### Author Topic: (Easy) Gold Coins [Solved]  (Read 6284 times)

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#### Sidoh

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##### (Easy) Gold Coins [Solved]
« on: August 09, 2005, 02:36:19 am »
You are considering buying 12 gold coins that look alike but have been told that one of them is a heavy counterfeit. How can you find the heavy coin in three weightings on a balance scale?
« Last Edit: August 09, 2005, 02:49:37 am by Sidoh »

#### Newby

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##### Re: (Easy) Gold Coins [Solved]
« Reply #1 on: August 09, 2005, 02:40:58 am »
Weigh 4 of the coins on one side, and 4 on the other.

Two possible results from this:
• One side is heavier. If this happens, take those 4 coins and throw the other 8 away.
• They are equal in weight. If so, throw those 8 coins away.

Next you have four coins. Do the same thing, but in groups of 2s. Which ever side is heavier, take those two coins. Weigh them. Which ever one is heavier, you have your counterfeit.
- Newby
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Quote
[17:32:45] * xar sets mode: -oooooooooo algorithm ban chris cipher newby stdio TehUser tnarongi|away vursed warz
[17:32:54] * xar sets mode: +o newby
[17:32:58] <xar> new rule
[17:33:02] <xar> me and newby rule all

I'd bet that you're currently bloated like a water ballon on a hot summer's day.

That analogy doesn't even make sense.  Why would a water balloon be especially bloated on a hot summer's day? For your sake, I hope there wasn't too much logic testing on your LSAT.

#### Sidoh

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##### Re: (Easy) Gold Coins [Solved]
« Reply #2 on: August 09, 2005, 02:43:37 am »
You're on the right track, but you're using more than three weighing sessions to determine the answer.

You're so close that if it was a snake it'd bite you!

#### Newby

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##### Re: (Easy) Gold Coins [Solved]
« Reply #3 on: August 09, 2005, 02:46:51 am »
I used three...

Let me rephrase. Seperate them into groups of 4.

1. 4 v 4 (which ever side is heavier, take those 4 coins for #2 weighing, otherwise throw all of 'em out and take the remaining 4 for #2)
2. 2 v 2 (take the side that is heavier for #3)
3. 1 v 1
- Newby
http://www.x86labs.org

Quote
[17:32:45] * xar sets mode: -oooooooooo algorithm ban chris cipher newby stdio TehUser tnarongi|away vursed warz
[17:32:54] * xar sets mode: +o newby
[17:32:58] <xar> new rule
[17:33:02] <xar> me and newby rule all

I'd bet that you're currently bloated like a water ballon on a hot summer's day.

That analogy doesn't even make sense.  Why would a water balloon be especially bloated on a hot summer's day? For your sake, I hope there wasn't too much logic testing on your LSAT.

#### Sidoh

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• Posts: 17668
• MHNATY ~~~~~
##### Re: (Easy) Gold Coins [Solved]
« Reply #4 on: August 09, 2005, 02:49:27 am »
I didn't read your answer right. That'll work. Nice job. This is the answer I have, which is why I thought yours was wrong:

Quote
1)   Split the 12 in half, placing 6 on each side of the balance.
2)   Remove the light 6. Then split the heavy 6 in half, placing 3 on each side of the balance.
3)   Remove the light 3. Then set a random one of the heavy 3 aside and split the remaining 2 in half, placing 1 on each side of the balance. If the balance is not “balanced” you will know which is heavy. If it balances, you know that these two are both the same, thus the 1 placed aside was the heavy coin