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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.
no one got an exact answer
3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.
how would u check your answer
That was stated on the link I gave you.
yea but .....nvm ill just go with what they got but link did help thnx for your support
I hate word problems lol
(sigh) IM DONE thnx again
Theorem: You'll need weights 3^0 through 3^N to cover values 1 through S(N) = sum(3^i) for i = 0 to N. Proof: You've given the base case where N = 1. Now assume this holds for N < M. For the case N = M we'll have weights 3^0=1 through 3^M which we already know covers values up to S(M-1). Consider that by trading sides for each weight on the scale we can express all negative values down to -S(M-1) with these same weights as well. This it will be sufficient to prove that we can express values S(M-1) + 1 through S(M) as 3^M + X where -S(M-1) <= X <= S(M-1). But S(M) = S(M-1) + 3^M so this is clear provided that S(M-1) + 1 >= 3^M - S(M-1). That is, if 3^M <= 1 + 2 * S(M-1) = 1 + sum(2 * 3^i) for i = 0 to M-1. This seems clear to me at the moment, but I've had a few wingspantails and a proof wasn't really what you were asking for anyway, so I'll leave this final step as an exercise to the reader. By induction, QED. From: http://stackoverflow.com/questions/2589986/puzzle-find-the-minimum-number-of-weights Seems the clearest explanation I can think of.