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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

### #21 nair

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Posted 09 May 2008 - 11:45 AM

The way I think, I use the solutions alredy given. I think in 6 weighings one can find the defective one.
1. Divide as mentioned 27 to 3 equal heaps of 9.
2. Weigh any two heaps. If they are equal , then the third heap contains the defective coin.
3. If unequal change the heap on one pan with the left over heap. If they are equal the left over contains defective coin. if the position of the scale is same then the heap on the opposite pan of the scale contains defective coin. If heap count is 1 goto 6.
4. Take the heqp with defective coin and make 3 equal heaps. Repeat steps 2 and 3. Goto 5 when the heap count is 3, else goto 2.
5. Find the defective haeap of 3coins from step 4 and repeat steps 2 and 3
6. Find the defective coin. Seeing the position of the defective coin against good coin in scale pan , one can find whether it is heavy or light
Total 6 weighings will be required. This logic can be extended to any number = 3 ^N.
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### #22 PuzzlerGuzzler

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Posted 04 June 2008 - 05:17 PM

I have a minimum of 1 and a maximum of 4. Begin by removing one coin, then weigh the two sets of 13. If they are equal then the removed coin is the culprit. If not, then take the set that does not weigh 130 g and repeat the procedure. Remove one, and weigh the two sets of 6. If one of the sets is not equal to 60g, then weigh each of the 3 coins in that set. For the set not weighing 30 g, remove one and weigh each of the other two, and you will find the over/underweight coin. Agree?
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Posted 14 June 2008 - 04:48 PM

I think I can do it with a best-case of three, and a worst case of four.

If you always partition the coins into three groups, and weigh two groups against each other
AND if you always get a result of "equals"
Then you always know that the coins you didn't weigh are flawed,
and so can achieve the result in THREE steps. (Best case).

For the other cases:

Let us say that the coins are numbered 0-26
If weighing 0-8 vs 9-17 gives us a value of unequal
we know that one of them must be flawed, but cannot tell which.

So, our next weighing is 0-2,9-11 against 12-14,3-5.
What we're doing is taking three coins from each pile and setting them aside, and then switching three coins from their pans.

If the pans don't change their balance, the flawed coin is in the group that didn't switch, i.e. 0-2, 12-14
If the pans change balance, the flawed coin is in the group that did switch, i.e. 3-5,9-11
If the pans balance, the flawed coin is in the group that was set aside, i.e. 6-8, 15-17

Either way, in 2 steps we've eliminated all but 6 coins.

From these six coins, if they were NOT the ones set aside, i.e. they are from {0-5, 9-14}
renumber them as 0-2 for the ones that came from the left pan, and 3-5 for the ones that came from the right pan.
weigh 0,3 against 1,4

if they tilt, the flawed coin is in 1,3, i.e. the coins that moved.
if they don't change, the flawed coin is in 0,4, i.e. the coins that didn't move
if they balance exactly, the flawed coin is in 2,5, i.e. the coins that weren't weighed.

It will take a further weighing against an ideal specimen to find the flawed coin, taking FOUR steps

If the coins WERE set aside, remember which side was lighter than the other,
do the swap/remove as above, and measure whether the pans tilt, stay the same, or balance.

Takes FOUR steps in all cases except for the best.

If we could somehow only do three-way comparisons, we could always do it in log3 (27) = 3 steps, but I guess 4 steps in general is good.

You can think of it as 3 steps to find out which coin it might be, and 1 step to figure out if it's heavier or lighter, which is not needed if we never use the coin.

Anyway, happy sorting!

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### #24 mohanrao2005

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Posted 15 June 2008 - 01:41 PM

Separate the coins into 13 each and one more. Put 13 on each side of balance, if equal, the odd one is the one with a different weight. If one is more or less than 130 Gms that bundle is taken for further scrutiny. The thirteen in that each are similarly divided into 6, 6 and one to find out if one left out is the culprit. If one of the "sixes" weighs more or less than 60 gms that bundle is divided into 3 each to find out where is the odd one. Out of the three with one odd, take out one and weigh the other two and we aill definitely get the odd one. Hejnce with a minimum of weighing 4 times we can get the odd coin out.
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### #25 janica

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Posted 28 June 2008 - 06:06 AM

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?

tah-dah!

right??

