Thread: problem of coins and fake coin

We have 12 coins. from them, only one is FAKE.
We do not know whether the FAKE coin is lighter or heavier than the others.
We have a scale also so we can put them on the scale and scale them.
How can we detect the FAKE coin, using only 3 times scaling?


Thank you in advance for helping me

We have 12 coins. from them, only one is FAKE.
We do not know whether the FAKE coin is lighter or heavier than the others.
We have a scale also so we can put them on the scale and scale them.
How can we detect the FAKE coin, using only 3 times scaling?


Thank you in advance for helping me

hi
divide them in 3 group of 4 coin each

first scaling: weigh any 2, you will get a group either heavier or lighter than other group

second scaling: take that group of 4 coin obtain above and divide in two parts of 2 coins each, you will get a group either heavier or lighter than other group

third scaling: take that group of 2 coin obtain above and divide in two parts of 1 coins each, you will get a coin either heavier or lighter than other coin

Thank you my friend for your solution.
But I do not understand some points there.

1- if in first scaling the 2 groups are equal then the fake is in the 3rd group right? if not then it is inside one of the current 2 groups. now can you please explain more, how do you understand that the fake one belongs to which group? the heavier one or the lighter one? since we do not know, whether the fake is lighter or heavier?

Thank you

Thank you my friend for your solution.
But I do not understand some points there.

1- if in first scaling the 2 groups are equal then the fake is in the 3rd group right? if not then it is inside one of the current 2 groups. now can you please explain more, how do you understand that the fake one belongs to which group? the heavier one or the lighter one? since we do not know, whether the fake is lighter or heavier?

Thank you

oh sorry , i forget that part
then i dnt hav idea how to solve

people, any other help? I worked on it and I almost have solved it but Its not 100%

Actually, I have a different answer. I've also tried solving this problem, and I found out that it is somehow possible at the same time impossible to locate the fake coin with only 3 weighing--there is always a double possible outcome. I suggest that minimum no. of weighing to locate the coin is 4. And here how it goes.......

Divide the coins into 3 groups (4 coins each)
Left side----------------right side
1st weighing:
4 coins---------------4 coin

If the coins weigh even, then the counterfeit coin is in the 3rd group of coins. Otherwise, if the weights are not even, the fake coin is on either side of the scale. We take the coins on either side of the scale, then replace the empty side with the 3rd group of coins.

2nd weighing:
4 coins----------------4 coins
If the group of coins weigh the same, then the fake coin is in the group of coins that was set aside. Otherwise, if they don't weigh the same--the fake coin is in the side of the scale before the actual 2nd weighing took place.

3rd weighing
1 coin---------------1 coin
2 possibilities

1st
If they weigh the same, then we set them aside. So we know that the fake coin is either of the 2 coins that were left.
2nd
If they don't. So we know that the fake is either of the 2 coins.

4th weighing
1 coin---------------1 coin
So know we are down we only 2 coins. To locate the fake coin, we set aside either of the two coins. Then we get any of the ten coins that were set aside in the 1st-3rd weighing. If the coins weigh the same, then the coin that was set aside is the counterfeit. Otherwise, if they don't--then we located the fake coin (if you know what I mean).