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Who is the most logical and intellectual on TSR??? Watch

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    Hi guys, i thought to make this thread because i think itll be interesting for some of you. basically here is a problem, solving which is very interesting!

    there are 12 coins. 11 of them are golden but 1 of them is fake - they all look the same. the 11 golden coins weigh exactly the same but the fake one has a different weight - either heavier or lighter - you dont know this. how can you know which one of the 12 is the fake, having a two sided scales (the one that doesnt show the weight in number, just weighs two items and tells which one is heavier/lighter) and having only three chances to weight any combination of the coins?

    This is a very interesting problem, i solved in in three days and nights I know not all of you will enjoy this but im sure some you will, so please do write you thoughts.

    P.S. - this is not a stupid problem, it really is a very clever problem with an extremely clever and logical answer. you dont have to know any specific knowledge of either mathematics or any other subject - its all logical thinking.
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    Hmm...

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    Put 5 coins on each side of the scale. If they balance, one of the two left is the fake.

    Then you just have to test one of the real coins against the first one that's left, and if that balances then it's the other one. I think.


    Right?
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    If you were smart you'd have just googled it.
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    (Original post by Aurora.)
    Hmm...

    Spoiler:
    Show
    Put 5 coins on each side of the scale. If they balance, one of the two left is the fake.

    Then you just have to test one of the real coins against the first one that's left, and if that balances then it's the other one. I think.


    Right?
    What if the 10 coins don't balance...?
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    (Original post by Aurora.)
    Hmm...

    Spoiler:
    Show
    Put 5 coins on each side of the scale. If they balance, one of the two left is the fake.

    Then you just have to test one of the real coins against the first one that's left, and if that balances then it's the other one. I think.


    Right?
    Yes true BUT what if the first five coins on each side didn't weigh the same. I thought I just worked out the answer to this question but you do not know if the coin is lighter or heavier.
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    Hmm. Okay I give up
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    This is a common question in investment bank and programming job interviews.
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    Place 6 coins on either side of the scales; we then know that the side that comes down the most contains the heavier coin. So, again, place 3 coins on either side of the scales, all 6 chosen specifically from the 6 coins that we know contain the heavier one; again, whichever side comes down most contains the heavier coin. Now place two coins on either side of the scale, chosen specifically from the 3 that we know contain the heavier one. If one side comes down more than the other, that's the coin we're looking for; if neither of them come down (i.e. the scales show the 2 coins to weigh the same), then the heavier coin must be the only coin from the 3 mentioned that wasn't placed on the scales.



    What do I win?
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    (Original post by jismith1989)
    Place 6 coins on either side of the scales; we then know that the side that comes down the most contains the heavier coin. So, again, place 3 coins on either side of the scales, all six chosen specifically from the 6 coins that contain the heavier one; again, whichever side comes down most contains the heavier coin. Now place two coins on the scale, chosen specifically from the 3 that we know contain the heavier one. If they one comes down the most, that's the coin we're looking for; if neither of them come down (i.e. the scales show the 2 coins to weigh the same), then the heavier coin must be the only coin from the 3 mentioned that wasn't placed on the scales.



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    I thought this but we don't know whether the fake coin is heavier or lighter, so it doesn't work...
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    Well, looks like I just failed to become an investment banker (assuming googling is not possible mid interview). I get as far as this:

    1. 6 coins each side, one side now heavier
    2. Binary split - switch 3 coins between each side, and weigh again. So you now know whether it is one of the 6 coins you moved or not
    3. Er, I only have one weigh in left, I'm stuck.
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    1) Weigh 1,2,3,4 vs 5,6,7,8

    If = then you have 2 more weighs to find which of 9,10,11,12 it is, which is easy (weigh 9,10 vs 11,Gold, if = then its 12, if < or > weigh 9vs10 (equal = 11, same as before = 9, different = 10))
    If < goto step 2,
    If > goto step 2 but in reverse.

    2) You now know that EITHER one of 1,2,3,4 is fake and heavy or one of 5,6,7,8 is fake and light. Weigh 1,5,6 against 2,7,8

    If =, fake is 3 or 4 and is heavy, weigh against each other.
    If < then either 1 is fake and heavy or one of 7,8 is fake and light, goto 3
    If > then either 2 is fake and heavy or one of 5,6 is fake and light, goto 3 in reverse and with 2,5,6 instead.

    3. Weigh 1 and 7 against two gold ones discarded earlier.

    If =, 8 is fake and light
    If <, 1 is fake and heavy
    If >, 7 is fake and light.


    This is I think the simplest (and therefore most elegant!) solution, requiring 4v4, 3v3 and 2v2 rather than 3 lots of 4v4.
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    I would say one of the post maths/computing degree guys might have a shot at most logical, I wouldn't like to guess on who is most intellectual.

    There are loads of ways of doing this,

    split the coins in to 2 lots of six, then two lot's of three, then two lots of one with an odd coin lying out, but I don't know if you can tell which one is heavier or lighter.


    I remember doing this a while ago with a really annoying method, it worked out a little like this but I don't think I can remember the solution.
    Label the coins 1 to 12.

