Teaser #6

[noseall]

split the coins into four equal piles of three each.

weigh two of those piles putting three on each tray.

if they balance then you know the other six contain the duff coin.

if they don't balance you know the other six are good.

so now after one weighing, you have a two piles of six coins, one good and one bad.

take three coins from the good pile and weigh them with three coins from the duff pile.

if they don't balance, then you now know you have three coins which contain the fake, and, wether it is heavier or lighter.

weigh two of those coins to determine which is the fake.

if they do balance then you have a pile of nine good coins and a pile of three with one dud.

DAMN!

 

[bethrob99]

Bit of trial and error here but I think I have it

Number the coins 1-12
Weigh 1 2 3 4 v 5 6 7 8 note result
weigh 1 2 3 5 v 4 9 10 11 note result
weigh 1 4 6 9 v 2 7 10 12 note result


Fake if Result
1 \\\ or ///
2 \\/ or //\
3 \\- or //-
4 \/\ or /\/
5 /\- or \/-
6 /-\ or \-/
7 /-/ or \-\
8 /-- or \--
9 -/\ or -\/
10 -// or -\\
11 -/- or -\-
12 --/ or --\

I have used \ , - and / to emulate scale position

 

[empip]

This was difficult.

Just messed up direct posting so here is my spreadsheet not that a s/s was required - just handy ...

12 coins numbered 1 - 12.
Break into 4 groups of 3 coins.

Grp A = coins 1,2,3

Grp B = coins 4,5,6

Grp C = coins 7,8,9

Of the three remaining coins, coin 10 is placed in the left tray and coin 11 in the right tray, they remain in the trays at each of the first two weighings, coin 12 is set to one side.

1st weighing.
Grp A + coin 10 weighed against Grp B + coin 11
2nd weighing
Grp C + coin 10 weighed against Grp A + coin 11
Tick off the results in the larger right hand tables above- where the ticks occur on the same row for weighings 1 & 2 the tables will point to a group A, B or C or a single coin being heavier or lighter in weight.

example
From the tables, A > B was ticked, as was A = C .. this points to Grp B being lightweight.
For the third weighing, check the balance between any two coins from Grp B .. If equal then the third coin is the lightweight otherwise the answer is obvious from the observed inequality.

In the case of A = B ticked followed by A = C ticked then coin 12 must be the odd one and at the third weighing will be measured against any of the remaining 11 coins. to determine if 'light' or 'heavy' weight.

Edit.. better mention this
Use the same process in the case of coins 10 and 11 where one is noted as 'light', the other will be noted as 'heavy' only one can be odd weighted compared to the rest ( measure either one against any coin 1-9 or 12 )

That is a tortuous method but I know it works - easier to actually carry out than explain.

 

[noodlz]

Boy, is this a stinker!

Ok...here goes...number coins 1 to 12.

1st weigh - 1 2 3 4 against 5 6 7 8

If balanced, fake must be from 9 10 11 12

2nd weigh - 9 10 11 against 5 6 7 (good coins)

If balanced, fake must be 12 and just need to determine whether heavier or lighter, so...

3rd weigh 1 against 12 will tell us

If, after second weigh, it is not balanced it tells us fake coin is 9 10 or 11 and whether heavier or lighter than the rest.

Then 3rd weigh would be 9 against 10

If balanced, 11 is the fake and we'll know from the second weigh if it was heavier or lighter

However, if 9 is heavier than 10 and 9 10 and 11 were heavier on the second weigh we know that 9 is the fake, which is heavier than the other coins. If 10 is heavier than 9 and 9 10 and 11 were also heavier on the second weigh then 10 is the fake, which is heavier than the other coins. The same principle applies if 9 10 and 11 were lighter.

Now the really tricky part If after first weigh 1 2 3 4 were not balanced with 5 6 7 8 we know the fake coin is one of these eight coins and it will tell us which group is heavier or lighter

Then the 2nd weigh is 1 2 5 against 3 6 12(a good coin - the stinker )

If balanced then we know 4, 7 or 8 is the fake,

So then 3rd weigh 7 against 8. If balanced 4 is the fake and from the 1st weigh we can tell if its heavier or lighter. Otherwise, if 7 is heavier than 8, the first weigh will have told us it if fake is heavier or lighter (for example, if 5 6 7 8 in first weigh was heavier and 7 is heavier than 8 in last then 7 is the fake, which is heavier than the good coins (you can see the other permutations on this one )

If however, 1 2 5 is heavier than 3 6 12 we know the fake is 1 2 5 3 or 6

If on 1st weigh 1 2 3 4 was heavier than 5 6 7 8 then 3rd weigh 1 against 2 whichever is heavier is fake and is heavier than the good coins. If balanced then 6 is the fake and is lighter

If 1 2 5 is heavier than 3 6 12, but 5 6 7 8 was heavier than 1 2 3 4 (on 1st weigh) then 5 or 3 must be the fake so weigh 5 against 1 (a good coin), if unbalanced its the fake and we're told whether heavier or lighter. If balanced 3 is the fake and from 1st weigh tells us if heavier or lighter

I'm sure this is probably a convoluted way of solving the problem and my explanations could be honed down, but is this right?

