### Brain Teasers

# Find the Fake Coin

There are 12 gold coins. One of the coins is fake. It is known that a fake coin weighs either slightly less or slightly more than a real coin. Using a balance scale, how can you find the fake coin, and determine if it weighs less or more than a real coin, in only three weighings?

### Answer

Divide the coins into three groups of 4 coins each. First, weigh two of the groups. If they balance, then the fake coin is in the third group, so take three of the coins, and weigh them against three of the coins that we know for sure are real (from the first 8 coins). If they balance, then the remaining coin is the fake one, but we still don't know if it is lighter or heavier, so weigh it against a real coin to determine that. But if it didn't balance, then we know that the fake coin is one of the three, and we also know if it is lighter or heavier, because we know which side had the real coins. So weigh two of the coins against each other, if they balance then the third coin is fake, if they don't balance we know which coin is fake, because we already know if it is lighter or heavier.Now let's go back to the first weighing; if the two groups didn't balance, we know that the fake coin is in one of the groups, but we don't know which group. Let's mark the group that was heavier as group A, and the group that was lighter group B. For the second weighing, weigh two coins of group A and one coin of group B, against the other two coins of group A and one coin of group B. If they balance, the fake coin must be one of the remaining two, and a weigh between them will tell us which one (they are both from group B so we know that it's lighter). If the second weigh didn't balance, look at the side that is heavier: Either the fake coin is one of the two group A coins, or it can the group B coin on the other side of the scale. Take the two group A coins, and weigh them against each other. If they balance, then it is the group B coin that is fake, and if they don't balance, then the heavier one is the fake coin.

And that's that. We have covered all possibilities.

Hide Answer Show Answer

## What Next?

View a Similar Brain Teaser...

If you become a registered user you can vote on this brain teaser, keep track of which ones you have seen, and even make your own.

### Solve a Puzzle

## Comments

Good brain teaser. But wouldn't it be easier if you split the coins in 2 groups of 6, weight them, then split the 6 in 2 groups of three, weigh them, the weigh the remaining 3, using only one coin on each scale? If it balances, then you're holding the fake coin.

Dividing the group into 2 groups of 6 would not let you find the fake coin every time. Suppose the last two coins that you weighed were not equal. Which one is the fake then? Good try Danny1979.

I got the same answer as Danny. 1 problem with it though: You don't know if a fake coin is heavier or lighter, therefore by weighting 6 against 6, you wouldn't know if the fake coin was in the light group or the heavier one

i'M STRUGGLING WITH THIS SOLUTION. i DON'T THINK IT WORKS. SINCE YOU DON'T KNOW IF THE ODD COIN IS LIGHTER OR HEAVIER, THIS SECOND STEP DOESN'T WORK (THE COIN COULD BE LIGHTER TOO!) A FEW YEARS AGO I SOLVED THIS USING (I THINK) GROUPS OF THREE AND UNFORTUNATELY DIDN'T WRITE THE SOLUTION DOWN...I'VE BEEN GOING CRAZY TRYING TO SOLVE AGAIN....HELP! I REALLY DON'T THINK YOUR POSTED SOLUTION WORKS....

Good one! Took all of 40min to figure out! Keep 'em coming :-)

but the answer is not correct cause in the teaser it says that you can only weigh 3 times and in the answer you weigh more than three times...right?

no aries those are the different ways you could do it u dont do them in order unless it says so.

That is thee most detailed answer i have ever read in my life.

That is thee most detailed answer i have ever read in my life.

impressive

tooooooo impressive. now I learn that one thing can lead to thousands of possibilties.

it's not the ultimate answer

this riddle is using the old weighing system of 1 3 and 9

each of the 12 balls can be numbered and translated into 1 3 and 9

ex: 12 = 9+3

7=9+1-3

thus

grouping each balls that have +1 in the left side on first weigh, and -1 on the right side, and 3 and 9

you'll get

1,4,7,11 - 2,5,8,11

2,3,4,11,12 - 5,6,7

5,6,7,8,9,10,11,12

now this is illogicall right?

so we shift the balls to make it even

1,4,7,8 2,5,10,11

2,3,4,11 5,6,7,12

5,6,7,11 8,9,10,12

so, when the weigh is heavier on the left side, you write +1, and -1 if heavier on the right

then multiply the data by 1 for the first, by 3 for the 2nd, and 9 for the third

the sum is the number of the ball, and the sign(+/-) is heavier or lighter, off course we have to change it for 8,9,10, and 12, because we shifted them

this riddle is using the old weighing system of 1 3 and 9

each of the 12 balls can be numbered and translated into 1 3 and 9

ex: 12 = 9+3

7=9+1-3

thus

grouping each balls that have +1 in the left side on first weigh, and -1 on the right side, and 3 and 9

you'll get

1,4,7,11 - 2,5,8,11

2,3,4,11,12 - 5,6,7

5,6,7,8,9,10,11,12

now this is illogicall right?

so we shift the balls to make it even

1,4,7,8 2,5,10,11

2,3,4,11 5,6,7,12

5,6,7,11 8,9,10,12

so, when the weigh is heavier on the left side, you write +1, and -1 if heavier on the right

then multiply the data by 1 for the first, by 3 for the 2nd, and 9 for the third

the sum is the number of the ball, and the sign(+/-) is heavier or lighter, off course we have to change it for 8,9,10, and 12, because we shifted them

1,4,7,8 - 2,5,10,11

2,3,4,11 - 5,6,7,12

5,6,7,11 - 8,9,10,12

2,3,4,11 - 5,6,7,12

5,6,7,11 - 8,9,10,12

UHHHHHHHHHHHHHHH...UMMMMMMMMMMMMMMMMMMMMMM...

To post a comment, please create an account and sign in.

## Follow Braingle!