Let's be the weights $a_1<a_2<a_3<a_4$ . Each weights has one position from these $+\;-\;0$ :

• $+$ means that we add this weight
• $-$ means that we substract this weight
• $0$ means that we don't use this weights

Now we have $4$ weights with $3$ positions, so $3^4=81$ , but we should see only positive sums, so we have $\cfrac{81-1\text{(zero)}}{2\text{(negative numbers)}}=40$ . Therefore with these weights we can calculate weigths only between $1$ and $40$ . Therefore:

$\begin{array}{c|c|c|c|c} a_1&a_2&a_3&a_4&\text{sum}\\ \hline +&+&+&+&40\\ 0&+&+&+&39\\ -&+&+&+&38\\ +&0&+&+&37\\ 0&0&+&+&36\\ -&0&+&+&35\\ +&-&+&+&34\\ \vdots&\vdots&\vdots&\vdots&\vdots \end{array}$

Now there are the main rows:

$\begin{array}{c|c|c|c|c} a_1&a_2&a_3&a_4&\text{sum}\\ \hline +&0&0&0&1\\ 0&+&0&0&3\\ 0&0&+&0&9\\ 0&0&0&+&27\\ \end{array}$

So the weights are: $1,3,9,27$ .

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Or:

It will be a numeral system: $(a_4a_3a_2a_1)_3$ . Therefore the solution is $27,9,3,1$ .

Consider the base-3 number system, with the first $4$ digits $(3^3, 3^2, 3^1, 3^0)$ which can represent from $0$ to $80$ .

• The number from $0$ to $80$ is represented by the $4$ digits, each digits with $3$ choices $(0,1,2)$
• Define a $1-1$ mapping $\color{#3D99F6}{h}$ by changing the $3$ choices $x \in \{0,1,2\}$ to $y=x-1 \in \{-1,0,1\}$
• This mapping is just changing the original number $m \in (0, 80)$ to $n = m-40 \in (-40, 40)$
• Now, each digit is regarded as a weights $a=3^0=1, b=3^1=3, c=3^2=9, d=3^3=27$
• and each choice represents how to operate the measuring weight to get the desire weigh
• $-ve$ means put it together with the rice (object) on the LHS , $0$ - means put it aside, $+ve$ - means put the weight to the RHS of the balance

for example to measure $\color{#3D99F6}{(15)}$ kg,

$15_{10} \rightarrow 15+40 = 55_{10} \rightarrow 2001_3 \rightarrow^{\color{#3D99F6}{h}} 1(-1)(-1)(0) = 27 - 9 - 3 + 0 = 15 \\ [ \begingroup\color{#3D99F6}(15)\endgroup + 9 + 3 ] = [27]$

The first piece of weight must be $1$

Then with the two weights $1$ and $39$ we can measure $1,38,39,40$ ( $1$ on the left pan and $39$ on the right pan of the balance will measure $38$ , while both the weights on the right pan will measure $40$ ). To measure $3$ and $37 (=40-3)$ , another piece of $3$ must be available. With these three, viz. $1,3,36$ , we can measure $1,2,3,4,32,33,34,35,36,37,38,39,40$ . To measure $5$ and $31 (40-9)$ , we need a third piece of weight $9$ . Thus the four pieces of weights $1,3,9,27$ measures all weights from $1$ to $40$ .

If we were only allowed to put weights on one side, the result should be written in binary, using 1, 2, 4, 8, 16 ... weights. (To put each weight or not).

Here we have one more option : to put weights on the other side. Hence we are dealing with ternary approach. To put a weight (+) or not (0), OR to put it on the other side (-).

As all weights sum up 40, a must be equal to 1 in order to sum up 39. (All weights without a)

38 is made by putting a on the other side. (All weights minus a)

37 is made by removing b which must be 3. (All weights without b)

And so forth ...

36 = all without a and b

As c + d = 36, 35 = c + d - a

34 = c + d - b + a

33 = c + d - b

32 = c + d - b - a

Now a and b have made their maximum to lighten c+d. We have to remove c 31 = d + b + a Then c = 9

And d = 27

Answer : a + d =28

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Other solution

As all weights sum up 40, a must be equal to 1 in order to sum up 39. (with all weights except a).

a = 1

with a the possible values are -1, 0 or 1.

Now let's try with b = 2:

If b = 2 : the possible weight values with b are -2, 0 or 2. The possible weight values with a and b are (by mixing -1, 0, 1 and -2, 0 2) -3, -2, -1, 0, 1, 2, 3

If b = 3 the possible weight values with b are -3, 0, 3 The possible weight values with a and b are -4, -3, -2, -1, 0, 1, 2, 3, 4 IT'S BETTER than with b=2

If b = 4, it's impossible to make a weight of 2 with the help of a

So b = 3

And so forth : 9 is the maximum possible value for c. With c = 10, it's impossible to make 5 with the help of a and b

c = 9

Hence d = 27