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# Weighing in a Harder Way

Best Answer bonanova, 18 January 2008 - 06:49 PM

Jkyle1980's solution is correct, and much less verbose than what follows.

You can distinguish among 3**N cases in N weighings.
There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
So three weighings [27 cases] won't do it, but four [81 cases] can.

Spoiler for Here's a very slightly modified approach
Go to the full post

83 replies to this topic

### #41 Prime

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Posted 26 January 2009 - 09:13 AM

You can get it in 3... every time... and its not hard.

First, divide the 27 into 3 piles of 9. Put 9 on one half of the scale and 9 on the other. If one of these groups of 9 weighs less, you will continue with these 9. If they are even, you will continue with the last group of 9 not yet weighed. So at these point we have it narrowed to 9.

Second, divide these 9 into 3 piles of 3. Put 3 on one half of the scale and 3 on the other. If one of these groups of 3 weighs less, you will continue with these 3. If they are even, you will continue with the last group of 3 not yet weighed. Now we have it narrowed down to 3.

Third and last, divide these 3 into 3 piles of 1. Put one on one half of the scale and 1 on the other. If one is lighter you have found your answer. If they are even it must obviously be the last one not weighed.

Solved in 3.

Per OP, you don't know whether counterfeit weighs more, or less than true coin. So if one group of 9 weighs less than the other, you are left with 18 coins.
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Past prime, actually.

### #42 trifecta

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Posted 29 January 2009 - 11:30 PM

Actually, the answer is 1. You make three piles, 13 coins, 13 coins and 1 coin. Weigh the two piles of 13 coins, and if they balance, the odd coin is the heavier/lighter one.
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### #43 riddle.riddle

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Posted 12 February 2009 - 07:46 PM

hey... the coin does not have to be light... it can be heavy too... so 3 will not work.... i think 4 tries are required....
1st and 2nd weighs> say first you divide it as a, b, c (9 coins each), and weigh a against b and a against c, the one with an odd coin (less or more weight can be found out) and we will also know if it has more weight or less weight.
3rd weigh> now divide that group into 3 of 3 coins each, say x, y, z. compare x and y. if x=y, z has odd coin. otherwise, according to our knowledge of the nature of the coin (heavy or light), we know which one of x and y contains odd coin.
4th weigh> take that group and weigh one coin against another one. the same argument goes to decide the odd coin (similar to the odd group in the previous step)
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### #44 riddle.riddle

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Posted 12 February 2009 - 08:06 PM

the method i mentioned will always work, whatever be the case.
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### #45 lovinduke

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Posted 13 February 2009 - 05:33 PM

My method would still work in 3 if the coin was heavier... I don't know what you're not getting dude. The answers is 3. Re-read my answer... it works every time, whether the coin is heavier or lighter.
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### #46 Prime

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Posted 13 February 2009 - 10:20 PM

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).
The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?
I have no answer to this question.
Do you?

My method would still work in 3 if the coin was heavier... I don't know what you're not getting dude. The answers is 3. Re-read my answer... it works every time, whether the coin is heavier or lighter.

I have highlighted the relevant parts of the OP (Original Post). Clearly, it implies that you don't know whether counterfeit coin lighter or heavier.
Here is a detailed explanation of how your method could fail:
1. Weigh 9 against 9 and one side is lighter. You took the lighter group of 9 and
2. Weigh 3 vs. 3 with 3 on the side. They weigh the same. So you take 3 on the side and
3. Weigh 1 vs. 1 with 1 on the side. They weigh the same, so you assume (wrongly) that the coin on the side is the counterfeit. Whereas, it was the one the 9 in the heavy group, which you have abandoned.

The answer has been posted throughout this topic many times over. And I have already noted that you can figure out counterfeit out of 39 coins with 4 weighings. But you are not guaranteed to get it out of 27 in less than 4. Three weighings is enough for 12 coins, where you don't know whether counterfeit is heavier, or lighter.
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Past prime, actually.

### #47 DOLEMITE027

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Posted 07 March 2009 - 01:41 AM

First you put all the coins on the scale your either gonna get 269G or 271G.
Then you keep taking off each coin which would be 10G less each time until you get the one of abnormal weight so I would have to say the answer is once.
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### #48 parikshit

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Posted 18 March 2009 - 09:46 AM

Assuming you have a simple scale, my answer is 9.

First divide the coins into 3 piles of 9.

Measure 2 piles at a time until you find the pair that is equal in mass. This can take up to 3 tries, but if you were lucky you could get it on the first try.

Put the piles of coins with equal mass to the side.

Repeat the process this time with 3 piles of 3 coins, maximum 3 measurements.

Repeat the process this time with the 3 remaining coins, find the two that are equal, the remaining coin has a different mass.

If you are a very luck person you coulp potentially only have to make 4 meaurements, 3 to rule out equal pairs and 1 to compare the final coin to the rest.

9 is the most you would have to do the efficient way, and at any point you can observe is the unequal pile has more or less mass.

Six is the minimum for exact weight of finding the coin

1. Divide 27 into 3 piles of 9 coins each.
Weigh two piles against each other.
If weight is same, the third pile has the coin of different weight.
2. If weight is different, take one pile off the weighing machine and put the other pile of 9 coins.
Again weigh against each other.
If weights are same, then the pile you took off has the coin. If not, then the pile on the machine has the coin
So in 1 or 2 weighings you got the pile of 9 coins.

3. Similarly, divide the 9 coins into 3 piles of 3 coins each
Weigh two piles against each other.
If weight is same, the third pile has the coin of different weight.
4. If weight is different, take one pile off the weighing machine and put the other pile of 3 coins.
Again weigh against each other.
If weights are same, then the pile you took off has the coin. If not, then the pile on the machine has the coin
So in 3 or 4 weighings you got the pile of 3 coins.

5. Similarly, divide the 3 coins into 3 piles of 1 coin each
Weigh two piles against each other.
If weight is same, the third pile has the coin of different weight.
6. If weight is different, take one pile off the weighing machine and put the other pile of 1 coins.
Again weigh against each other.
If weights are same, then the pile you took off has the coin. If not, then the pile on the machine has the coin
So in 5 or 6 weighings you got the coin with different weight.
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### #49 Nik

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Posted 18 March 2009 - 10:27 AM

ok guys...heres my point of view on this one... though this sounds difficult to perform...
Spoiler for ta da

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### #50 kylor

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Posted 30 April 2009 - 08:05 AM

It only takes two weighing s!!!

Here is how- First divide the coins into the three piles of nine which I will call A,B, and C. Then weigh the first two piles, piles A and B. The scale will either be 1.-even or 2.-uneven

1. If the scale is even then you take out pile B and put pile C in. If the scale on pile c's side goes down then the coin weighs 11g and if it goes up it only weighs 9g.

2. If the scale is uneven then take out pile B and put pile in pile C. If the scale went down for pile B and it goes down for pile C then we know that pile A contains the coin weighing 9g. If the scale goes up for pile B and C then we know pile A contains a coin weighing 11g. If you put pile C in and the scale becomes even then you know that pile B has a coin weighing either 9 or 11 g depending on which way it tipped on the initial weighing.
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