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

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.

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.

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.

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

They are proberly higher than yours judging by your spelling

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

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

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.

Hmm...



Right?

Good old TSR. Neg rep for giving it a try and getting the wrong answer

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

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.