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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
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83 replies to this topic

### #31 Prime

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Posted 04 August 2008 - 08:40 PM

... You can distinguish among 3**N cases in N weighings. ...

I’ll take up an issue with that statement. I’d say it depends on the kind of possibilities that you are trying to distinguish. The puzzle required to find the counterfeit coin AND tell whether it is heavier or lighter. While you can definitely solve 27 in 4 weighings, I don’t see how you can solve 40 in 4. Although, there are only 80 possibilities (less than 3**4). You can find one counterfeit coin out of 40 in 4 weighings, but not necessarily tell whether it is heavier, or lighter.
For the sake of saving space, solve 13 coins in 3 weighing. I can see how you can find counterfeit coin there, but I can’t ensure determining whether it is heavier, or lighter.
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Past prime, actually.

### #32 Prime

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Posted 04 August 2008 - 09:43 PM

In fact the answer is 2!!!!

You need a minimun of 2 weighs to find the bad coin IF YOU ARE LUCKY!

1- You take 2 coins, you weight them and you get an unbalanced situation.
2- You take one of them and you compare to an outside coin and you still find an unbalanced situstion ==> The bad coin is X and its 9g or 11g

So the minimum ways needed are 2.

First, I’d like to complement the clever redundant use of the factorial sign. (2!!!! = 2).
However, your answer is wrong. The original question was: “… how many minimum ways …”. Whereas, you pointed out only one such way. Indeed, the minimum number of weighings to determine counterfeit coin is 2, if you get lucky. But there are 27!/(25!*2!)=351 ways to compare 2 coins out of 27. And then there are 25*2 ways to compare one of those two to the 25 on the side. Or you could choose to drop 13 coins onto one side of the scale and 13 onto another, and get lucky finding that they weigh the same. And there are C(27,13)*C(14,13) ways, or 27!/(14!*13!)*14 of doing that. Then there are 26 ways to compare the counterfeit coin with one of the true ones. Simplified, the correct answer is: 27! / 24! + 27! * 26 / (13! * 13!). That seems like a very large number, which may not fit into the space allotted by this forum.
(Also, see my previous post for further discussion of the puzzle as it was meant.)
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Past prime, actually.

### #33 Prime

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Posted 07 August 2008 - 06:17 AM

For an in-depth discussion and a maximized version of weighing problem, see my "Ultimate weighing problem" topic on this forum.
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Past prime, actually.

### #34 Niraj Ram

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Posted 31 August 2008 - 07:21 PM

Actually 3 weighs are enough.. Its very simple..
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### #35 Niraj Ram

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Posted 31 August 2008 - 07:23 PM

First split it into 3 parts of 9-9-9 each.. weigh any two, if they weigh same other part of 9 contains faulty one, else the one which weighs different contains faulty ball..
so
first step--> 9-9-9.
second step-->3-3-3.
thord step--> 1-1-1.
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### #36 gdell

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Posted 17 September 2008 - 09:52 PM

The answer is 4 times. split the pile into two piles of 13 if the scale is even than you have the light coin in your hand. if one side is light than you split that side into two piles of six. If the scale is even than the light one is in your hand. If one side is light than you split that side into two piles of 3 if the scale is even than the one in your hand is the light one. if the scale is light on one side than you take two of the 3 coins left and weigh them. If the scale is even than the coin you have in your hand is the light one. if one side is light than you have your light coin
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### #37 Blasphemer

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Posted 28 September 2008 - 05:16 AM

4 tries, after which I can also tell you whether the coin is heavier or lighter than the other coins...

Here's an interesting variation. Suppose there were two odd coins and that one odd coin weighs half as much as an ordinary coin, and the other odd coin weighs half again more than a regular coin (e.g., regular coins weigh 10, the lighter odd coin weighs 5, and the heavier odd coin weighs 15).

If I had more time I'd write you a shorter description of the solution.
Spoiler for my solution

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Posted 28 November 2008 - 03:27 AM

I got a minimum of 4 and maximum of 7

Spoiler for Minimum and Maximum

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### #39 Scarr

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Posted 15 January 2009 - 12:46 AM

I got it in 4, a different way, and with some optimism, possibly 2.

27 coins.

Split into two groups. Remainder 1. Weigh the two groups of 13 against eachother. If they balance, the remainder is the coin. (be that the case, remove 1 coin out of one of the groups weighed, and replace it with the remainder coin to determine heavier or lighter)

13 and 13 narrows down to one heavier or lighter group

Split it into two, and take another remainer.

6 and 6 remainder 1

Continue the process

3 and 3 no remainder

1 and 1 remainder 1 <- in this case, you are down to the last 3 coins. If they balance, the remainder is the one. If not, the one that is heavier or lighter, depending on what was determined earlier, is the one.

So the weighing is:

13 v 13
6 v 6
3 v 3
1 v 1 4 times.
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### #40 lovinduke

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Posted 26 January 2009 - 06:13 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.
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