You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.

Question

anonymous · Answer

If you did denominations as powers of two, then you'd need one of each denomination.

anonymous · Answer

And the number of denominations would be $\log_2(1000)$

anonymous · Answer

Doing it as powers of three, you'd need 2 of each denomination, and the number of denominations would be $\log_3(1000)$

tkhunny · Answer

However, we've already seen that 1-4 takes only 1 and 3.  This is a better solution than 1, 2, 4.  Using powers of 2 will be a solution, but not a mimimum solution, as it considers only additive relationships.

anonymous · Answer

oh wait...

anonymous · Answer

If you did it as powers of three, you'd only need 1 of each denomination...

anonymous · Answer

@tkhunny  Do you think powers of 3 is optimal then?  That's what I seem to be getting.
Though I haven't considered inconsistent denominations.

tkhunny · Answer

Probably not.  Too simple.  I'll have to give it some thought.

anonymous · Answer

Well, $ \lceil \log_3(1000) \rceil = 7$ so you'd have to get like 6 weights to beat it.

anonymous · Answer

It's the optimal 'powers of' solution.
So you'd need a solution which has some other method of allotting denominations.

tkhunny · Answer

Okay, I give.  I can't prove it, yet, but I was thinking that 1,3 being better than 1,2,4 for the values 1-4 was just an odd coincidence.

It is clear to me now that $3^{k}$ is the maximum prime base that will cover the low values.

It is clear to me now that values have exactly three states.
1) Plus Side
2) Minus Side
3) Omitted

I no longer believe that 3 is a coincidence.  I still can't quite prove it, so officially I'm holding out for some exotic solution.  At the moment, I'm not seeing such a solution.