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anonymous
 one year ago
!!THIS IS WHY I HATE MATH++!!
anonymous
 one year ago
!!THIS IS WHY I HATE MATH++!!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/53944eafe4b01c619df12792

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no one got an exact answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would u check your answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was stated on the link I gave you.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea but .....nvm ill just go with what they got but link did help thnx for your support

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I hate word problems lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(sigh) IM DONE thnx again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Theorem: 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(M1). Consider that by trading sides for each weight on the scale we can express all negative values down to S(M1) with these same weights as well. This it will be sufficient to prove that we can express values S(M1) + 1 through S(M) as 3^M + X where S(M1) <= X <= S(M1). But S(M) = S(M1) + 3^M so this is clear provided that S(M1) + 1 >= 3^M  S(M1). That is, if 3^M <= 1 + 2 * S(M1) = 1 + sum(2 * 3^i) for i = 0 to M1. 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/puzzlefindtheminimumnumberofweights Seems the clearest explanation I can think of.
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