 

[jbonding]

Boy, is this a stinker!

Ok...here goes...number coins 1 to 12.

1st weigh - 1 2 3 4 against 5 6 7 8

If balanced, fake must be from 9 10 11 12

2nd weigh - 9 10 11 against 5 6 7 (good coins)

If balanced, fake must be 12 and just need to determine whether heavier or lighter, so...

3rd weigh 1 against 12 will tell us

If, after second weigh, it is not balanced it tells us fake coin is 9 10 or 11 and whether heavier or lighter than the rest.

Then 3rd weigh would be 9 against 10

If balanced, 11 is the fake and we'll know from the second weigh if it was heavier or lighter

However, if 9 is heavier than 10 and 9 10 and 11 were heavier on the second weigh we know that 9 is the fake, which is heavier than the other coins. If 10 is heavier than 9 and 9 10 and 11 were also heavier on the second weigh then 10 is the fake, which is heavier than the other coins. The same principle applies if 9 10 and 11 were lighter.

Now the really tricky part If after first weigh 1 2 3 4 were not balanced with 5 6 7 8 we know the fake coin is one of these eight coins and it will tell us which group is heavier or lighter

Then the 2nd weigh is 1 2 5 against 3 6 12(a good coin - the stinker )

If balanced then we know 4, 7 or 8 is the fake,

So then 3rd weigh 7 against 8. If balanced 4 is the fake and from the 1st weigh we can tell if its heavier or lighter. Otherwise, if 7 is heavier than 8, the first weigh will have told us it if fake is heavier or lighter (for example, if 5 6 7 8 in first weigh was heavier and 7 is heavier than 8 in last then 7 is the fake, which is heavier than the good coins (you can see the other permutations on this one )

If however, 1 2 5 is heavier than 3 6 12 we know the fake is 1 2 5 3 or 6

If on 1st weigh 1 2 3 4 was heavier than 5 6 7 8 then 3rd weigh 1 against 2 whichever is heavier is fake and is heavier than the good coins. If balanced then 6 is the fake and is lighter

If 1 2 5 is heavier than 3 6 12, but 5 6 7 8 was heavier than 1 2 3 4 (on 1st weigh) then 5 or 3 must be the fake so weigh 5 against 1 (a good coin), if unbalanced its the fake and we're told whether heavier or lighter. If balanced 3 is the fake and from 1st weigh tells us if heavier or lighter

I'm sure this is probably a convoluted way of solving the problem and my explanations could be honed down, but is this right?

YEH and i done it ages ago,look back.

 

[noodlz]

YEH and i done it ages ago,look back.

I did......and your answers confused me more than the original question (and that's saying something )

BAS, is this method right?

Can't quite work out Bethrob99's answer but is it possible to figure it out with three set weighings like this regardless of how each one turns out?

 

[markie]

Bas put them out of their misery, before they self-harm them selfs,

 

[jbonding]

YEH and i done it ages ago,look back.

I did......and your answers confused me more than the original question (and that's saying something )

BAS, is this method right?

Can't quite work out Bethrob99's answer but is it possible to figure it out with three set weighings like this regardless of how each one turns out?

 

[ban-all-sheds]

Bas put them out of their misery, before they self-harm them selfs,

Sorry chaps - been busy - I'll dig out the answer and see if the one(s) above work...

 

[noseall]

erm......................................










still waiting bas

 

[markie]

He might of made it up it could be all fiction not fact,

 

[keyplayer]

Bas put them out of their misery, before they self-harm them selfs,

And who else could they self harm?

 

[markie]

Bas put them out of their misery, before they self-harm them selfs,

And who else could they self harm?

Yes the ( them selfs ) bit doesn't sound right, i must admit. Never mind i'm sure everone knew what i ment.

 

[jbonding]

maybe someone thought of a way of doing it in the least possible weigh ins ie 2 and it re-wrote the puzzle books

split into 2 sets of 5 if they weigh the same its one of the other two, done in one weigh in,your left with 2 coins, take a coin out of the five and add one of the coins so its 5 against 5 if its equal the coin youve got left is fake ive done it in 2 weigh ins"
see

 

[ban-all-sheds]

ive done it in 2 weigh ins" see

JB - you have to come up with a method that works all of the time, no matter what the outcome of any of the weighings...

And a repeat apology to the others - I will find the answer I have and post it here soon....