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

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    Use the fake anyway. Sorted!
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    You bite the gold coins - the fake will be softer and leave teeth prints.
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    I just wrote a silly response
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    Edit.
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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.
    Nice. +Rep.
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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.

    Am I being very silly, or is this wrong?

    1) All in A weigh the same
    2) Half of B is equal to half of A, these are discarded.
    3) Out of your remaining three, which is the fake, seeing as you do not know if the fake is heavy or light?

    Sorry If I am being really stupid. Can't find a mistake in my solution YET.....
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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.
    If in step 2, the two sets you weight against each other are the same weight, then you haven't found out whether the fake is heavier or lighter in the final step.


    My solution (I don't think I've missed anything)
    split the coins into 3 groups of 4 a,b,c.
    weight a vs b.
    either (a = b
    in which case the fake is in c, then weigh 3 coins from c, against if the weighing is equal then the last coin is the fake, if the weighing is unequal, say the 3 in c are lighter than the 3 in a (symmetrical case for heavier) then weigh two of the weighed coins in c, either the lighter one is fake, or if equal the third coin is fake)

    otherwise (a<b (symmetrical case for a>b) either fake is in a and is lighter, or fake is in b and is heavier. weigh two coins from c, plus 1 from a, plus 1 from b, against two from a, and two from b. if equal (then weigh last a against coin from c, if != then that a is fake, otherwise, last b is fake) otherwise (2c + b + a side < 2a + 2b side (or symmetrical case) then fake is amongst a on left, or 2bs on right. weigh the 2 b's against each other, if equal then the a is fake, otherwise the heavier of the 2 bs is fake)

    hope that makes sense, I'll try to type up a clearer version if I get time.
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    omg youre all so brailleint waht are your iq scores/?
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    (Original post by fat_hampster)
    If in step 2, the two sets you weight against each other are the same weight, then you haven't found out whether the fake is heavier or lighter in the final step.


    My solution (I don't think I've missed anything)
    split the coins into 3 groups of 4 a,b,c.
    weight a vs b.
    either (a = b
    in which case the fake is in c, then weigh 3 coins from c, against if the weighing is equal then the last coin is the fake, if the weighing is unequal, say the 3 in c are lighter than the 3 in a (symmetrical case for heavier) then weigh two of the weighed coins in c, either the lighter one is fake, or if equal the third coin is fake)

    otherwise (a<b (symmetrical case for a>b) either fake is in a and is lighter, or fake is in b and is heavier. weigh two coins from c, plus 1 from a, plus 1 from b, against two from a, and two from b. if equal (then weigh last a against coin from c, if != then that a is fake, otherwise, last b is fake) otherwise (2c + b + a side < 2a + 2b side (or symmetrical case) then fake is amongst a on left, or 2bs on right. weigh the 2 b's against each other, if equal then the a is fake, otherwise the heavier of the 2 bs is fake)

    hope that makes sense, I'll try to type up a clearer version if I get time.

    Yay someone agrees with me about the other method. Well done this does actually work.
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    (Original post by RupertTheBear)
    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
    3^3 is 27, so your system in theory can distinguish between 27 cases ( I haven't checked whether the weighing you provide actually works, but if not there exists a similar weighing which does). In the given problem there are 24 cases (coin 1 is heavy, coin 1 is light, coin 2 is heavy etc.) you could deal with 13 coins, plus the possibility that there is no fake.
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    (Original post by Eloades11)
    They are proberly higher than yours judging by your spelling
    hahahaha, this coming from captain "proberly"

    it's 'probably' retard.
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    (Original post by jolteon)
    hahahaha, this coming from captain "proberly"

    it's 'probably' retard.
    thanks, didnt mention my IQ though, because i'm not stupid enough to care about it
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    I take 10 of the coins, decreasing the chances of the fake one being among them, even if it is there there'd be 9 real gold coins.

    Then I'll have 10 coins (hopefully al Gold) I'll have them smelted back into raw, unshaped gold and proceed to sell them to the highest bidder on Amazon or E-bay, some high paying sell-your-gold company and or to a foreign sheikh that wants gold or something.
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    (Original post by Eloades11)
    thanks, didnt mention my IQ though, because i'm not stupid enough to care about it
    i think someone's mad. your iq low bro? you think you were a 170 and you got handed a 112? sucks man. but don't worry, youll look cool at the pub quiz when you get an answer or two more than the builders.
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    (Original post by yokabasha)
    You can put 6 on but gradually do it so put 1 on each side at the same time until you you get an imbalance the 2 that imbalance it are the ones that are suspected to be the fakes ones now you take the ones that you know to be real and test if they balance against this one if one does the other must be the fake one, all depends if you count that as 3 moves, but since everyone thinks putting six is on each side then putting it slowly surely counts as well
    That's not three moves.
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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.
    Sorry if im just being plain stupid, but in 3) im presuming you select 2 from the 3(with fake in) to compare with 2 from the 9 you have established are not fake. I accept that IF the 2 you select from the set of 3, weigh the same as the 2 you select from the 9 gold ones then the unweighed one is fake. But I don't understand how you would find the fake one if the 2 you select from the 3 would weigh differently to the 2 real ones :s again sorry if im being silly
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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?
    Good old TSR. Neg rep for giving it a try and getting the wrong answer :rolleyes:
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    (Original post by Aurora.)
    Good old TSR. Neg rep for giving it a try and getting the wrong answer :rolleyes:
    Just to be a **** I'm going to neg rep this post.
    BTW I'm not the same person who neg repped your other post.
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    (Original post by shamim2k9)
    Sorry if im just being plain stupid, but in 3) im presuming you select 2 from the 3(with fake in) to compare with 2 from the 9 you have established are not fake. I accept that IF the 2 you select from the set of 3, weigh the same as the 2 you select from the 9 gold ones then the unweighed one is fake. But I don't understand how you would find the fake one if the 2 you select from the 3 would weigh differently to the 2 real ones :s again sorry if im being silly
    His solution doesn't work. In fact I'm reasonably confident that it impossible to solve the problem with six coins and two weighings (even when you have standard coins to which to compare yours), so you need to do more on the first step (either get down to a smaller group or get some information about whether it's heasvier or lighter).
    (Original post by fat_hampster)
    3^3 is 27, so your system in theory can distinguish between 27 cases ( I haven't checked whether the weighing you provide actually works, but if not there exists a similar weighing which does). In the given problem there are 24 cases (coin 1 is heavy, coin 1 is light, coin 2 is heavy etc.) you could deal with 13 coins, plus the possibility that there is no fake.
    Edit: Misunderstood the specific implementation of the idea. The approach can be adapted to 13. /Edit. In fact while the approach is valid it won't work with 12 as written (both 2 and 12 have been put on both sides at the same time).
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    How does this determine who is the most intellectual?

    I'd just go out out and buy some digital scales...
 
 
 
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