There are still two weighings and each one has three modes: balance (=), left pan heavier (>), left pan lighter ( <

(2,3,4)/(5,6,8) = = > < = < >

The table is read as follows: if the results of the 5th and 6th are weighed (), it is numbered 4 … etc.

5) The four weighings are unbalanced. We get three configurations in which the 9th piece is the type R ,H and L

5-1) A = (H, H) , B = (H, L), C = (H, L), D = (H, L) and 9th coin = R

5-2) A = (H, H), B = (H, L), C = (H, L), D = (R, L) and 9th coin = H

5-3) A = (H,H), B = (H, L), C = (H, L), D = (H, R) and 9th coin = L

Given the permutations of A, B, C and D, there are a total of 1 + 4 + 4 = 9 possible configurations.

There are still two weighings and each one has three cases: balance (=), left pan heavier (>), left pan lighter ( <

(2,3,4)/(5,6,8) = = > < = < >

The table is read as follows: if the results of the 5th and 6th weighings are (), it is numbered 4 … etc.

It is possible to find the real coin in six weighings

We share the 9 coins in five lots, four lots A, B, C, D of 2 coins each + the ninth coin aside. Four successive weighings are carried out with a coin of each lot A, B, C, D against the second coin of the same lot.

If any of the weighings is balanced then the corresponding two coins of the lots are necessarily false and are of the type HH or LL

If instead one weighing is unbalanced, then the lot is of type HL or HR or RL

The four successive weighings provide five possible outcomes:

1) The four weightings are balanced. Therefore the real coin is the ninth one. Four weighings are enough.

2) Three weighings are balanced (A, B, C,WLOG) and the fourth is unbalanced (D). D is of the type (H,R) or (R,L) and the ninth coin is H or L. D necessarily contains the real coin and one weighing is enough to determine R. Five weighings in total.

3) Two weighings are balanced (A and B,WLOG) and the other two are unbalanced (C and D). We have A = (H,H) and B = (L, L) or vice versa. Therefore the possible configurations of C, D with the ninth coin are the following:

Lot C Lot D 9th coin

3-1) (H, R) (H, L) L

3-2) (H, L) (H, L) R

3-3) (H, L) (H, R) L

3-4) (H, L) (R, L) H

3-5) (R, L) (H, L) H

In the fourth weighing, the heavier coin of C is weighed against the 9th coin.

If the weighing is balanced, this is the 3-4 case and a fifth weighing provides the real coin in D.

If the weighing is unbalanced with RL or H> R, a fifth weighing is performed by comparing the lighter coin of D to the 9th coin and according to the three possible outcomes, the real coin is in C or is the 9th coin or is in D.

4) A weighing is balanced (A, WLOG) and three unbalanced weights are (B, C and D).

There are two possible cases:

4-1) A = (H, H), B = (H, L), C = H,L), D =(R,L) and 9th coin = L with two permutations of B, C and D

4-2) A = (L, L), B = (H, L), C =(H,L), D =(H, R) and 9th coin = H with two permutations of B, C and D

A fifth weighing allows to identify 4-1 or 4-2.Indeed in the case 4-1 the lighter coins of B, C and D and the 9th coin are all L while in the case 4-2 the lighter coins and the 9th coin are of the types H,L and R. So we take these 4 coins and we compare two coins against the other two. If equal, this is the 4-1 case. If unbalanced, this is the case 4-2

A sixth weighing allows to identify R.Indeed:

– In the case 4-1, we weigh the heavier coin of any of the three lots B, C or D against the heavier coin of another lot. If equal, the two coins are of type H and the real coin is in the third lot where it is the heavier. If unbalanced, the real coin is the lighter.

– In the case 4-2, we weigh the lighter coin of any of the three lots B, C or D against the lighter coin of another lot. If equal, the two coins are the type L and the real coin is in the third lot where it is the lighter. If unbalanced, the real coin is the heavier.

In both cases six weighings are enough.

5) The four weighings are unbalanced. We get three configurations in which the 9th piece is the type R ,H and L

5-1) A = (H, H) , B = (H, L), C = (H, L), D = (H, L) and 9th coin = R

5-2) A = (H, H), B = (H, L), C = (H, L), D = (R, L) and 9th coin = H

5-3) A = (H,H), B = (H, L), C = (H, L), D = (H, R) and 9th coin = L

Given the permutations of A, B, C and D, there are a total of 1 + 4 + 4 = 9 possible configurations.

There are still two weighings and each one has three modes: balance (=), left pan heavier (>), left pan lighter ( <

(2,3,4)/(5,6,8) = = > < = < >

The table is read as follows: if the results of the 5th and 6th weighings are (), it is numbered 4 … etc.

