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Challenge:Puzzle

Discussion in 'Competitions, Contests and Challenges' started by Darkstar, Oct 22, 2014.

  1. Darkstar

    Darkstar Member

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    The answer is on the internet, so there's no prize (& strictly speaking no challenge) apart from the warm feeling that you get from getting the answer for yourself.
    Some of you may have seen this before "elsewhere" too.

    Inspired by the marbles/ball bearings thread:

    You have 12 marbles, all with different markings. All of them, apart from one, are exactly the same weight.
    I will let you use my scales (see picture) but only 3 times. After that you must tell me which marble is not the same weight as the others & whether it is heavier or lighter.
    (For clarification, the difference in weight is too small for you to feel & the description of the solution must work for all eventualities)
    [​IMG]
     
    #1
  2. Saint-Just

    Saint-Just Moderator Staff Member

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    It's easy in 4, or in 3 if we know lighter or heavier. But in 3 without knowing, must still research the subject :)
     
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  3. Saint-Just

    Saint-Just Moderator Staff Member

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  4. MaC

    MaC Moderator Staff Member

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    Divide in three.

    Four on each pan and four off.

    If the pans balance, then the four off must contain the heavy one.
    If the pans don't balance then the one that's lowest must contain the heavy one.

    Take the four that contains the heavy one and seperate into twos.

    Weigh again.
    Lowest pair contain the heavy one.

    Seperate those two and weigh again.
    Lowest pan contain the heavy one.

    Three uses of the scales. Haven't worked out how to do it if it's a lighter one though.

    MaC
     
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  5. Darkstar

    Darkstar Member

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    Light would be the same as heavy (just reverse the way you choose the groups)... if you knew if the marble that you wanted was either heavy or light.

    You aren't told which of the 2 scenarios you are dealing with, so must plan for both options (i.e. the odd-one-out could be light OR heavy) within the 3 weighings.
    FWIW I don't think that you can do it with 2 weighings even if you are told if it's heavy or light.

    I like the start though.
     
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  6. Saint-Just

    Saint-Just Moderator Staff Member

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    Yes my solution starts like Mary's.
     
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  7. Emrys

    Emrys Member

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    If you know the odd one is heavier, or lighter it is theoretically possible in two weighings.

    Start with 5 on each side.

    If both sides are equal then one of the remaining must be the odd one. put one on either side. sorted!

    If both sides are not equal you have two goes to find heaviest/lightest of the 5
     
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  8. Saint-Just

    Saint-Just Moderator Staff Member

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    In theory it can be done in one, if you just put one in each tray and the odd one is one of the 2.

    But the point here is that you need to do it in 3 max, without knowing if the odd one is heavier or lighter.
     
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  9. Renton

    Renton Subscribed Member

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    If you managed to pick two first go that didn't match you wouldn't know which of the two was the odd one out - you'd have to check one of them against another ball.
     
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  10. Saint-Just

    Saint-Just Moderator Staff Member

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    No. I was replying to Emrys, who had assumed he knew whether the odd ball was lighter or heavier.

    The solution of the problem is different.
     
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  11. happy camper

    happy camper Member

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    I can do it in 3, but not for all eventualities (I still seem to need 4 weighings if the heavy/light one happens to be in one of the first groups to be weighed).
     
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  12. Darkstar

    Darkstar Member

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    There are different answers, depending on how each trial goes. You've got the first branch sorted by the sounds of it, now you have the information from the first weighing & 2 more trials left to deal with the branch where the scales don't balance the first time.
    Obviously, the result of the 2nd weighing must leave you able to deal with those branches so that you complete your analysis with the last trial.
     
    #12

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