Puzzle....Math...Head hurts...please help me solve this puzzle

[masmdu]

There are 12 coins, they look the same, one coin is a fake and it can be determined by it's weight compare to a real coin.

It is unknown if the fake coin weighs more or less than a real coin.

you may use a balance scale (the pivot kind of scale with a weighing plate on each side of the pivot) only three times to determine which coin is the fake.

How do you go about determining which coin is the fake?

TIA

[Richardo]

1. Removed due to dumbness

"unknown if fake is lighter or heavier"

[meegbear]

2. Genius!!!

Good call.

[masmdu]

5. Not right...the fake may be part of the 6 set aside as the lighter group

keep trying

[masmdu]

3. NOPE...

remember you don't know if the fake weighs more or less than a real coin.

[pnb]

4. Same here

I had an entire solution in before I remembered that line.

[el_gato]

6. four groups of 3

weigh the first of
3 and 3

if they are the same toss em

if not the coin is in this group of 6

now divide the 6 into 3 groups of 2

compare the first set of 2 and 2

if they are the same your coin is in the last group of 2

you know from the comparison of the first 2 and 2 group that
these are good coins

now just take your suspect two and compare one of them to a good
coin if it weighs the same your other is the coin you want.

is this clear or clear as mud?


edited for shitty spelling

[Richardo]

9. Yeah but if the first 2-and-2 are NOT equal, you've got four suspects

and only one weigh left. Is there a way around that?

[el_gato]

13. i see

doh!

[THUNDER HANDS]

7. make three piles

of four coins each

weigh them

One of the piles will weigh different than the other two.

Then take that pile and ... well, I guess you'd have to make an educated guess as to which of the four coins is the fake one.

Then, at least, you'd have a 1/4 chance of getting it right.

[Tyler Durden]

8. solution.

measure 6 and 6. discard the 6 that are heavier or lighter (the fake makes the other six weigh more, in which case the other six will be light, or less, in which case the other six will weigh more.)

from the retained six, weigh 3 and 3, act as before.

with the 3 left, balance any two. if they balance, the one left off is the fake, if not, the one that is lighter or heavier is the fake.

[THUNDER HANDS]

11. you don't know if it's heavier or lighter

[Speck Tater]

10. Couple of solutions from the web...

http://www.greylabyrinth.com/Puzzles/answer056.htm

http://users.aol.com/AmazingMazeMan/scales/Scale301.html

http://puzzles.nigelcoldwell.co.uk/one.htm

I found about 250 solution pages on the web but in fadct they are all the same solution because there is only one solution.

Here's one version:

First, weigh 4 coins againt 4 coins. If the balance tips, we have 4 coins we know are correct, 4 coins which might be too heavy but cannot be too light, and 4 coins which might be too light but cannot be too heavy.

Second, (assuming we tipped the balance on the first weighing) place 2 possibly light coins and 1 possibly heavy coin on each side. If the balance tips, we have 9 known correct coins, 2 possibly light coins, and 1 possibly heavy coin.

Third, weigh the two possibly light coins againt each other. If the balance tips, the lighter coin is counterfeit and too light. Otherwise the possibly heavy coin is counterfeit and is too heavy.

If the balance in the second weighing does not tip, we have 10 known correct coins and 2 possibly heavy coins. Weigh those two coins against each other, the heavier coin is counterfeit and too heavy.

If the balance in the first weighing does not tip, we have 8 known correct coins and 4 possibly heavy or light coins. For the second weighing, weigh 3 of the possibly bad coins againt 3 known correct coins. If the balance does not tip, we have 11 known correct coins and one possibly light or heavy coin.

Third, weigh the counterfeit coin against a correct coin to determine if the counterfeit is too heavy or light.

If the balance in this second weighing tips, we have 9 known correct coins and either 3 possibly heavy coins or 3 possibly light coins. For the third weighing, weigh one of the possibly counterfeit coins against another, and from that we know which coin is counterfeit. (From the last weighing we already knew whether it would be too heavy or light.)

[Seggie]

12. The problem states the coins look the same....

but, it says nothing about the coins sounding the same. 11 of the coins will sound the same if you drop them one by one on a hard surface. The coin that sounds different is the fake coin.

Then you need to use the scale only once to determine if the fake is heavier or lighter. Place the fake opposite of a genunie coin on the balance and watch it rise or fall!

[northzax]

14. simple answer.

get thee to a vending machine. Attempt to buy a snickers bar. repeat. the coin rejected bythe machine is the fake one.eat your candy.

[cmf]

15. I wasted my entire lunch period with this puzzle

And even longer to write it out. But I love stuff like this, so here is my answer:

Split the coins into 3 groups of 4. Weigh two of the groups of 4 against each other.

If the scales balance, then you know those original 8 are real:
The fake coin is in the 3rd group of 4.
Split that group into 2 groups of 2 each and weigh one of those groups against 2 of the coins we know are real (from the 8 we weighed already) for the second weigh.
If the scales balance, the fake is in the last two. Pick one and weigh it against one of the known real coins for the 3rd weigh. If it doesn't balance in the 3rd weigh, then that one you just picked is fake. If it does, then the last one you never weighed is fake. This is the only scenario where you won't know if the fake coin is lighter or heavier.
If the scales don't balance in the 2nd weighing, then the fake is in that new group of 2 you just weighed. Take one coin off of each side and weigh again for the 3rd weigh. If it doesn't balance, then the one left on the scale is fake, if it does, then the one you took off is fake.

If the scales don't balance, then you know the 4 you didn't weigh are real: This scenario is a little trickier. All 8 original coins are suspect. 4 may be fake and lighter, 4 may be fake and heavier. Now you have to mix it up and see if the scales change at all. Make all new groups of the following composition:
Set aside 2 possibly lighter coins and one possibly heavier coin.
Group #1 will be one group of 2 possibly lighter coins + 2 possibly heavier coins (4 coins total)
Group #2 will be one group of 1 possibly heavier coin + 3 known real coins (4 coins total)
There will be one known real coin left over
Now weigh group 1 vs group 2 for the 2nd weigh.
If 1=2, then one of the coins you set aside will be the fake, weigh the possibly light coins against each other. If one is lighter than the other, that one is fake, if they equal then the possibly heavier coin you set aside is the fake.
If group 1 is lighter than group 2, then either the 2 possibly lighter coins in group 1 are fake or the one possibly heavier coin in group 2 is fake. Weigh the 2 possibly lighter coins against each other for the 3rd weigh. If one is lighter than the other, that one is fake, if they equal then the possibly heavier coin is the fake.
If group 1 is heavier than group 2, then one of the 2 possibly heavier coins in group 1 is the fake. Weigh them against each other for weigh #3 and the heavier one is the fake.

Anyway, my way may be much too complicated, but it works and gives you the answer in 3 weighs.