The Magnetized Coin

[TimeSpaceLightForce]

One of the 12 identical coins was magnetized so that it weighs
heavier on one face and lighter on the other when placed on
a certain weighing scale. But it weighs normal like the other coins
when it is standing balanced on its side or edge.

Find out which coin is it using the weighing scale 3 times.

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#68

[plainglazed]

...if the normal unit weight is known:

1) weigh seven coins
if they weigh seven units the odd coin must be one of the other five coins
2) weigh three of those five coins
if they weigh three units the odd coin must be one of the last two.
3) weigh one of those two coins to see which is the odd.
if they weigh more (or less) than three units one of the three must be the odd coin.
3) of those three; flip one, put one on edge and leave one as is to find the odd coin.
if they weigh more (or less) than seven units the odd coin must be one of those seven.
2) weigh four of those coins flipping two of the four over.
if the four coins are more (or less) than four units, the odd coin must be one of the two not flipped.
if the four coins are less (or more) than four units, the odd coin must be one of the two flipped.
3) weigh one of those two coins to see which is odd.
if the four coins weigh four units the odd coin must be one of the three of the seven not weighed a second time.
3) of those three; flip one, put one on edge and leave one as is to find the odd coin.

[phil1882]

place 4 coins on the weighing scale. take the total, divide by 4.

take another 2 from the remaining coins. divide by 2. if the totals are equal then we know that the magnetized coin is in the last group of 6. either way we have 1 coin that can be be light or heavy, and 6 coins to search through.

NNNNNX

take three coins from the 6 coins we know the magnetized coin to be in, weigh them, divide by 3. if still equal, then the magnetized coin must be in the last group of 3, else it in this group of 3.

NNX

two more wieghings will identify the magnetized coin.

Edited February 7, 2014 by phil1882

[TimeSpaceLightForce]

place 4 coins on the weighing scale. take the total, divide by 4.

take another 2 from the remaining coins. divide by 2. if the totals are equal then we know that the magnetized coin is in the last group of 6. either way we have 1 coin that can be be light or heavy, and 6 coins to search through.

NNNNNX

take three coins from the 6 coins we know the magnetized coin to be in, weigh them, divide by 3. if still equal, then the magnetized coin must be in the last group of 3, else it in this group of 3.

NNX

two more wieghings will identify the magnetized coin.

Thanks for the math idea phil...lets say the coins normally weigh 1 unit or slightly heavy or lighter if magnetized..can you give it another go for just 3 weighing?

[phil1882]

hmm, 4 is possible not sure about 3.

weigh 6 coins. if total is 6, we know the coin is in the other 6.

weigh 3 coins. if total is 3, odd coin in other 3.

weigh 1 coin, weigh 1 coin. done.

[TimeSpaceLightForce]

..using the weighing scale 3 times includes determining which face of the magnetized coin is heavier or lighter.

[plasmid]

For each of the 12 coins, hold it up on edge and flick it so it's spinning on the balance. Turn the balance on, and the weight will be the weight of 12 normal coins because the magnetized coin is spinning and on its edge. The number on the balance will change when the magnetized coin falls to one side. Whichever coin stops spinning and falls flat when the balance changes is the magnetized coin, and you will also know which side makes it heavier and which makes it lighter.

[TimeSpaceLightForce]

@plasmid : Thanks for the solution..

That is pure lateral thought with dexterity!

The scale reads the weight 2 times only with the following assumptions:

1. the scale has on off button

2. the spinning coins do not collide with each other

3. the coins do not fall off or bumps on tray edge

4. the the tray has enough flat surface

5. the coins that stop spinning did not remain on its edge

6. the coin stop spinning at the same time with another coin

7. the spinning (left/right) magnetized coin on existing magnetic

field around the scale has no effect on its weight?

..with uncertainty but it might work.

Hint : The 3 weighing solution is very easy..

[Brainy Binary]

@plasmid : Thanks for the solution..

That is pure lateral thought with dexterity!

The scale reads the weight 2 times only with the following assumptions:

1. the scale has on off button

2. the spinning coins do not collide with each other

3. the coins do not fall off or bumps on tray edge

4. the the tray has enough flat surface

5. the coins that stop spinning did not remain on its edge

6. the coin stop spinning at the same time with another coin

7. the spinning (left/right) magnetized coin on existing magnetic

field around the scale has no effect on its weight?

