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Math Help - Coins

  1. #1
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    Coins

    You have 10 boxes with 10 coins each. All the coins from 9 of those boxes(all 90 of them) all weigh 1g. However the other box has coins which all weigh 1.1g. You have a scale that measures the weight of the load you put on. How can you determin which box has the 1.1g coins just by weighing one load of any combination of these 100 coins?
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  2. #2
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    Quote Originally Posted by usagi_killer View Post
    You have 10 boxes with 10 coins each. All the coins from 9 of those boxes(all 90 of them) all weigh 1g. However the other box has coins which all weigh 1.1g. You have a scale that measures the weight of the load you put on. How can you determin which box has the 1.1g coins just by weighing one load of any combination of these 100 coins?
    Put one coin of the first box, two coins of the second box,..., ten coins of the tenth box on the scale. You should easily see why this works.
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  3. #3
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    Sorry can you explain why it works?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by usagi_killer View Post
    Sorry can you explain why it works?
    Can you make in into an equation?

    The weight of this pile will be w=1*x_1+2*x_2+ \ldots + 10*x_{10}, with x_i being the weight of one coin from the i^{th} box. So, what will the weight be precisely? Substitute in the x_i's, and what do you get?
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  5. #5
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    Quote Originally Posted by usagi_killer View Post
    You have 10 boxes with 10 coins each. All the coins from 9 of those boxes(all 90 of them) all weigh 1g. However the other box has coins which all weigh 1.1g. You have a scale that measures the weight of the load you put on. How can you determin which box has the 1.1g coins just by weighing one load of any combination of these 100 coins?
    I would have thought with one weighing its impossible (except if you get lucky)

    CB
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    I would have thought with one weighing its impossible (except if you get lucky)

    CB
    This is also an interesting question:
    The Ultimate Puzzle Site - Brain-Teasers
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  7. #7
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    Quote Originally Posted by Laurent View Post
    Put one coin of the first box, two coins of the second box,..., ten coins of the tenth box on the scale. You should easily see why this works.
    Ahh.. it weighs the coins, I was reading what I expected to read .. that it was a balance.

    CB
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by malaygoel View Post
    This is also an interesting question:
    The Ultimate Puzzle Site - Brain-Teasers

    My solution is a bit different from theirs:

    Spoiler:
    Split the coins into three sets of 4 coins, and compare two of these on your balance. Either one set which you weigh and is heavier, or the two sets you weigh are equal and so the other set has the odd coin. That is one weigh, and you know the odd coin is in a specific set of 4 coins. Solving the problem for 4 coins and two weighs is arbitrary, and so we are done.
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    Smile thanx

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