Weighing Trouble

Note that $7$ weights work. Let the weights be $1,3,9,27,81,243,729$ --the powers of $3$ . We can represent the weight we want to measure in balanced ternary --a positional numeral system that uses the digits $-1,0,1$ instead of the usual $0,1,2$ for ternary. Then, for every digit $1$ , we put the appropriate weight on the opposite side, while for every digit $-1$ , we put the appropriate weight on the same side.

For example, $23_{10} = 10TT_{bt}$ , where $bt$ stands for "balanced ternary" and $T$ stands for the digit $-1$ . Thus to measure a weight of $23$ units, we put the $1$ -unit weight on the same pan (because the first digit from the right is $T$ ); we put the $3$ -unit weight also on the same pan (the second digit is $T$ ); we don't use the $9$ -unit weight (the third digit is $0$ ); and we put the $27$ -unit weight on the opposite pan (the fourth digit is $1$ ). Checking that this is a correct measurement is left to the reader.

The largest weight that can be measured is $1111111_{bt} = 1093_{10}$ , so we can measure up to $1093$ units.This is more than enough.

Now, suppose that we have only $6$ weights. Then there are at most $3^6 = 729$ cases that we can distinguish: each weight has $3$ places to put (on the same pan, on the opposite pan, or not on the balance), and there are $6$ weights. But we need to distinguish $1000$ cases, impossible. Thus $6$ weights don't suffice, and clearly neither do a smaller number of weights. Thus we have proven that $\boxed{7}$ weights is the minimum.

Further reading: Wikipedia

The weights are

1,3,9,27,84,243,729

So a total of 7

The weights intrestingly come out to be a GP of common ratio 3 with the first term as 1

We can generalize the problem for $1$ to $n$ , then have the solution to the generalized problem as follows.

Consider the recurrence relation $f_p=3f_{p-1}+1$ , for $p\geq1$ , with the initial condition $f_0=0$ .

Then it can be proved that

The minimum number of weights that you would need to be able to measure all (integral) weights from 1 kg to $n$ kg is the minimum positive integer $p$ such that $f_p\geq n$ ; suggesting the $a$ th weight (in increasing order) is $\left({2f_{a-1}+1}\right)$ kg, for ( $a\geq1$ ).

As here $n=1000$ , with $3^6<n<3^7$ , I didn't bother to find a closed form of the recurrence relation.

Now, $f_0=0$ , $f_1=1$ , $f_2=4$ , $f_3=13$ , $f_4=40$ , $f_5=121$ , $f_6=364$ , $f_7=1093$ ... ... ...

$\boxed{7}$ is the answer.

And the weights are: $w_1=1$ , $w_2=3$ , $w_3=9$ , $w_4=27$ , $w_5=81$ , $w_6=243$ , $w_7=729$ , but for $n=1000$ , any value of $w_7$ with $517\leq w_7 \leq 729$ works.