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Take 1 coin from one barrel 2 from next .....hope you got it
take 1 coin from first barrel, 2 coins from second, 4 coins from third, 8; 16; 32; 64; 128; 256; 512. weight will be from 1023 to 2046 included. if the answer is more than (2046+1+1023)/2=1535 there is 2 gram coins in tenth barrel, if less there will be 1 gram coins. for example it was more, now we have answer from 1535 to 2046. if the answer is more than (2046+1+1535)/2=1791 there is 2 gram coins in ninth barrel, if less 1 gram coins and so on...
You don't need any weighting, just look into these barrels. The 2-gram one will have half the coins.
I don't understand this.... Do the barrels have the same value?
The simplest solution would be to grab 4 or 5 coins from each barrel. Once each set of 4 or 5 coins has been removed compare them by placing separate sets in each hand. Compare 1&2, 3&4, ..., 9&10. Among the 5 comparisons, which set had felt unbalanced. From this set, choose the respective barrel which contained the heavier coin, this barrel is filled with 2 gram coins. This way you do not need the machine at all, it is important to consider your resources even if they are not explicitly stated in the question, ie- your hands.
I don't get it. Is there information missing from the question? I don't understand the limitations or what we're trying to do here..
More Info It seems, it will approach.
If it's faster to take the coins from the barrels and you want to reduce the calculations needed to find out the end result, it would be easier to take 1 coin form then first barrel, 10 from the second barrel and so on.When you see the result, which would be something like this: 2122121112Then you know that the ones with a 2 have 2 gram coins in it and the ones with 1 have 1 gram coins in it.
@JyRKS - good luck trying to take 1,000,000,000 from the 10th barrel...Giorgi Leonidze is correct.
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