]]>It is possible to find the real coin in six weighings

We share the 9 coins in five lots, four lots A, B, C, D of 2 coins each + the ninth coin aside. Four successive weighings are carried out with a coin of each lot A, B, C, D against the second coin of the same lot.

If any of the weighings is balanced then the corresponding two coins of the lots are necessarily false and are of the type HH or LL

If instead one weighing is unbalanced, then the lot is of type HL or HR or RL

The four successive weighings provide five possible outcomes:

1) The four weightings are balanced. Therefore the real coin is the ninth one. Four weighings are enough.

2) Three weighings are balanced (A, B, C,WLOG) and the fourth is unbalanced (D). D is of the type (H,R) or (R,L) and the ninth coin is H or L. D necessarily contains the real coin and one weighing is enough to determine R. Five weighings in total.

3) Two weighings are balanced (A and B,WLOG) and the other two are unbalanced (C and D). We have A = (H,H) and B = (L, L) or vice versa. Therefore the possible configurations of C, D with the ninth coin are the following:

Lot C Lot D 9th coin

3-1) (H, R) (H, L) L

3-2) (H, L) (H, L) R

3-3) (H, L) (H, R) L

3-4) (H, L) (R, L) H

3-5) (R, L) (H, L) H

In the fourth weighing, the heavier coin of C is weighed against the 9th coin.

If the weighing is balanced, this is the 3-4 case and a fifth weighing provides the real coin in D.

If the weighing is unbalanced with RL or H> R, a fifth weighing is performed by comparing the lighter coin of D to the 9th coin and according to the three possible outcomes, the real coin is in C or is the 9th coin or is in D.

4) A weighing is balanced (A, WLOG) and three unbalanced weights are (B, C and D).

There are two possible cases:

4-1) A = (H, H), B = (H, L), C = H,L), D =(R,L) and 9th coin = L with two permutations of B, C and D

4-2) A = (L, L), B = (H, L), C =(H,L), D =(H, R) and 9th coin = H with two permutations of B, C and D

A fifth weighing allows to identify 4-1 or 4-2.Indeed in the case 4-1 the lighter coins of B, C and D and the 9th coin are all L while in the case 4-2 the lighter coins and the 9th coin are of the types H,L and R. So we take these 4 coins and we compare two coins against the other two. If equal, this is the 4-1 case. If unbalanced, this is the case 4-2

A sixth weighing allows to identify R.Indeed:

– In the case 4-1, we weigh the heavier coin of any of the three lots B, C or D against the heavier coin of another lot. If equal, the two coins are of type H and the real coin is in the third lot where it is the heavier. If unbalanced, the real coin is the lighter.

– In the case 4-2, we weigh the lighter coin of any of the three lots B, C or D against the lighter coin of another lot. If equal, the two coins are the type L and the real coin is in the third lot where it is the lighter. If unbalanced, the real coin is the heavier.

In both cases six weighings are enough.

5) The four weighings are unbalanced. We get three configurations in which the 9th piece is the type R ,H and L

5-1) A = (H, H) , B = (H, L), C = (H, L), D = (H, L) and 9th coin = R

5-2) A = (H, H), B = (H, L), C = (H, L), D = (R, L) and 9th coin = H

5-3) A = (H,H), B = (H, L), C = (H, L), D = (H, R) and 9th coin = L

Given the permutations of A, B, C and D, there are a total of 1 + 4 + 4 = 9 possible configurations.

There are still two weighings and each one has three modes: balance (=), left pan heavier (>), left pan lighter ( <

(2,3,4)/(5,6,8) = = > < = < >

It’s impossible to find the the real coin for certain in 6 weighings using your solution. The worst-case scenario would still be 7 weighings.

]]>First you divide in groups of three and find out their relative weights in 3 weigings. If two groups are equal then the third group is either LLR or HHR and the coin can be easily found. If not, in the next three weighings you can find out the relative weight of the coins in the middle group. If one of them is real fine. If not then the middle group is LLH (or HHL) and you know that. Then the real coin is in the lighter/heavier group and can be found in one weighing.

]]>Pick up any two random coins. Depending on the result heavier or lighter….we can again pick another coin and keep eliminating. A rough sketch might be

Let’s say I had picked up ( assume L = Lighter, H = Heavier , F = Fair)

a–> L and b–>F…first weighing I know b > a…

Still this doesn’t give me enough info…again I pick another one….say H..and do two weighings

Case 1…if I had picked H….compare against a. If greater than both….I surely know I picked up H…and fair coin is b. Cant be both L…otherwise first weighing should have been equal.

Case 2. …if I had picked up L….will give a = L…and again I pick up another coin…to compare against b. Remember at this point I eliminated 2 coins. Both L’s. In max 6 moves…i can reach the solution

Might have missed something…..will verify again…but I think 6 cases are enough.

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