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anonymous
 4 years ago
Divide 40 kg mass into 4 piece such that you can weigh from 1 to 40 kg from those 4 masses
anonymous
 4 years ago
Divide 40 kg mass into 4 piece such that you can weigh from 1 to 40 kg from those 4 masses

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vishal_kothari
 4 years ago
Best ResponseYou've already chosen the best response.8what's the right answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ans to nahi pata to he post kara he lekin 1,2,19,18 nahi he

vishal_kothari
 4 years ago
Best ResponseYou've already chosen the best response.8ho bhi nahi sakta?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i am not really understand wat the question want >.>

vishal_kothari
 4 years ago
Best ResponseYou've already chosen the best response.8himanshu explain the question..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yaar 40kg ka weight ko 4 piece me torna he jisse 1kg se 40 kg ke sabhi mass 1 ya 1 se jada piece ka use karke sabhe ko weigh karna he

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0hmm don't think this is right either.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i assure u it is. the problem is an old one from Bachet's 'Problems plaisants et delectables'

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0well then tell me how with this combination u can get a weight of 2,5,7,8....... ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you have a point there, the version i remember did not place a restriction of 4 weights to the solution.

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0ya i guess this restriction is just the problem here

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.0You can weigh a 2 by putting a three on one side, and a one on the other. Ditto for 5 (9 vs 3+1)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's an old problem from pre revolution France. The conventional way of thinking was that the mass being weighed on the balance had to be measured with a number of the weights on the opposite side. The solution here is to realise that you can also add weights to the side being weighed. So 1 is weighed conventionally but 2 is weighed with a 3 on one side and a 1 being added to the side being weighed, etc 1, 1 2, 3, 1 3, 3 4,3,1 5,9,3,1 6,9,3 7,9,1,3 8,9,1 9,9
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