" behavior, getting the necessary flexibility to get all the digits in the middle of a large target number. The actual problem requires that we get all numbers, though, which means in particular making sure that all the small numbers are accessible. For this we will have to break the pattern a bit at the large end (because our algorithm will push errors into the higher digits).
Put the unknown weight on the short arm. Let $c$ be the amount of half-pounds we have sitting on the short arm (or alternatively the number of pounds needed to go on the long arm to balance the scale). At the start this will be the number of half-pounds of the target, but it may be negative if we have too much weight on the long arm. Fix $k$; we have weights for $7^i$ and $5\cdot 7^i$ half-pounds for each $0\le i<k$, as well as two large weights of $7^k$ and $7^k/2$ half-pounds. (Yes, our weights are in half-pounds, and we have one quarter-integer weight. The rules don't say anything about this being disallowed.) Calculate $c\bmod 7$ to determine the use of the smallest weights:
If $c\equiv0\pmod 7$, don't use the $1$ and $5$ weights.
If $c\equiv1\pmod 7$, put $1$ and $5$ on the short arm (and carry 1, see below).
If $c\equiv2\pmod 7$, put $1$ on the long arm.
If $c\equiv3\pmod 7$, put $5$ on the long arm (and borrow 1).
If $c\equiv4\pmod 7$, put $5$ on the short arm and $1$ on the long arm (and carry 1).
If $c\equiv5\pmod 7$, put $1$ and $5$ on the long arm (and borrow 1).
If $c\equiv6\pmod 7$, put $1$ on the short arm (and carry 1).
After these placements, we can add the weight on the short arm to $c$ and subtract double what was put on the short arm, and then it will be divisible by $7$ and we can repeat the process with the next higher place. Many of the cases involve either carrying a 1 into the next position (meaning that we made the target weight heavier and so $c$ got larger in the next higher place), or borrowing 1 (meaning we overcompensated and now $c$ has gotten smaller and maybe gone negative).
We will aim to support up to $3\cdot 7^k$ by this method. Once we get to the highest place, $c$ will be one of $\{-1,0,1,2,3\}$ times $7^k$ half-pounds sitting on the short arm, and we finish as follows:
If $c=-7^k$, put the $7^k$ weight on the short arm.
If $c=0$, we're done.
If $c=7^k$, put the $7^k/2$ weight on the long arm.
If $c=2\cdot 7^k$, put the $7^k$ weight on the long arm.
If $c=3\cdot 7^k$, put the $7^k/2$ and $7^k$ weights on the long arm.
Since the carries and borrows only push $1$ or $-1$ into the next higher place, we can handle any number for $c$ in the range $0,\dots,3\cdot 7^k-1$; and we can also cover $3\cdot 7^k$ directly by using only the large weights, but the method breaks down at that point.
To summarize: This method can weigh any integer number of half-pounds up to $3\cdot 7^k$ using $2k+2$ weights. In particular, using $k=2$ we can weigh up to $73.5$ pounds in increments of one-half pound using $6$ weights.
Actually, there is some hope to disprove $k=4$, because the bound there is tight - there are $81$ distinct values to hit (including $0$ which you can't avoid hitting by putting nothing on the scale), and $81$ possible ways to arrange 4 weights. This means that every arrangement of weights must occur exactly once for a target weight, so in particular you have to measure the largest amount, 40 pounds, by putting all the weights on the long arm (so the weights sum to $20$ pounds). The 39.5 pound target must involve removing the smallest weight (so the smallest weight is $1/4$ pound), and then putting the smallest weight on the short arm will measure $39+3/8$ pounds, which is not a target. Thus $4$ weights is impossible.
It is much harder to disprove $5$ weights, because we now have a lot of freedom to choose which weights to put where for each measurement. However, we can at least show that no weight can be irrational, or ridiculously large. One characteristic of these type of solutions is that they are part of a family - we can perturb one of the test weights, possibly changing others to compensate, and we will still have a solution. One way to look at this is that the solution is a point in a 5-dimensional space, and we have solutions along a line. But if we had such a thing, then we could change the parameter until one of the weights was $0$, and then we would have a 4-weight solution which we have already shown is impossible.
This argument works generally - any optimal solution cannot be part of a continuous family, because you can use that freedom to eliminate one of the weights.
More precisely, what this shows is that every weight must be a quarter-integer of size no more than 80 pounds, because this is the requirement for the weight to participate in two different targets. (It is unlikely that it is not at least a half-integer or larger than 40 pounds, though, because these weights can only participate on one of the balances, which cuts down their usefulness.)
