Magnetic Coin

[TimeSpaceLightForce]

One of the 12 identical coins was magnetized so that it weighs (Z or Y) heavier on one face and lighter on the other face when
place on a certain weighing scale. But it weighs X like the other normal coins when it is standing balanced on its side or edge.
Using the weighing scale just 3 times ..find that odd coin and determine its heavy or light face.

[imatfaal]

This is completely analogous to not knowing if the coin is heavy or light due to mis-manufacture

 

To aid explanation I have named the coins a->m (missing out i because it gets capitalised to I and looks like lower case L). I have weighed all the time with the same face down. This does not require any markings if you think about it - all you need to do is carefully put your coins in a nice line, keep them in this order in the weighing pan, always return them to the same place, and never turn a coin over in the weighing. This would be a faff - but is quite easy if a bit obsessive. This method would even work if you were to mix up the coins - but the explanation would be tedious and too long to type

 

this red columns show the test and the columns to the right of the test show the three possible results < = > the final column shows the inference.

 

mgnetic coin.pdf

[TimeSpaceLightForce]

The coins shall be weighed by digital weighing scale not the balance scale..

[imatfaal]

The coins shall be weighed by digital weighing scale not the balance scale..

 

No chance

[John Cuthber]

TimeSpaceLightForce

I'm inclined to agree with Imatfaal, can you provide a solution please?

[TimeSpaceLightForce]

@Imatfaal

It is easier than the classic 12 balls with odd one to find by balance scale in 3 trials..

@John Cuthber
Yes, It can be spoiled later.

[Delta1212]

I can do it in three weighings if I know the weight of a normal coin ahead of time, or in four weighings if I don't.

 

I don't see a way to do it in three weighings without any foreknowledge about the weight of the coins.

 

Edit: Actually, scratch that. I think I need another weighing for each. So I don't think you can do it in three even if you know the weight of a normal coin to start with.

Edited July 30, 2015 by Delta1212

[TimeSpaceLightForce]

@Delta 1212
You're on the right track!
The given weight of normal coin is X ..usually varies from 2.1 - 8 grams , but for the purpose of the puzzle assume 1 gram.
( I edited this OP w/ "1 unit like the other .." but instead had double posting .. and that was deleted )

[TimeSpaceLightForce]

 

 

[John Cuthber]

Would you care to explain what you mean by that diagram?

[imatfaal]

With the added information that you know the correct weight of a good coin and that the scales actually have a read out (not normal for these sorts of puzzles and should have been clearer) - then I think that might work

 

Would you care to explain what you mean by that diagram?

 

Seems to work it's very cryptic - but ...

 

 

test 1. weigh 9 coins face down and check to see if equal to expected weight (=G), or heavier (>G) or lighter (<G)

inference 1 if 1=G then odd coin must be in remaining three (10,11,12)

if 1>G then face down one of the 9 coins must be heavy

if 1<G then dace down one the 9 coins must be light

 

test 2a if 1=G so weigh those three remaining coins all face down for result of either >G or <G

inference 2a One of (10,11, or 12) face down is heavier if 2a>G or lighter if 2a<G

 

test 3a. then weigh with one coin face down (10), one face up (12) and one side on (11) for either >G, <G, or =G

inference 3a if 2a=G then side on coin is magnetic and you can tell heavier side by result of 2a

if 2a >G and 3a>G then the face down coin (10) is magnetic and face down it is heavier

if 2a>G and 3a<G then the coin changed to face up (12) is magnetic and face down it is heavier

and vice versa for 2a<G

 

test 2b if 1>G or 1<G weigh the original 9 coins with three face down (4,5,6) three flipped face up (7,8,9) and

three sideways (1,2,3)

inference 2b the same logic as inference 3a will tell you which group of three contains the magnetic coin

 

3b. The same test as 3a for a set of three coins which you know contains one magnetic and you know whether

heavier or lighter when facedown from test 1

 

 

 

[John Cuthber]

So, the original problem wasn't properly described.

No wonder some of us thought it impossible.