Edited by janica, 28 June 2008 - 06:13 AM.

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### #26 JoycefromRedmond

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Posted 19 July 2008 - 01:14 AM

Step 1) You take two piles of 13 coins each (leaving one out) and weigh them against each other, if one weighs more/less than the other you know which one has the odd coin, if they weigh the same then the coin that was not weighed is the odd coin.

Step 2) You take the stack of 13 coins that had the odd weight and separate into two piles of 6, leaving one out, and do the same as above.

Step 3) You take the stack of 6 coins and separate into two piles of 3 and determine which pile the odd coin is in

Step 4) You take the stack of 3 coins and take one out - if the 2 you weigh are the same then the one you did not weigh is the odd coin, if they do not weigh the same then the one that weighed differently is the odd coin.
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### #27 imsaurabh

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Posted 20 July 2008 - 05:16 PM

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### #28 sagekid

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Posted 20 July 2008 - 06:43 PM

Spoiler for Hmmmm

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### #29 sparkling1

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Posted 22 July 2008 - 04:53 AM

Spoiler for Hmmmm

I get it in 5

First weigh: 13 13 and 1 not weighed; if 2 are equal then the 1 not weighed is the odd one; otherwise of the 13 which is lighter (or heavier)
Second weigh: 6 6 and 1 not weighed; if 2 are equal then the 1 not weighed is the odd one; otherwise of the 6 which is lighter (or heavier)
Third weigh: 3 3; the 3 which are lighter (or heavier) contains the odd one
Fourth weigh: 1 1 and 1 not weighed; if 2 are equal then the 1 not weighed is the odd one; otherwise the 1 which is not weighed replaces either of those which were (noting if the one left on the scale is lighter or heavier - do not know if the odd one is heavier or lighter do we?)
Fifth weigh: if the one which was replaced is the same as the one removed the other is the odd one; if the two are the same then the 1 removed is the odd one
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### #30 akhil

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Posted 26 July 2008 - 12:40 AM

COINS TO BE DIVIDED INTO THREE GROUPS.

1ST ITERATION (27/3=9 COINS); 2ND ITERATION (9/3=3 COINS); 3RD ITERATION (3 INDIVIDUAL COINS)

MINIMUM MOVES = 3 (IF UR LUCKY)

1)make 3 groups of 9, if on 1st try u weigh equal weights, then 3rd group has odd coin.
2)make 3 groups of 3 coins from the 3rd group i.e from 9 coins which u separated after 1st iteration.
if u lucky and weigh equal weights again, then 3rd group of 3 coins has the odd coin.
3)if u r again lucky enuf to weigh equal coins, then the coin left out is the odd coin.

MAXIMUM MOVES = 4 (EVEN UR NOT SO LUCKY)

1)make 3 groups of 9, if d weights are unequal, then replace one pan with 3rd group,

so it takes 2 TRIES, if in 1st try left pan was heavier and after 2nd try both become equal then u come to know that the group of coins in left pan has the odd coin OR if d left pan is still heavier u obviosly know that is d one having odd coin and also very importantly we can get to know if .... the coin is of 11 g (if pan is heavier)... else of 9g (if pan is lighter) when compared with the other pan.

2)make 3 groups of 3 coins from the 3rd group i.e from 9 coins which u separated after 1st iteration.
now u very clearly know whether the coin is light or heavy ..... so u need not go for 2 tries.
in just one try. if u weigh equal, then 3rd group else which ever is (heavier or lighter) as u already know the weight of odd coin.
so just 1 TRY

3) again just 1 try needed, weigh any 2 coins and can clearly get d odd coin in just 1st try.

THUS ONLY 4 MOVES ARE ENOUGH, IF UR NOT LUCKY ENOUGH TO GET IT IN 3 MOVES.
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