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What is the smallest number of weights needed for weighing objects with masses 1g, 2g, ..., 35g?

I went this way: for 1g one needs 1g. For 2g one has to add 2g. For 3g no adding is needed. But for 4g, there are two ways: adding another 2g weight or adding 4g weight. Both ways ultimately lead to 4 weights for weighing 1g,2g,...,8g. Going on from that point up to being able to weigh all 1g, 2g, ..., 35g objects with the minimum number of weights, each time results in 7 objects. I was wondering if

1- There is a general trick to solve such problems because it is a problem in 7th grade math olympiad with ~100sec/problem.

2- Is 7 weights the ultimate minimum or less is possible?

Options to choose are 35,10,7,6,none.

2 Answers2

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Extending the answer of bjcolby15:

It is assumed that all of the weights are only allowed on one side of the scale and the object to be weighed is on the other side of the scale.

Any positive integer can be expressed in base $2$ format. In such a format, each digit in the expression will be either $1$ or $0$. This bijects to determining which weights to use to weigh the object.

For example, $7$ equals $111$, written in base $2$. This corresponds to using the three weights of $4$, $2$, and $1$ to equal a weight of $7$. Note, that under the assumption that each weight is either used on a specific scale or not, there are only $2$ choices for each weight. Either use it or not.

This is why base $2$ is so relevant. Because for each digit in a base $2$ representation, either the digit is $0$ (which corresponds to not using the weight) or $1$ which corresponds to using the weight.

See also, the Addendum below, which discusses an alternative assumption re the use of the weights.

Addendum
Changing the assumption of how the weights are to be used:

Suppose instead, that you place an object on the left scale, and that the weights may be placed on either the left scale or the right scale. This means that instead of having only $2$ choices for how to use each weight, you have $3$ choices:

  • Use the weight on the left scale, along with the object to be weighed. Imagine assigning the number $-1$ to the weight here.

  • Use the weight on the right scale, opposite the object to be weighed. Imagine assigning the number $+1$ to the weight here.

  • Don't use the weight at all. Imagine assigning the number $0$ to the weight here.

So, you have $3$ choices. This roughly corresponds to expressing numbers in a variant of the normal base $3$ format: instead of using the digits $0,1,2$, you use the digits $0,+1,-1$.

This is why (for example) $4$ weights are sufficient to weigh any object up to $40$. Because the weights of $1,3,9,27$ are used. Note that $\displaystyle 40 = \frac{3^4 - 1}{2}.$

user2661923
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Using base two (binary) as the weight, i.e. $1 = 2^0, 2 = 2^1, 4 = 2^2, 8 = 2^3, 16 = 2^4, 32 = 2^5$, we can measure any weights we want. With $5$ weights, you can only weigh up to $31$g, but with $6$ weights, you can measure up to $63$g. Hence the minimum number of weights you need to measure up to $35$g is $6.$

bjcolby15
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  • I still don't know WHY powers of two make the minimum number? I.e. why any other collection may not result in less number? –  Jan 17 '22 at 20:41
  • I used Geometry also to understand it but failed –  Jan 17 '22 at 20:44
  • Your answers are $0, 6, 7, 10$, and $35$. By process of elimination, $0, 10$ and $35$ are incorrect, which leaves us with $6$ and $7$. The only way you could make those weights from $1$ to $35$g is if you used a power of $2$.You cannot make anything up to $35$g because the total of all five weights is $31 (1 + 2 + 4 + 8 + 16)$ so you need a $32$g weight to do this.

    You could also incorporate a $64$g weight to make $7$, but since we're looking for the minimum amount of weights, $6$ is correct answer.

     – bjcolby15 Jan 18 '22 at 00:13
  • Thank you. Btw 0 is not an answer! None means none of the above answers are correct. –  Jan 18 '22 at 11:08
  • My mistake - I thought zero and none were interchangeable :-) – bjcolby15 Jan 18 '22 at 11:42