KCC's Quizzes AQQ292 about weighing using 2-plates scale

I think you need 6 weights: 1, 2, 3, 5, 10 and 20 grams.

In simple binary system we can make any number with combination of 2's powers so it could be 1,2,4,8,16,32 with these combination we can achieve any weight. 
If we want minimum we can try for 3 as well like 1,3,9,27 or may be 4 also 1,4,16,32 but with 4 we can't make 3 or other so no need to try more
Then 1,3,9,27 will be the minimum possible numbers to achieve the weight example 26 - (27-1) one with 27 and other with 1 example 40 then will keep all four at one side. 
Maximum achievable weight with this combination will be 40

My guess is you need 6 weights of 1, 2, 4, 8, 16, 32 gr each.  

I hope that the weights can also be placed on a dish with a weighed object.
The weights must have an odd value, so their sum can reach both odd and even numbers. The first two are definitely 1g and 3g. The first higher value above their sum is 5g, so this can be achieved by the combination 9g-3g-1g. And again the first highest unattainable for these 3 weights is the value 1g+3g+9g+1g = 14g, so the next weight value is these 14g and the sum of everything = 27g.

The question is whether we can create all the combinations now. I don't have to examine even numbers and the same goes for the exact weight values.

• 5g +3g+1g= 9g
• 7g +3g = 9g+1g
• 11g +1g = 9g+3g
• 13g = 9g+3g+1g
• 15g +3g+9g = 27g
• 17g +1g+9g = 27g
• 19g +9g = 27g+1g
• .
• .
• 39g = 27g+9g+3g
  
  All combinations are available. Thus we need 4 counterweights: 1g, 3g, 9g, 27g

 Concept: Balanced Ternary System

In a two-dish scale, you can place counterweights on both sides of the scale. This allows you to simulate subtraction as well as addition. So instead of just summing weights, you can balance them by placing some on the same side as the substance and some on the opposite side.

Balanced ternary uses digits:

• −1 (weight on the same side as the substance)
• 0 (weight not used)
• +1 (weight on the opposite side)

This system allows you to represent any integer using powers of 3.

 Solution

a. Minimum number of counterweights needed:

We need to find the smallest set of weights such that any integer from 1 to 40 can be represented using combinations of these weights on either side of the scale.

Let’s use powers of 3:

Now check the maximum representable value using balanced ternary with weights up to 3^3=27:

• Maximum value =
  1+3+9+27=401
  (when all weights are placed on the opposite pan)

So, 4 counterweights are sufficient.

b. Weight values:

The counterweights should be:

• 1g
• 3g
• 9g
• 27g

Thanks MTF_Walker for your nice development. But there is a solution with less counterweights! One needs also to rethink the way how items are put on the plates... May be you want to retry?

Excellent Sona22 , you get it! Congratulations!

Thanks brianb0722 . Your solution will work, but there is a solution with even less elements! May be you can retry?

YES, the minimum set is with 4 elements 1, 3, 9 and 27. Congratulations! But why you said even weights cannot be determined? Look, for example X=20 can be obtained by putting X + 9 + 1 in one plate and 3 + 27 in the other plate !

Well done RevSmeeAgain , you solve the quiz! Congratulations!