So, it's a finite problem, and by a computer search I've determined that there are no solutions for $k=5$, although it gets very close - I found the near-solution $1,4,7,32,40$, which hits every target except 27. Thus the solutions on this page with $k=6$ are optimal.
And to round this answer off, I also have the results of a computer search for the best solution with $k=6$ (meaning a solution which can cover the largest contiguous range):
$1,5,9,13,28,140$ pound weights can cover up to 88 pounds in half-pound increments. (I contest gnasher729's solution, which only reaches 75 pounds, not 101 as claimed.)
I have found the answer as:
6 weights
As example:
$2,4,8,16,24,25$
which is also the lowest total weight you can get. There are more than one answer for weights.
I have improved the lowest total weight required to
20 pound in total is enough to weigh all weights from 0 to 40 with 0.50 increment.
with the weights of
$0.25||0.25||1.00||2.50||6.25||9.75$
The solution from 0 to 40 is like below:
0.5 -> 0.25 |0.25 |1 | = 2.5 |
1 -> 0.25 |0.25 | =
1.5 -> 0.25 | = 1 |2.5 |
2 -> 0.25 |1 | = 0.25 |6.25 |
2.5 -> 0.25 |1 | =
3 -> 0.25 |0.25 |1 | =
3.5 -> 0.25 | = 0.25 |1 |6.25 |
4 -> 0.25 |0.25 |1 |2.5 | = 6.25 |9.75 |
4.5 -> 0.25 |2.5 | = 1 |
5 -> 0.25 |0.25 |2.5 | = 6.25 |9.75 |
5.5 -> 0.25 |0.25 |2.5 | = 1 |6.25 |9.75 |
6 -> 0.25 |0.25 |2.5 | =
6.5 -> 0.25 |0.25 |1 | = 6.25 |9.75 |
7 -> 0.25 |2.5 |6.25 | = 0.25 |1 |9.75 |
7.5 -> 0.25 |0.25 | = 6.25 |9.75 |
8 -> 0.25 |0.25 |1 |2.5 | =
8.5 -> = 1 |6.25 |9.75 |
9 -> 0.25 | = 2.5 |6.25 |9.75 |
9.5 -> 0.25 |6.25 | = 1 |2.5 |
10 -> 0.25 |0.25 |6.25 | = 1 |2.5 |
10.5 -> 0.25 |6.25 | = 2.5 |
11 -> 0.25 |0.25 |6.25 | = 2.5 |
11.5 -> 6.25 | = 1 |
12 -> 0.25 |6.25 | = 1 |
12.5 -> 0.25 |0.25 |6.25 | = 1 |
13 -> 0.25 |0.25 |1 |6.25 | = 2.5 |
13.5 -> 0.25 |0.25 |6.25 | =
14 -> 1 |6.25 | = 0.25 |0.25 |
14.5 -> 1 |6.25 | =
15 -> 0.25 |1 |6.25 | =
15.5 -> 0.25 |0.25 |1 |6.25 | =
16 -> 9.75 | = 1 |2.5 |
16.5 -> 0.25 |9.75 | = 1 |2.5 |
17 -> 0.25 |0.25 |9.75 | = 1 |2.5 |
17.5 -> 0.25 |0.25 |2.5 |6.25 | = 1 |
18 -> 0.25 |0.25 |9.75 | = 2.5 |
18.5 -> 0.25 |0.25 |2.5 |6.25 | =
19 -> 0.25 |9.75 | = 1 |
19.5 -> 0.25 |0.25 |9.75 | = 1 |
20 -> 0.25 |0.25 |1 |9.75 | = 2.5 |
20.5 -> 0.25 |0.25 |1 |2.5 |6.25 | =
21 -> 1 |9.75 | = 0.25 |0.25 |
21.5 -> 1 |9.75 | =
22 -> 0.25 |1 |9.75 | =
22.5 -> 0.25 |0.25 |1 |9.75 | =
23 -> 2.5 |9.75 | = 0.25 |0.25 |1 |
23.5 -> 2.5 |9.75 | = 1 |
24 -> 0.25 |2.5 |9.75 | = 1 |
