Find the Counterfeit Coin

[hyeok]

Since brain teasers seem to be becoming popular, here is my offering! Have fun:

You have a stack of 12 coins. One of the coins is counterfeit because it is either heavier or lighter than the other 11 coins. How can you find this coin with ONLY 3 measurements on a balance? The balance will only show you which side is heavier.

:twisted: This is fun. I have even a harder one on the way.

[Ehren]

here's how i'd do it -

I'd put 6 coins on each side of the balance. Keep the heavier side, set the other 6 aside.

With the six remaining coins, i'd put 3 on each side of the balance. I'd keep the heavier side, ditch the other 3.

With the remaining 3 coins, i'd put 2 of them ... pick any two ... on the balance. if one is heavier, that's the coin. If they are equal in weight, the coin you left out is the culprit.

The end.

[Taurus]

Yeah, thats how to do it, and it works!

[Spacehog]

Not necessarily... simply because of the line "because it is either heavier or lighter"... if you knew it was heavier or lighter, then you'd be alright, but you cannot solve this one without doing a fourth measurement to ascertain whether it's heavier than the others or lighter than the others (to be fair, the riddle normally specifies that it is ligher, or that it is heavier).

Martin

[hyeok]

yup.. you don't know if it is heavier or lighter. Your answer assumes that the counterfeit coin is heavier! :lol: And it IS doable with only 3 measurements without knowing whether the counterfeit is heavier or lighter! It's hard... :twisted:

[ImaX]

Mmh, that's hard... here's my way:

- divide the coins into 3 stacks of 4 coins

1) take two and measure them

2) put the heavier one aside (or the lighter one, doesn't matter) and measure the third stack together with the lighter one from 1)

- if they are equal the coin must be in the stack you put aside and is heavier. If not, the coins must be in the stack you measured 2 times (the lighter one from 1) and is lighter than the other ones.

3) divide this stack into 2:2 and measure. The coin must be one of the 2 coins on the lighter side (or on the heavier side, depending on what you found out about the weight).

So I'd need one more measurement :roll: perhaps someone get's an idea through this, I like this idea because you find out if the copin is lighter or heavier.

[kepsilon]

i almost solved it but i hit a brick like everybody else

1)Divide into 3 groups of 4(group 1,group2,group3)
2)Measure group 1 and 2 , if they are equal move to (4), else move to (7)


4)Split group3 into four coins(1,2,3,4) and measure for example '1' & '2', if they are equal move to (5), if not move to (6)

5)Measure '3' with '1' or '2', if they are equal '4' is your coin, otherwise '3' is your coin. END

6)Measure '1' with either '3' of '4', if they are equal '2' is your coin, otherwise '1' is your coin. END


7)Combine group 1 and 2 (1,2,3,4,5,6,7,8 ) and measure '1,2,3' against '4,5,6' and stack '7' and '8' aside,
if they are equal move to step (8 ) else to step (9)

8 )Measure any coin from '1' to '6' with '7', if they are equal '8' is your coin,else '7' is your coin. END

9)NO IDEA


hope that wasnt too complex :P

[Enigma™]

I HAD figured it all out, but then the stupid brower *after typing my ass off* decides to make a popup and I hit backspace, which made my browser go back.
In any case, I lost everything I was typing, so here's the answer:

First run: 1 2 3 4 || 5 6 7 8
If the pan tilts right or left, label the heavier side coins "H", then lighter ones "L", and 9-12 are "N". It doesn't matter which way it tilts, so we'll assume that it tilts to the left. (If it doesn't, then relabel your stupid coins.) Now, we have the following: 1H 2H 3H 4H 5L 6L 7L 8L 9N 10N 11N 12N
Second run: 1 2 5 || 3 6 9
If the pan balances, then we have 4H 7L 8L; the rest are normal.
Third run: 7 || 8
If it tilts left, then 7 is light. If it tilts right, then 8 is light. It will not balance.
If the pan tilts left, we have 1H 2H 6L; the rest are normal.
Third run: 1 || 2
If it tilts left, then 1 is heavy. If it tilts right, then 2 is heavy. If it balances, then 6 is light.
If the pan tilts right, we have 3H 5L; the rest are normal.
Third run: 3 || 1
If it tilts left, then 3 is heavy. If it balances, then 5 is light. It will not tilt right.
If the pan balances, then we have the following: 1-8 are "N", 9-12 are "U".
Second run: 9 10 || 11 1
If the pan balances, then we have 12HL remaining.
Third run: 12 || 1
If it tilts left, then 12 is heavy. If it tilts right, then 12 is light. It will not balance.
If the pan tilts left, then we have 9H 10H 11L; the rest are normal.
Third run: 9 || 10
If it tilts left, then 9 is heavy. If it tilts right, then 10 is heavy. If it balances, then 11 is light.
If the pan tilts right, then we have 11H 9L 10L; the rest are normal.
Third run: 9 || 10
If it tilts left, then 9 is light. If it tilts right, then 10 is light. If it balances, then 11 is heavy.