    Take the following weights (Or something similar if this is wrong)

    Right------------------Left



    [3][5][6][11] vs [1][4][8][12]
    [2][3][5][12] vs [7][8][10][12]
    [1][2][10][11]vs[2][5][8][9]
    L means that the left pan goes down, R that the right pan goes
    down, and – means they stay balanced.


    Lets assume the fake is heavy (if it is light, it is the oposite to this, swap the R's with L's).
    Ok the results of these three weighings should be unique to every possibility. Say if 1 is the fake you would get (R,-,L)

    2: (-,L,L)
    3: (L,L,-)

    etc...

    The idea is that by getting the sequence you can see which coin is the fake.
    I think this does actually work. There we go, with a capital cool
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    (Original post by jismith1989)
    Place 6 coins on either side of the scales; we then know that the side that comes down the most contains the heavier coin. So, again, place 3 coins on either side of the scales, all 6 chosen specifically from the 6 coins that we know contain the heavier one; again, whichever side comes down most contains the heavier coin. Now place two coins on either side of the scale, chosen specifically from the 3 that we know contain the heavier one. If one side comes down more than the other, that's the coin we're looking for; if neither of them come down (i.e. the scales show the 2 coins to weigh the same), then the heavier coin must be the only coin from the 3 mentioned that wasn't placed on the scales.



    What do I win?
    A/Q you don't know whether the fake coin is heavier or lighter :fyi:

    You win a D- for effort.
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    (Original post by Aurora.)
    I thought this but we don't know whether the fake coin is heavier or lighter, so it doesn't work...
    Hmm, I could work it out in that case if we had four goes, but not with three. I'll have a think about it, these kind of things interest me.

    EDIT: I got there after cogitating over it as I was in the shower when I came off the computer this morning. However, "HistoryRepeating" found the answer much faster than any of us! Well done him/her.
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    From another thread a long time ago:
    http://www.thestudentroom.co.uk/show...3&postcount=51
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    Or the first person to google it...
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    (Original post by jismith1989)
    Place 6 coins on either side of the scales; we then know that the side that comes down the most contains the heavier coin. So, again, place 3 coins on either side of the scales, all 6 chosen specifically from the 6 coins that we know contain the heavier one; again, whichever side comes down most contains the heavier coin. Now place two coins on either side of the scale, chosen specifically from the 3 that we know contain the heavier one. If one side comes down more than the other, that's the coin we're looking for; if neither of them come down (i.e. the scales show the 2 coins to weigh the same), then the heavier coin must be the only coin from the 3 mentioned that wasn't placed on the scales.



    What do I win?


    Doesnt work as you dont know the fake is heavier, it could be lighter. Thus at the second stage the scales could balance, showing that the fake is lighter but wasting a measurement.

    See my solution.
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      (Original post by vahik92)
      Hi guys, i thought to make this thread because i think itll be interesting for some of you. basically here is a problem, solving which is very interesting!

      there are 12 coins. 11 of them are golden but 1 of them is fake - they all look the same. the 11 golden coins weigh exactly the same but the fake one has a different weight - either heavier or lighter - you dont know this. how can you know which one of the 12 is the fake, having a two sided scales (the one that doesnt show the weight in number, just weighs two items and tells which one is heavier/lighter) and having only three chances to weight any combination of the coins?

      This is a very interesting problem, i solved in in three days and nights I know not all of you will enjoy this but im sure some you will, so please do write you thoughts.

      P.S. - this is not a stupid problem, it really is a very clever problem with an extremely clever and logical answer. you dont have to know any specific knowledge of either mathematics or any other subject - its all logical thinking.
      arbitrarily isolate 4, 8 coins into sets S_1, S_2 respectively. First we weigh S_2, pitting any 4 vs any 4. This will uniquely determine which group of 4 the fake coin is in (if the first weighing is equal, then the fake coin is in S_1). Now, you have 2 turns to distinguish between 4 coins; that's enough, just pit 2 vs 2, then 1 vs 1. WRONG.

      That took all of 2 minutes.
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      (Original post by Saichu)
      then 1 vs 1.

      That took all of 2 minutes.
      but you dont know if it is the heavier one or the lighter one if you do 1 v 1.
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      (Original post by HistoryRepeating)
      Divide the coins in half, 6 coins in pile A and 6 in pile B

      1) Split Pile A in half and weigh 3 on each side of the scale. If they weigh the same they are all gold and the fake is in B, if they weight differently the fake is in A. Lets say the fake is in B, so A is all gold (note: this doesnt make a difference to the proof, if the fake is in A just swap A and B in all the below reasoning, I've just done it for ease)

      2) Take the 6 B (with the fake in them), split it in half. weigh one half against any 3 of the A, gold coins. If they weigh the same they are all gold and can be added to A and the fake is in the other half, if they weigh differently one of them is fake AND you know whether the fake is heavier or lighter than gold.

      3) Take the half (3 coins) with the fake in and pick any two of them to weight against each other. If they weigh the same the third (unweighed one) is the fake, if they weigh differently, the one that is heavier or lighter (as found out in step 2 above) is the fake.

      Coming up with this solution took me 4 minutes with a pen and paper from first principles, I've never done a problem like this before and I didnt cheat.
      This guy has it.
     
     
     
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