24.5 -> 0.25 |0.25 |2.5 |9.75 | = 1 |
25 -> 0.25 |2.5 |9.75 | =
25.5 -> 0.25 |0.25 |2.5 |9.75 | =
26 -> 1 |2.5 |9.75 | = 0.25 |0.25 |
26.5 -> 1 |2.5 |9.75 | =
27 -> 0.25 |1 |2.5 |9.75 | =
27.5 -> 0.25 |0.25 |1 |2.5 |9.75 | =
28 -> 6.25 |9.75 | = 0.25 |0.25 |1 |2.5 |
28.5 -> 6.25 |9.75 | = 1 |2.5 |
29 -> 0.25 |6.25 |9.75 | = 1 |2.5 |
29.5 -> 0.25 |0.25 |6.25 |9.75 | = 1 |2.5 |
30 -> 0.25 |6.25 |9.75 | = 2.5 |
30.5 -> 0.25 |0.25 |6.25 |9.75 | = 2.5 |
31 -> 6.25 |9.75 | = 1 |
31.5 -> 0.25 |6.25 |9.75 | = 1 |
32 -> 0.25 |0.25 |6.25 |9.75 | = 1 |
32.5 -> 0.25 |0.25 |1 |6.25 |9.75 | = 2.5 |
33 -> 0.25 |0.25 |6.25 |9.75 | =
33.5 -> 1 |6.25 |9.75 | = 0.25 |0.25 |
34 -> 1 |6.25 |9.75 | =
34.5 -> 0.25 |1 |6.25 |9.75 | =
35 -> 0.25 |0.25 |1 |6.25 |9.75 | =
35.5 -> 2.5 |6.25 |9.75 | = 0.25 |0.25 |1 |
36 -> 2.5 |6.25 |9.75 | = 1 |
36.5 -> 0.25 |2.5 |6.25 |9.75 | = 1 |
37 -> 0.25 |0.25 |2.5 |6.25 |9.75 | = 1 |
37.5 -> 0.25 |2.5 |6.25 |9.75 | =
38 -> 0.25 |0.25 |2.5 |6.25 |9.75 | =
38.5 -> 1 |2.5 |6.25 |9.75 | = 0.25 |0.25 |
39 -> 1 |2.5 |6.25 |9.75 | =
39.5 -> 0.25 |1 |2.5 |6.25 |9.75 | =
40 -> 0.25 |0.25 |1 |2.5 |6.25 |9.75 | =
Weights of
1, 3, 9, 27
are enough. You are told that the thing to be weighed is a multiple of 0.5. So e.g. a weight of 8.5 can be recognised as being heavier than 8 but lighter than 9.
Case 1: Object is only on the long arm:
7 weights: 1,2,4,8,16,32,64
Case 2: Object is only on the short arm:
7 weights: 0.25,0.5,1,2,4,8,16
Case 3: Object could be on either arm:
9 weights: 0.25,0.5,1,2,4,8,16,32,64
Five weights are not enough, but with weights 4, 11, 17, 18 and 19 we can weigh anything from 0.5 to 38. So close, but not quite there. Finding the right way to weigh each weight is a puzzle in itself. For example, to weigh 1/2, put 4 + 19 on the short side, and 11 + 1/2 on the long side.
Since five weights are almost enough, six weights should cover a much larger range than 0.5 to 40 pounds. Actually,
We can weigh any weight from 0.5 to 101 pounds using weights 0.5, 2, 2.5, 4, 22 and 4.
Computer is checking right now whether there are solutions covering a larger range, but currently I have only checked integer and half pound weights. As another answer has shown, quarter pound weights might go even further. There is a better solution:
We can weigh any weight from 0.5 to 102 pounds using weights 0.5, 2, 2.5, 4, 37 and 44.5. On the long arm of the scales, they count as 1, 4, 5, 8, 74 and 89.
