# Hard MATH PROBLEM HELP PLEASSE!!!!!!!!!?????

THere are twelve coins that are numbered 1 through 12. Eeleven weigh the same and one is either lighter or heavier than the others. Using just three weighings with a balance scale, devise a scheme that will find the counterfeit coin AND determine whether it is lighter or heavier.

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Four coins are put on each side. There are two possibilities:

1. One side is heavier than the other. If this is the case, remove three coins from the heavier side, move three coins from the lighter side to the heavier side, and place three coins that were not weighed the first time on the lighter side. (Remember which coins are which.) There are three possibilities:

1.a) The same side that was heavier the first time is still heavier. This means that either the coin that stayed there is heavier or that the coin that stayed on the lighter side is lighter. Balancing one of these against one of the other ten coins will reveal which of these is true, thus solving the puzzle.

1.b) The side that was heavier the first time is lighter the second time. This means that one of the three coins that went from the lighter side to the heavier side is the light coin. For the third attempt, weigh two of these coins against each other: if one is lighter, it is the unique coin; if they balance, the third coin is the light one.

1.c) Both sides are even. This means the one of the three coins that was removed from the heavier side is the heavy coin. For the third attempt, weigh two of these coins against each other: if one is heavier, it is the unique coin; if they balance, the third coin is the heavy one.

2. Both sides are even. If this is the case, all eight coins are identical and can be set aside. Take the four remaining coins and place three on one side of the balance. Place 3 of the 8 identical coins on the other side. There are three possibilities:

2.a) The three remaining coins are lighter. In this case you now know that one of those three coins is the odd one out and that it is lighter. Take two of those three coins and weigh them against each other. If the balance tips then the lighter coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is lighter.

2.b) The three remaining coins are heavier. In this case you now know that one of those three coins is the odd one out and that it is heavier. Take two of those three coins and weigh them against each other. If the balance tips then the heavier coin is the odd one out. If the two coins balance then the third coin not on the balance is the odd one out and it is heavier.

2.c) The three remaining coins balance. In this case you know that the unweighed coin is the odd one out. Weigh the remaining coin against one of the other 11 coins and this will tell you whether it is heavier or lighter.

I needed a 4th weigh, maybe you can figure a way around it

Make 3 piles of 4 - A, B, C (we’ll call the odd coin "n")

(1) A vs B

__a)they are the same, meaning that n is in pile C

____(2) 3 from pile A vs 3 from pile C

_______a)they are the same, meaning that n is the one remaining from pile C

__________(3)1 frome pile A vs the remaining one from pile C

_____________a)the remaining one from pile C is n and it is heavier

_____________b)the remaining one from pile C is n and it is lighter

_______b)pile C is heavier, meaning that n is heavier

_________(3)1 from pile C (from the original 3) vs 1 from pile C (from the original 3)

____________a)they are the same, meaning that n is the one remaining from the 3 from pile C and n is heavier

____________b)one is heavier, this is n

_______c)pile C is lighter, meaning that n is lighter

_________(3)1 from pile C (from the original 3) vs 1 from pile C (from the original 3)

____________a)they are the same, meaning that n is the one remaining from the 3 from pile C

____________b)one is lighter, this is n

__b)A is heavier, meaning that n is in A or n is in B (C is normal and will be our "test" pile)

____(2)A vs C

_______a)they are the same, meaning that n is in B and n is lighter

_________(3)3 from C vs 3 from B

___________a)they are the same, meaning that the one left from B is n

___________b)B is lighter, meaning that n is one of the 3 from B

_____________(4)1 from the 3 from B vs 1 from the 3 from B

________________a)they are the same, meaning that n is the one left from the 3 from B

________________b)they are different, n is the lighter one

________b)A is heavier, meaning that n is in A and n is heavier

__________(3)3 from C vs 3 from A

_____________a)they are the same, meaning that the one left from A is n

_____________b)A is heavier, meaning that n is one of the 3 from A

_______________(4)1 from the 3 from A vs 1 from the 3 from A

________________a)they are the same, meaning that n is the one left from the 3 from fA

________________b)they are different, n is the heavier one

__b)A is lighter, meaning that n is in A or n is in B (C is normal and will be our "test" pile)

***same as c)

ooo. I see. 1st Answer is a good plan.

In three weighings (X, Y, and Z) determine which of 12 coins (A through L) is of different weight. The "-" represents the center of a balance or scale. When coin I is used in weighing Y2 and Z3, it could be I, J, K, or L.

1. Weighing X, ABCD — EFGH set aside I, J, K, and L

12. X balanced, weighing Y1, AI — JK set aside L

12a. X balanced, Y1 balanced, the odd coin is L and weight is determined by weighing L against any other coin

12a. X balanced, Y1 tilted, the odd coin is I, J, or K

12a3. Weighing Z1, J — K set aside I

12a3a. Z1 balanced, the odd coin is I and weight is determined by Y1

12a3b. Z1 tilted opposite Y1, the odd coin is J and weight is determined by Y1 or Z1

12a3c. Z1 tilted same as Y1, the odd coin is K and weight is determined by Y1 or Z1

12b. X tilted, weighing Y2, ABG — CEI set aside D, F, and H

12ba. X tilted, Y2 balanced, the odd coin is D, F, or H

12ba3. Weighing Z2, F — H set aside D

12ba3a. Z2 balanced, the odd coin is D and weight is determined by X

12ba3b. Z2 tilted opposite X, the odd coin is F and weight is determined by X or Z2

12ba3c. Z2 tilted same as X, the odd coin is H and weight is determined by X or Z2

12bb. X tilted, Y2 tilted opposite as X, the odd coin is C or G

12bb3. Weighing Z3, C — I set aside G

12bb3a. Z3 balanced, the odd coin is G and weight is determined by X or Y2

12bb3b. Z3 tilted, the odd coin is C and weight is determined by X, Y2, or Z3

12bc. X tilted, Y2 tilted same tilt as X, the odd coin is A, B, or E

12bc3. Weighing Z4, A — B set aside E

12bc3a. Z4 balanced, the odd coin is E and weight is determined by X or Y2

12bc3b. Z4 tilted opposite X and Y2, the odd coin is B and weight is determined by X, Y2, or Z4

12bc3c. Z4 tilted same as X and Y2, the odd coin is A and weight is determined by X, Y2, or Z4