[TimeSpaceLightForce]

 

 

Lets number the coins according to stack up position.. #1 top - #12 bottom
1st trial: weigh 9 coins all facing down ( #1 ,#2...#9)
If weighing heavier or lighter (purple)..see diag.
2nd trial: weigh them again with 3 standing ,3 facing down, 3 facing up (#1#2#3,#4#5#6,#7#8#9)
a) if weight is 9X or 9 grams, one of the standing or balanced coins is magnetized. (#1#2#3)
b) if weight is the same as 1st trial ..one of the coins facing down is magnetized. (#4#5#6)
c) If weight is different from 1st trial..one of the coins facing up is magnetized. (#7#8#9)

3rd trial: a) weigh (#1#2#3) - facing down (red), facing up (blue) ,balanced (black)
3rd trial: b) weigh (#4#5#6) - facing down (red), facing up (blue) ,balanced (black)
3rd trial: c) weigh (#7#8#9) - facing down (red), facing up (blue) ,balanced (black)

Summary:
1st 2nd 3rd coin
a)
w>9 w=9 w> 3 #1
w>9 w=9 w< 3 #2
w>9 w=9 w= 3 #3
w<9 w=9 w< 3 #1
w<9 w=9 w> 3 #2
w<9 w=9 w= 3 #3
b)
w>9 w>9 w> 3 #4
w>9 w>9 w< 3 #5
w>9 w>9 w= 3 #6
w<9 w<9 w< 3 #4
w<9 w<9 w> 3 #5
w<9 w<9 w= 3 #6
c)
w>9 w<9 w> 3 #7
w>9 w<9 w< 3 #8
w>9 w<9 w= 3 #9
w<9 w>9 w< 3 #7
w<9 w>9 w> 3 #8
w<9 w>9 w= 3 #9

Else: Weight is 9X or 9 grams (black) -none of them is magnetized..
2nd trial: weigh the remaining 3 coins facing down to check if lighter or heavier (purple) (#10,#11,#12)
3rd trial: weigh (#10,#12,#11) facing down (red), facing up (blue) , balanced (black)
if weight is the same as 2nd trial ..the coin still facing down is magnetized.
if weight is 3 grams, the balanced coin is magnetized.
if weight is different from 2nd trial..the coin facing up is magnetized.

w=9 w>3 >3 #10
w=9 w>3 =3 #11
w=9 w>3 <3 #12

w=9 w<3 <3 #10
w=9 w<3 =3 #11
w=9 w<3 >3 #12

 

 

[Commander]

This is a simpler version of the 39 Ball Puzzle I had created and put up [solved by imatfaal]

 

http://www.scienceforums.net/topic/86946-solve-this-if-you-can/

 

 

The same Principle will be applied and Solution for 39 Coins can be found in 4 weighing on a digital scale with 1 gm per coin weight.

 

Other Conditions apply just like the 12 Coin Puzzle.

Edited August 10, 2015 by Commander

[TimeSpaceLightForce]

 

 

1st : [9T,18H] 12

if w<>27

2nd: [9T,9T,9S]

3rd: [3H,3T,3S]

4th: [1H,1T,1S]

 

if 1st : w = 27

same as 12 for 3 trials

 

Edited August 12, 2015 by TimeSpaceLightForce

[Commander]

 

 

1st : [9T,18H] 12

if w<>27

2nd: [9T,9T,9S]

3rd: [3H,3T,3S]

4th: [1H,1T,1S]

 

if 1st : w = 27

same as 12 for 3 trials

 

 

Yes, Absolutely right !

You may draw your Diagram Solution too while I try here to elaborate.

 

1st : Weigh [27] 12 kept aside.

if 1st : w = 27

same as 12 for 3 trials !

if w < 27 then the defective Coin is among these 27 in Lighter Mode Position

2nd: Weigh the same 27 Coins again with [9 on Edge ,9 Inverted ,9 Same Position] ; If w = 27 then pick 9 on Edge; If w > 27 then pick 9 Inverted.; If w < 27 as before then pick 9 Same Position for the next weighing.

if w > 27 then the defective Coin is among these 27 in Heavier Mode Position

2nd: Weigh the same 27 Coins again with [9 on Edge ,9 Inverted ,9 Same Position] ; If w = 27 then pick 9 on Edge; If w < 27 then pick 9 Inverted.; If w > 27 as before then pick 9 Same Position for the next weighing.

Similarly in the 3rd and 4th weighing the Defective Coin will be identified !

3rd: [3 on Edge ,3 Inverted ,3 Same Position]

4th: [[1 on Edge ,1 Inverted ,1 Same Position]

Just make sure while arranging the coins while inverting or keeping on edge etc that the information already known is preserved and used appropriately in the next weighing !

Edited August 12, 2015 by Commander