2 + 4 = 2 * 0.5 + 5
1 + 4 = 5
1.5 + 2.5 = 4
4 = 2 * 2
0.5 + 2 + 2.5 = 2 * 2.5
2 + 4 = 2 * 3
3.5 + 2 + 2.5 = 8
4 = 4
4.5 + 0.5 = 5
5 = 5
5.5 + 2.5 = 8
6 + 2 = 8
6.5 + 2.5 = 1 + 8
7 + 2 = 1 + 8
7.5 + 0.5 = 8
8 = 8
8.5 + 0.5 = 4 + 5
9 = 4 + 5
9.5 + 2.5 = 4 + 8
37 = 2 * 10 + 4 + 5 + 8
10.5 + 0.5 + 2 = 5 + 8
11 + 2 = 5 + 8
11.5 + 0.5 = 4 + 8
37 = 2 * 12 + 5 + 8
37 = 2 * 12.5 + 4 + 8
13 = 5 + 8
37 = 2 * 13.5 + 1 + 4 + 5
37 = 2 * 14 + 4 + 5
37 = 2 * 14.5 + 8
2 + 37 = 2 * 15 + 1 + 8
2 + 37 = 2 * 15.5 + 8
37 = 2 * 16 + 5
37 = 2 * 16.5 + 4
2 + 37 = 2 * 17 + 5
2.5 + 44.5 = 2 * 17.5 + 4 + 8
4 + 37 = 2 * 18 + 5
37 = 2 * 18.5
2 + 4 + 37 = 2 * 19 + 5
2.5 + 44.5 = 2 * 19.5 + 8
0.5 + 2.5 + 37 = 2 * 20
4 + 37 = 2 * 20.5
21 + 2 + 2.5 + 4 + 44.5 = 74
2 + 4 + 37 = 2 * 21.5
0.5 + 2.5 + 4 + 37 = 2 * 22
0.5 + 44.5 = 2 * 22.5
23 + 2.5 + 4 + 44.5 = 74
2.5 + 44.5 = 2 * 23.5
24 + 2.5 + 4 + 44.5 = 1 + 74
2 + 2.5 + 44.5 = 2 * 24.5
25 + 2 + 2.5 + 44.5 = 74
25.5 + 4 + 44.5 = 74
26 + 2 + 2.5 + 44.5 = 1 + 74
2 + 2.5 + 4 + 44.5 = 2 * 26.5
27 + 2.5 + 44.5 = 74
27.5 + 2 + 44.5 = 74
28 + 0.5 + 2 + 4 + 44.5 = 5 + 74
28.5 + 2 + 4 + 44.5 = 5 + 74
29 + 0.5 + 44.5 = 74
29.5 + 44.5 = 74
30 + 0.5 + 4 + 44.5 = 5 + 74
30.5 + 4 + 44.5 = 5 + 74
31 + 2.5 + 44.5 = 4 + 74
31.5 + 4 + 44.5 = 1 + 5 + 74
32 + 0.5 + 2 + 44.5 = 5 + 74
32.5 + 2 + 44.5 = 5 + 74
33 + 2 + 2.5 + 44.5 = 8 + 74
33.5 + 44.5 = 4 + 74
34 + 0.5 + 44.5 = 5 + 74
34.5 + 44.5 = 5 + 74
35 + 2.5 + 44.5 = 8 + 74
35.5 + 2 + 44.5 = 8 + 74
2.5 + 37 + 44.5 = 2 * 36 + 4 + 8
0.5 + 37 + 44.5 = 2 * 36.5 + 4 + 5
0.5 + 37 + 44.5 = 2 * 37 + 8
37.5 + 44.5 = 8 + 74
2.5 + 37 + 44.5 = 2 * 38 + 8
38.5 + 44.5 = 4 + 5 + 74
2 + 2.5 + 37 + 44.5 = 2 * 39 + 8
0.5 + 2 + 37 + 44.5 = 2 * 39.5 + 5
2.5 + 37 + 44.5 = 2 * 40 + 4
40.5 + 2 + 44.5 = 5 + 8 + 74
0.5 + 37 + 44.5 = 2 * 41
41.5 + 44.5 = 4 + 8 + 74
2.5 + 37 + 44.5 = 2 * 42
42.5 + 44.5 = 5 + 8 + 74
2 + 2.5 + 37 + 44.5 = 2 * 43
43.5 + 2 + 2.5 + 4 + 37 = 89
2.5 + 4 + 37 + 44.5 = 2 * 44
2 + 2.5 + 4 + 37 + 44.5 = 2 * 44.5 + 1
2 + 2.5 + 4 + 37 + 44.5 = 2 * 45
45.5 + 2.5 + 4 + 37 = 89
46 + 2 + 4 + 37 = 89
46.5 + 44.5 = 4 + 5 + 8 + 74
47 + 0.5 + 2 + 2.5 + 37 = 89
47.5 + 2 + 2.5 + 37 = 89
48 + 4 + 37 = 89
48.5 + 2 + 2.5 + 37 = 1 + 89
49 + 0.5 + 2.5 + 37 = 89
49.5 + 2.5 + 37 = 89
50 + 2 + 37 = 89
50.5 + 0.5 + 2 + 4 + 37 = 5 + 89
51 + 2 + 4 + 37 = 5 + 89
51.5 + 0.5 + 37 = 89
52 + 37 = 89
52.5 + 0.5 + 4 + 37 = 5 + 89
53 + 4 + 37 = 5 + 89