..with uncertainty but it might work.

Hint : The 3 weighing solution is very easy..

From plasmid idea:

1st weighing: Put the 12 coins standing on the edges in such way like Dominoes to Topple= 12 units reading

2nd weighing: Push one coin to topple one by one . Watch out for the reading till more or less is registered.

The heavy/light side of the coin is also determined.

[TimeSpaceLightForce]

@plasmid : Thanks for the solution..

That is pure lateral thought with dexterity!

The scale reads the weight 2 times only with the following assumptions:

1. the scale has on off button

2. the spinning coins do not collide with each other

3. the coins do not fall off or bumps on tray edge

4. the the tray has enough flat surface

5. the coins that stop spinning did not remain on its edge

6. the coin stop spinning at the same time with another coin

7. the spinning (left/right) magnetized coin on existing magnetic

field around the scale has no effect on its weight?

..with uncertainty but it might work.

Hint : The 3 weighing solution is very easy..

From plasmid idea:

1st weighing: Put the 12 coins standing on the edges in such way like Dominoes to Topple= 12 units reading

2nd weighing: Push one coin to topple one by one . Watch out for the reading till more or less is registered.

The heavy/light side of the coin is also determined.

Nice elimination of uncertainty but that can be using the scale12 times if the last to fall is the odd one.

For every event (falling or stop spin) you check the reading..

[TimeSpaceLightForce]

...if the normal unit weight is known:

1) weigh seven coins

if they weigh seven units the odd coin must be one of the other five coins

2) weigh three of those five coins

if they weigh three units the odd coin must be one of the last two.

3) weigh one of those two coins to see which is the odd.

if they weigh more (or less) than three units one of the three must be the odd coin.

3) of those three; flip one, put one on edge and leave one as is to find the odd coin.

if they weigh more (or less) than seven units the odd coin must be one of those seven.

2) weigh four of those coins flipping two of the four over.

if the four coins are more (or less) than four units, the odd coin must be one of the two not flipped.

if the four coins are less (or more) than four units, the odd coin must be one of the two flipped.

3) weigh one of those two coins to see which is odd.

if the four coins weigh four units the odd coin must be one of the three of the seven not weighed a second time.

3) of those three; flip one, put one on edge and leave one as is to find the odd coin.

That is it!

You may not be able to determine the polarity but it is not asked in the OP..

[dgreening]

I may be missing something. some of these schemes seem to be based on knowing the weight of a regular coin.

I didn't read the problem that way.

I believe that we only know that 11 coins weigh some unknown amount and that the 12th coin weighs slightly more or less than thatt unknown amount depending on how it is positioned..

Our task is to find the magnetized coin without the additional information of what a regular coin weighs.

[LVan Toren]

I may be missing something. some of these schemes seem to be based on knowing the weight of a regular coin.

I didn't read the problem that way.

I believe that we only know that 11 coins weigh some unknown amount and that the 12th coin weighs slightly more or less than thatt unknown amount depending on how it is positioned..

Our task is to find the magnetized coin without the additional information of what a regular coin weighs.

Indeed dgreening, I understand the problem the same way as you do. If you need to know the exact weight of a regular coin, you need a supplementary weighing. I think also about a balance with two scales, that only tells you whether what is in the left scale is heavier or lighter than what is in the right scale.

[TimeSpaceLightForce]

I may be missing something. some of these schemes seem to be based on knowing the weight of a regular coin.

I didn't read the problem that way.

I believe that we only know that 11 coins weigh some unknown amount and that the 12th coin weighs slightly more or less than thatt unknown amount depending on how it is positioned..

Our task is to find the magnetized coin without the additional information of what a regular coin weighs.

Indeed dgreening, I understand the problem the same way as you do. If you need to know the exact weight of a regular coin, you need a supplementary weighing. I think also about a balance with two scales, that only tells you whether what is in the left scale is heavier or lighter than what is in the right scale.

my sincere apologies..i tried to edit the info about the 1unit weight when i realized its too late..The puzzle was originally for the balance scale..but the solution for the classic odd out of 12 in 3 is applicable so i shifted to such weighing scale..With unknown unit of each, the puzzle would look like a simple problem as presented by phil..thanks all for your interest