[Georges]

I have got another easier riddle.

I can prove mathematically that 1 = 2.

Here we go:

1 ) 1 = 1
2 ) a = a
3 ) a² = a²
4 ) a² - a² = a² - a²
5 ) a (a-a) = (a+a) (a-a)
6 ) a = a+a
7 ) a = 2a
8 ) 1 = 2


Surprised ? Well, didn't you know that 1 is actually 2 ???
.
.
.

P.S.: of course, there is a flaw, ok, let's narrow it down: there is only ONE single flaw in the whole demonstration here above. Tell us which one and why you think that it may THE flaw.

[goo-goo]

I have got another easier riddle.

I can prove mathematically that 1 = 2.

Here we go:

1 ) 1 = 1
2 ) a = a
3 ) a² = a²
4 ) a² - a² = a² - a²
5 ) a (a-a) = (a+a) (a-a) <-----FLAW (a(a-aa))
6 ) a = a+a
7 ) a = 2a
8 ) 1 = 2


Surprised ? Well, didn't you know that 1 is actually 2 ???
\.

a(a-a) = 0
a-a = 0, and the product of a*0 = 0

[Liquid Shadow]

The answer is 21.

[Lyngs]

The answer is 21.

:lol:

[Lyngs]

Hrhmm... ok - I'll surpass George and proove a little more sophisticated, that 2 actually equals 1.

Here we go...

1 ) a = b

Ok, get it so far? Now we multiply with a on both sides:

2 ) a² = ab

Now we add a² - 2ab on both sides:

3 ) a² + a² - 2ab = ab + a² - 2ab

We simplify this to:

4 ) 2(a² - ab) = a² - ab

And finally we divide with a² - ab on both sides, which gives us:

5 ) 2 = 1


Okay, so there's a minor, but important mistake in this equation - who will be the first? :)

[Georges]

5 ) a (a-a) = (a+a) (a-a) <-----FLAW (a(a-aa))

Not quite, actually multiplying by zero would not be the problem. The problem is indeed that (a-a) equals zero. The error however lies between the two lines:

a (a-a) = (a+a) (a-a) / divide both sides by (a-a)=0
a = a+a

You may multiply by zero, add by it, subtract by it but not DIVIDE BY ZERO. That's the most supreme principle of mathematics.

[hyeok]

I HAD figured it all out, but then the stupid brower *after typing my ass off* decides to make a popup and I hit backspace, which made my browser go back.
In any case, I lost everything I was typing, so here's the answer:

First run: 1 2 3 4 || 5 6 7 8
If the pan tilts right or left, label the heavier side coins "H", then lighter ones "L", and 9-12 are "N". It doesn't matter which way it tilts, so we'll assume that it tilts to the left. (If it doesn't, then relabel your stupid coins.) Now, we have the following: 1H 2H 3H 4H 5L 6L 7L 8L 9N 10N 11N 12N
Second run: 1 2 5 || 3 6 9
If the pan balances, then we have 4H 7L 8L; the rest are normal.
Third run: 7 || 8
If it tilts left, then 7 is light. If it tilts right, then 8 is light. It will not balance.
If the pan tilts left, we have 1H 2H 6L; the rest are normal.
Third run: 1 || 2
If it tilts left, then 1 is heavy. If it tilts right, then 2 is heavy. If it balances, then 6 is light.
If the pan tilts right, we have 3H 5L; the rest are normal.
Third run: 3 || 1
If it tilts left, then 3 is heavy. If it balances, then 5 is light. It will not tilt right.
If the pan balances, then we have the following: 1-8 are "N", 9-12 are "U".
Second run: 9 10 || 11 1
If the pan balances, then we have 12HL remaining.
Third run: 12 || 1
If it tilts left, then 12 is heavy. If it tilts right, then 12 is light. It will not balance.
If the pan tilts left, then we have 9H 10H 11L; the rest are normal.
Third run: 9 || 10
If it tilts left, then 9 is heavy. If it tilts right, then 10 is heavy. If it balances, then 11 is light.
If the pan tilts right, then we have 11H 9L 10L; the rest are normal.
Third run: 9 || 10
If it tilts left, then 9 is light. If it tilts right, then 10 is light. If it balances, then 11 is heavy.

::applause:: congratulations!

[Georges]

Hrhmm... ok - I'll surpass George and proove a little more sophisticated, that 2 actually equals 1.

Here we go...

1 ) a = b

We simplify this to:

4 ) 2(a² - ab) = a² - ab

And finally we divide with a² - ab on both sides, which gives us:

5 ) 2 = 1

Okay, so there's a minor, but important mistake in this equation - who will be the first? :)

Minor mistake: a=b, but may pass as an assumption

Major mistake: division by a²-ab, because a²=ab and therefore a²-ab equals 0, division by zero is not allowed

[goo-goo]

5 ) a (a-a) = (a+a) (a-a) <-----FLAW (a(a-aa))

Not quite, actually multiplying by zero would not be the problem. The problem is indeed that (a-a) equals zero. The error however lies between the two lines:

a (a-a) = (a+a) (a-a) / divide both sides by (a-a)=0
a = a+a

You may multiply by zero, add by it, subtract by it but not DIVIDE BY ZERO. That's the most supreme principle of mathematics.


ohh i know dude. i was just trying to prove the unequality of the left side.

[The Wizard]

That's the most supreme principle of mathematics.

Yeah, and why is it the MOST SUPREME?

[Georges]

ohh i know dude. i was just trying to prove the unequality of the left side.

There is no unequality, both sides are perfectly equal (because both sides equal zero), no other flaw, besides the division by zero.

Yeah, and why is it the MOST SUPREME?

Maths is the science of reasoning but like any science it has its own set of basic rules which make it work. If you allow division by zero, many demonstrations become useless and you could prove many things which may not make sense in the mathematical world, not to mention the repercussions on the other (human and/or natural) sciences such as physics, economics, etc. Compare it to the supreme principle of physics (namely that the energy is constant within the whole universe, while only energy transfers happen between its components), throw it away and you can put 99% of contemporary physics into question.

[The Wizard]

ohh i know dude. i was just trying to prove the unequality of the left side.

There is no unequality, both sides are perfectly equal (because both sides equal zero), no other flaw, besides the division by zero.

Yeah, and why is it the MOST SUPREME?

Maths is the science of reasoning but like any science it has its own set of basic rules which make it work. If you allow division by zero, many demonstrations become useless and you could prove many things which may not make sense in the mathematical world, not to mention the repercussions on the other (human and/or natural) sciences such as physics, economics, etc. Compare it to the supreme principle of physics (namely that the energy is constant within the whole universe, while only energy transfers happen between its components), throw it away and you can put 99% of contemporary physics into question.

Yeah, I see all that, but I still see no reason to call it the most supreme principle of mathematics. Take any mathematical axiom, throw it away, and you can put a large percent of mathematics into question.

[Georges]

Take any mathematical axiom, throw it away, and you can put a large percent of mathematics into question.

E.g.?

[The Wizard]

Well, pretty much anything here (http://mathworld.wolfram.com/Axiom.html).

For instance, Peano's axioms:

1. Zero is a number.
2. If a is a number, the successor of a is a number.
3. zero is not the successor of a number.
4. Two numbers of which the successors are equal are themselves equal.
5. (induction axiom.) If a set S of numbers contains zero and also the successor of every number in S, then every number is in S.