53.5 + 2.5 + 37 = 4 + 89
54 + 4 + 37 = 1 + 5 + 89
54.5 + 0.5 + 2 + 37 = 5 + 89
55 + 2 + 37 = 5 + 89
55.5 + 2 + 2.5 + 37 = 8 + 89
56 + 37 = 4 + 89
56.5 + 0.5 + 37 = 5 + 89
57 + 37 = 5 + 89
57.5 + 2.5 + 37 = 8 + 89
58 + 2 + 37 = 8 + 89
58.5 + 2.5 + 37 = 1 + 8 + 89
59 + 2 + 37 = 1 + 8 + 89
59.5 + 0.5 + 37 = 8 + 89
60 + 37 = 8 + 89
60.5 + 0.5 + 37 = 4 + 5 + 89
61 + 37 = 4 + 5 + 89
61.5 + 2.5 + 37 = 4 + 8 + 89
62 + 37 = 1 + 4 + 5 + 89
62.5 + 0.5 + 2 + 37 = 5 + 8 + 89
63 + 2 + 37 = 5 + 8 + 89
63.5 + 0.5 + 37 = 4 + 8 + 89
64 + 37 = 4 + 8 + 89
64.5 + 0.5 + 37 = 5 + 8 + 89
65 + 37 = 5 + 8 + 89
65.5 + 2 + 2.5 + 4 = 74
66 + 37 = 1 + 5 + 8 + 89
66.5 + 2 + 2.5 + 4 = 1 + 74
67 + 0.5 + 2.5 + 4 = 74
67.5 + 2.5 + 4 = 74
68 + 2 + 4 = 74
68.5 + 0.5 + 37 = 4 + 5 + 8 + 89
69 + 37 = 4 + 5 + 8 + 89
69.5 + 2 + 2.5 = 74
70 + 4 = 74
70.5 + 2 + 2.5 = 1 + 74
71 + 0.5 + 2.5 = 74
71.5 + 2.5 = 74
72 + 2 = 74
72.5 + 0.5 + 2 + 4 = 5 + 74
73 + 2 + 4 = 5 + 74
73.5 + 0.5 = 74
74 = 74
74.5 + 0.5 + 4 = 5 + 74
75 + 4 = 5 + 74
75.5 + 2.5 = 4 + 74
76 + 4 = 1 + 5 + 74
76.5 + 0.5 + 2 = 5 + 74
77 + 2 = 5 + 74
77.5 + 2 + 2.5 = 8 + 74
78 = 4 + 74
78.5 + 0.5 = 5 + 74
79 = 5 + 74
79.5 + 2.5 = 8 + 74
80 + 2 = 8 + 74
80.5 + 2 + 2.5 + 4 = 89
81 + 2 = 1 + 8 + 74
81.5 + 0.5 = 8 + 74
82 = 8 + 74
82.5 + 2.5 + 4 = 89
83 + 2 + 4 = 89
83.5 + 2.5 = 4 + 8 + 74
84 + 0.5 + 2 + 2.5 = 89
84.5 + 2 + 2.5 = 89
85 + 4 = 89
85.5 + 0.5 = 4 + 8 + 74
86 = 4 + 8 + 74
86.5 + 2.5 = 89
87 = 5 + 8 + 74
87.5 + 0.5 + 2 + 4 = 5 + 89
88 + 2 + 4 = 5 + 89
88.5 + 0.5 = 89
89 = 89
89.5 + 0.5 + 4 = 5 + 89
90 + 4 = 5 + 89
90.5 + 2.5 = 4 + 89
91 = 4 + 5 + 8 + 74
91.5 + 0.5 + 2 = 5 + 89
92 + 2 = 5 + 89
92.5 + 2 + 2.5 = 8 + 89
93 = 4 + 89
93.5 + 0.5 = 5 + 89
94 = 5 + 89
94.5 + 2.5 = 8 + 89
95 + 2 = 8 + 89
95.5 + 2.5 = 1 + 8 + 89
96 + 2 = 1 + 8 + 89
96.5 + 0.5 = 8 + 89
97 = 8 + 89
97.5 + 0.5 = 4 + 5 + 89
98 = 4 + 5 + 89
98.5 + 2.5 = 4 + 8 + 89
99 = 1 + 4 + 5 + 89
99.5 + 0.5 + 2 = 5 + 8 + 89
100 + 2 = 5 + 8 + 89
100.5 + 0.5 = 4 + 8 + 89
101 = 4 + 8 + 89
101.5 + 0.5 = 5 + 8 + 89
102 = 5 + 8 + 89
Not the complete answer but the idea would be:
For the integer valued weights ( 1, 2, 3, . . . 40 ) We can use left or right side. But for the fractioned ones we only can use the right side ( or else balancing weight would be .25 , which we don't have ) So for the fractioned ones we need 1, 3, 5, 7, 9, . . . 79, for others we can have .5 or 2, 1 or 4, 1.5 or 6 . . . 20 or 80.
