How to find a different weighing ( higher/lower) coin amongst 12 coins which look the same?

Question

There is 12 coins which look the same but one of them weigh differently (heavier or lighter). How can you determine in only three weightings which coin is different and if it is heavier or lighter?

Answer 1

Divide the 12 coins into 3 groups of 4 coins, and then divide each of the 4 coin groups into a single coin and a pile of 3 coins. Place one 4 coin group on each pan of the balance, and the third one on the table.

Observe the condition of the balance. This is the first weighing.

Rotate the groups of 3 coins, moving the one from the right pan to the table, the one from the left pan to the right pan and the one from the table onto the left pan.

Observe the condition of the balance. If it changes, it will identify the group of 3 coins that has the odd coin and its relative weight. This is the second weighing.

In that case, you have 3 coins one of which is known to be heavier (or lighter). Clear the balance and use the same method as above to identify the odd coin. That would be the third weighing.

If the condition of the balance does not change, the odd coin is one of the single coins and can be identified and its relative weight determined by rotating them as with the groups of 3. That would be the third weighing.

Answer 2

The complication here is that you don't know if the counterfeit coin is heavier or lighter. If that piece of infomation were known the weighings would be trivial. You need to not only figure out which coin weighs differently, you need to figure out if it is heavier or lighter, and do this all in 3 weighings.

If it has not already been done, label the coins 1, 2, 3,..., 10, 11, 12 so that we can distinguish between and identify them using these labels.

A basic solution that is usually given runs along the following lines. Weigh 1, 2, 3, 4 against 5, 6, 7, 8

1) If they balance, so 9, 10, 11, 12 contain the odd coin. Weigh 6, 7, 8 against 9, 10, 11.

1a) If they balance, therefore 12 is the odd coin and so weigh 12 against any other to discover whether it is heavy or light.

1b) Otherwise if 9, 10, 11 are heavy and so they contain an odd heavy coin. Weigh 9 against 10. If they balance, 11 is the odd heavy coin, otherwise the heavier of 9 and 10 is the odd coin.

1c) Otherwise, If 9, 10, 11 are light, we use the same procedure to reach the same conclusion for the odd light coin.

2) 5, 6, 7, 8 are heavy and so either they contain an odd heavy coin or 1, 2, 3, 4 contain an odd light coin. Weigh 1, 2, 5 against 3, 6, 10.

2a) If they balance, so the odd coin is 4 (light) or 7 or 8 (heavy). Thus weigh 7 against 8. If they balance 4 is light, otherwise the heavier of 7 and 8 is the odd heavy coin.

2b) 3, 6, 10 are heavy, so the odd coin can be 6 (heavy) or 1 or 2 (light). Thus weigh 1 against 2. If they balance 6 is heavy, otherwise the lighter of 1 and 2 is the odd light coin.

2c) 3, 6, 10 are light, so the odd coin is 3 and light or 5 and heavy. We thus weigh 3 against 10. If they balance 5 is heavy, otherwise 3 is light.

3) If 5, 6, 7, 8 are light we use a similar procedure to that in 2.

This solution, which requires different courses of action depending on the outcomes of previous weighings, is not particularly elegant or easy to remember.

In a better method, four specified coins are weighed against four other specified coins in each of the three weighings and the results are noted. If we observe say the left-hand side of the balance, then for an individual weighing there are three possible alternatives: the left-hand side is heavy (H), light (L) or equal (E) as compared with the right-hand side of the balance.

If we use all twelve coins in the three weighings, and ensure that no particular coin appears on the same side of the balance in all three weighings, the outcomes HHH, LLL, EEE are not possible. We thus have only 24 possible outcomes

Thus we would weigh 1, 2, 3, 4 against 5, 6, 7, 8; then 7, 8, 9, 11 against 1, 5, 6, 10; and finally 2, 5, 8, 9 against 3, 7, 11, 12. The results of these three weighings as observed on the left of the balance are noted.

The mapping of the results from the 3 weighings will uniquely identify the coin and whether it is heavier or lighter.

HLE --> Coin 1 heavy
HEH --> Coin 2 heavy
HEL --> Coin 3 heavy
HEE --> Coin 4 heavy
LLH --> Coin 5 heavy
LLE --> Coin 6 heavy
LHL --> Coin 7 heavy
LHH --> Coin 8 heavy
EHH --> Coin 9 heavy
ELE --> Coin 10 heavy
EHL --> Coin 11 heavy
EEL --> Coin 12 heavy
LHE --> Coin 1 light
LEL --> Coin 2 light
LEH --> Coin 3 light
LEE --> Coin 4 light
HHL --> Coin 5 light
HHE --> Coin 6 light
HLH --> Coin 7 light
HLL --> Coin 8 light
ELL --> Coin 9 light
EHE --> Coin 10 light
ELH --> Coin 11 light
EEH --> Coin 12 light

Try it out to test it. Let's imagine that coin 7 was a light coin. First weighing of 1, 2, 3, 4 against 5, 6, 7, 8:
Left side would appear heavier --> H

Second weighing of 7, 8, 9, 11 against 1, 5, 6, 10:
Left side would appear lighter --> L

Third weighing of 2, 5, 8, 9 against 3, 7, 11, 12:
Left side would appear heavier --> H

HLH corresponds to coin 7 being lighter. Try it for other combinations to confirm this.

Answer 3

1. place a group of 6 coins on each side of the scale
2. which ever group is lighter, put them aside
3. so take the 6 coins from the heavier group and put those in groups of 3 on each side of the scale
4. put the lighter group aside
5. take any 2 of the heavier group of 3, and balance those. if coin A sinks, that is the heavier coin. same with coin B. if both coin A and coin B remain at the same height, coin C, the one you did not balance, is the heaviest

that took a little thinking before i figured out that its on a balance, not an electronic scale =]

Answer 4

This question was asked some time ago - the problem is very well-known - please follow this link for the complete solution:
http://answers.yahoo.com/question/index;_ylt=AuDYlipQpuc1971zIF5wqkPty6IX;_ylv=3?qid=20070907002600AAMlyIt&show=7#profile-info-oEK1QuZTaa